For more information, please see full course syllabus of Multivariable Calculus

For more information, please see full course syllabus of Multivariable Calculus

### Further Examples with Gradients & Tangents

[(x

^{2})/4] + y

^{2}+ [(z

^{2})/9] = 3 and y

^{2}+ z

^{2}= 10 at (2, − 1,3).

- Recall that the gradient of a function at a point →P is a vector perpendicular to the line tangent to the function passing through →P. See image.
- Let f(x,y,z) = [(x
^{2})/4] + y^{2}+ [(z^{2})/9] − 3 and g(x,y,z) = y^{2}+ z^{2}− 10. By finding ∇f(2, − 1,3) and ∇g(2, − 1,3) we can obtain a vector →v passing through →P and perpendicular to both ∇f(2, − 1,3) and ∇g(2, − 1,3). - Now, ∇f(x,y,z) = ( [x/2],2y,[2z/9] ) and ∇g(x,y,z) = (0,2y,2z). Hence ∇f(2, − 1,3) = ( 1, − 2,[2/3] ) and ∇g(2, − 1,3) = (0, − 2,6).
- Let →v = (v
_{1},v_{2},v_{3}), since ∇f(2, − 1,3) and →v are prependicular we have ∇f(2, − 1,3) ×→v = 0. So ∇f(2, − 1,3) ×(v_{1},v_{2},v_{3}) = ( 1, − 2,[2/3] ) ×(v_{1},v_{2},v_{3}) = v_{1}− 2v_{2}+ [2/3]v_{3}= 0. - Similarly ∇g(2, − 1,3) ×→v = 0. So ∇g(2, − 1,3) ×(v
_{1},v_{2},v_{3}) = (0, − 2,6) ×(v_{1},v_{2},v_{3}) = − 2v_{2}+ 6v_{3}= 0. - We now have a system of equations

which yieldv _{1}− 2v_{2}+ [2/3]v_{3}= 0− 2v _{2}+ 6v_{3}= 0

where k > 0 is a constant.v _{1}= [16/3]v_{3}v _{2}= 3v_{3}v _{3}= k - Setting v
_{3}= 3 we obtain v_{2}= 9 and v_{1}= 16. Our vector →v = (16,9,3). Note that the value of v_{3}is arbitrary. - This vector →v has the same direction of the line tangent to the curve of intersection between the surfaces [(x
^{2})/4] + y^{2}+ [(z^{2})/9] = 3 and y^{2}+ z^{2}= 10 at (2, − 1,3). - Hence we can construct a parametric representation of the line using →x(t) = →P + t→v.
- Substituting yields →x(t) = →P + t→v = (2, − 1,3) + t(16,9,3).

^{2})/4] + y

^{2}+ [(z

^{2})/9] = 3 is an ellipsoid and y

^{2}+ z

^{2}= 10 is a cylinder along the x - axis. The image below depicts the yz - plane view of our surfaces. Where would the curves of intersection be?

^{2}− y

^{2}} = 3 and 4x

^{2}+ y

^{2}+ z = 5 at (0,√3 ,2).

- Recall that the gradient of a function at a point →P is a vector perpendicular to the line tangent to the function passing through →P. See image.
- Let f(x,y,z) = √{6z − x
^{2}− y^{2}} − 3 and g(x,y,z) = 4x^{2}+ y^{2}+ z − 5. By finding ∇f(0,√3 ,2) and ∇g(0,√3 ,2) we can obtain a vector →v passing through →P and perpendicular to both ∇f(0,√3 ,2) and ∇g(0,√3 ,2). - Now, ∇f(x,y,z) = ( [( − x)/(√{6z − x
^{2}− y^{2}} )],[( − y)/(√{6z − x^{2}− y^{2}} )],[3/(√{6z − x^{2}− y^{2}} )] ) and ∇g(x,y,z) = (8x,2y,1). Hence ∇f(0,√3 ,2) = ( 0, − [(√3 )/3],1 ) and ∇g(0,√3 ,2) = (0,2√3 ,1). - Let →v = (v
_{1},v_{2},v_{3}), since ∇f(0,√3 ,2) and →v are prependicular we have ∇f(0,√3 ,2) ×→v = 0. So ∇f(0,√3 ,2) ×(v_{1},v_{2},v_{3}) = ( 0, − [(√3 )/3],1 ) ×(v_{1},v_{2},v_{3}) = − [(√3 )/3]v_{2}+ v_{3}= 0. - Similarly ∇g(0,√3 ,2) ×→v = 0. So ∇g(0,√3 ,2) ×(v
_{1},v_{2},v_{3}) = (0,2√3 ,1) ×(v_{1},v_{2},v_{3}) = 2√3 v_{2}+ v_{3}= 0. - We now have a system of equations

which yield− [(√3 )/3]v _{2}+ v_{3}= 02√3 v _{2}+ v_{3}= 0

where k > 0 is a constant.v _{1}= kv _{2}= 0v _{3}= 0 - Setting v
_{1}= 1 our vector →v = (1,0,0). Note that the value of v_{1}is arbitrary. - This vector →v has the same direction of the line tangent to the curve of intersection between the surfaces √{6z − x
^{2}− y^{2}} = 3 and 4x^{2}+ y^{2}+ z = 5 at (0,√3 ,2). - Hence we can construct a parametric representation of the line using →x(t) = →P + t→v.

^{2}− y

^{2}+ z

^{2}= 4 and 9x

^{2}− 4y

^{2}+ z

^{2}= 9 at (1,1,2).

- Recall that the gradient of a function at a point →P is a vector perpendicular to the line tangent to the function passing through →P. See image.
- Let f(x,y,z) = x
^{2}− y^{2}+ z^{2}− 4 and g(x,y,z) = 9x^{2}− 4y^{2}+ z^{2}− 9. By finding ∇f(1,1,2) and ∇(1,1,2) we can obtain a vector →v passing through →P and perpendicular to both ∇f(1,1,2) and ∇g(1,1,2). - Now, ∇f(x,y,z) = ( 2x, − 2y,2z ) and ∇g(x,y,z) = (18x, − 8y,2z). Hence ∇f(1,1,2) = ( 2, − 2,4 ) and ∇g(1,1,2) = (18, − 8,4).
- Let →v = (v
_{1},v_{2},v_{3}), since ∇f(1,1,2) and →v are prependicular we have ∇f(1,1,2) ×→v = 0. So ∇f(1,1,2) ×(v_{1},v_{2},v_{3}) = ( 2, − 2,4 ) ×(v_{1},v_{2},v_{3}) = 2v_{1}− 2v_{2}+ 4v_{3}= 0. - Similarly ∇g(1,1,2) ×→v = 0. So ∇g(1,1,2) ×(v
_{1},v_{2},v_{3}) = (18, − 8,4) ×(v_{1},v_{2},v_{3}) = 18v_{1}− 8v_{2}+ 4v_{3}= 0. - We now have a system of equations

which yield2v _{1}− 2v_{2}+ 4v_{3}= 018v _{1}− 8v_{2}+ 4v_{3}= 0

where k > 0 is a constant.v _{1}= [3/8]v_{2}v _{2}= kv _{3}= 0 - Setting v
_{2}= 8 we obtain v_{1}= 3 and v_{3}= 0. Our vector →v = (3,8,0). Note that the value of v_{2}is arbitrary. - This vector →v has the same direction of the line tangent to the curve of intersection between the surfaces x
^{2}− y^{2}+ z^{2}= 4 and 9x^{2}− 4y^{2}+ z^{2}= 9 at (1,1,2). - Hence we can construct a parametric representation of the line using →x(t) = →P + t→v.

^{2}− xy, g(x,y) = y

^{2}− x

^{2}and C(0) = (1,1).

Suppose [d/dt]f(C(0)) = 2 and [d/dt]g(C(t)) = − 3. Find C′(0).

- Recall that [d/dt]F(C(t)) = ∇F(C(t)) ×C′(t). We can find the gradients of both f and g and use our available information to solve for C′(t).
- Now, ∇f(x,y) = (4x − y, − x) and ∇g(x,y) = ( − 2x,2y). Note that since C(0) = (1,1) we can compute ∇f(C(0)) and ∇g(C(0)) by letting x = 1 and y = 1.
- We now obtain a system of equations

or by substitution[d/dt]f(C(0)) = ∇f(C(0)) ×C′(0) [d/dt]g(C(0)) = ∇g(C(0)) ×C′(0)

where C′(0) = (u,v).2 = (3, − 1) ×(u,v) − 3 = ( − 2,2) ×(u,v) - Simplifying yields

which has solutions u = [1/4] and v = − [5/4].2 = 3u − v − 3 = − 2u + 2v

- Recall that [d/dt]F(C(t)) = ∇F(C(t)) ×C′(t). We can find the gradients of both f and g and use our available information to solve for C′(t).
- Now, ∇f(x,y) = ( − sin(x + y), − sin(x + y) − cos(y)) and ∇g(x,y) = (ycos(xy),xcos(xy)). Note that since C(p) = ( 0, − [3p/2] ) we can compute ∇f(C(p)) and ∇g(C(p)) by letting x = 0 and y = − [3p/2].
- We now obtain a system of equations

or by substitution[d/dt]f(C(p)) = ∇f(C(p)) ×C′(p) [d/dt]g(C(p)) = ∇g(C(p)) ×C′(p)

where C′(p) = (u,v).− [1/2] = (1,1) ×(u,v) [2/5] = ( − [3p/2],0 ) ×(u,v) - Simplifying yields

which has solutions u = − [4/15p] and v = − [1/2] + [4/15p].− [1/2] = u + v [2/5] = − [3p/2]u

- Recall that [d/dt]F(C(t)) = ∇F(C(t)) ×C′(t). We can find the gradients of both f and g and use our available information to solve for C′(t).
- Now, ∇f(x,y) = ( [1/(2√{x − 1} )],[1/(2√{y − 1} )] ) and ∇g(x,y) = ( [y/(2√{xy} )],[x/(2√{xy} )] ). Note that since C(1) = ( 2,5 ) we can compute ∇f(C(1)) and ∇g(C(1)) by letting x = 2 and y = 5.
- We now obtain a system of equations

or by substitution[d/dt]f(C(1)) = ∇f(C(1)) ×C′(1) [d/dt]g(C(1)) = ∇g(C(1)) ×C′(1)

where C′(1) = (u,v).4 = ( [1/2],[1/4] ) ×(u,v) − 2 = ( [5/(2√{10} )],[1/(√{10} )] ) ×(u,v) - Simplifying yields

which has solutions u = − 32 − 4√{10} and v = 80 + 8√{10} .4 = [1/2]u + [1/4]v − 2 = [5/(2√{10} )]u + [1/(√{10} )]v

^{2x − 3y}, g(x,y) = xe

^{ − y2}and C( − 1) = ( − 1,1). Suppose [d/dt]f(C( − 1)) = 7 and [d/dt]g(C( − 1)) = 12. Find C′( − 1).

- Now, ∇f(x,y) = (2e
^{2x − 3y}, − 3e^{2x − 3y}) and ∇g(x,y) = (e^{ − y2}, − 2xye^{ − y2}). Note that since C( − 1) = ( − 1,1) we can compute ∇f(C( − 1)) and ∇g(C( − 1)) by letting x = − 1 and y = 1. - We now obtain a system of equations

or by substitution[d/dt]f(C( − 1)) = ∇f(C( − 1)) ×C′( − 1) [d/dt]g(C( − 1)) = ∇g(C( − 1)) ×C′( − 1)

where C′( − 1) = (u,v).7 = (2e ^{ − 5}, − 3e^{ − 5}) ×(u,v)12 = (e ^{ − 1},2e^{ − 1}) ×(u,v) - Simplifying yields

which has solutions u = [36/7]e + 2e7 = 2e ^{ − 5}u − 3e^{ − 5}v12 = e ^{ − 1}u + 2e^{ − 1}v^{5}and v = [24/7]e − e^{5}.

^{5},[24/7]e − e

^{5}).

^{2}+ 2z

^{2}− 6x − y = − 2 and x

^{2}+ 4z

^{2}− y = 6 at (1, − 1,1).

- To find the angle θ between two surfaces, it suffices to find the normal vectors N
_{1}and N_{2}to the surfaces at the point and apply cos θ = [(N_{1}×N_{2})/(|| N_{1}|||| N_{2}||)]. - Recall that ∇F is the normal vector to a surface at a point P. Let f(x,y,z) = x
^{2}+ 2z^{2}− 6x − y + 2 and g(x,y,z) = x^{2}+ 4z^{2}− y − 6. - Then ∇f(x,y,z) = (2x − 6, − 1,4z) and ∇g(x,y,z) = (2x, − 1,8z), so that N
_{1}= ∇f(1, − 1,1) = ( − 4, − 1,4) and N_{2}= ∇g(1, − 1,1) = (2, − 1,8).

_{1}×N

_{2})/(|| N

_{1}|||| N

_{2}||)] = [(( − 4, − 1,4) ×(2, − 1,8))/(|| ( − 4, − 1,4) |||| (2, − 1,8) ||)] = [25/(3√{253} )] and so θ = cos

^{ − 1}( [25/(3√{253} )] ) ≈ 58.4

^{o}

^{2}− 5y

^{2}− z = 9 and y

^{2}+ xy − z = 5 at (2,1, − 2).

- To find the angle θ between two surfaces, it suffices to find the normal vectors N
_{1}and N_{2}to the surfaces at the point and apply cos θ = [(N_{1}×N_{2})/(|| N_{1}|||| N_{2}||)]. - Recall that ∇F is the normal vector to a surface at a point P. Let f(x,y,z) = 3x
^{2}− 5y^{2}− z − 9 and g(x,y,z) = y^{2}+ xy − z − 5. - Then ∇f(x,y,z) = (6x, − 10y, − 1) and ∇g(x,y,z) = (y,2y + x, − 1), so that N
_{1}= ∇f(2,1, − 2) = (12, − 10, − 1) and N_{2}= ∇g(2,1, − 2) = (1,4, − 1).

_{1}×N

_{2})/(|| N

_{1}|||| N

_{2}||)] = [((12, − 10, − 1) ×(1,4, − 1))/(|| (12, − 10, − 1) |||| (1,4, − 1) ||)] = [( − 9)/(7√{10} )] and so θ = cos

^{ − 1}( [( − 9)/(7√{10} )] ) ≈ 114

^{o}

^{2}+ yz

^{2}= − 3 and x + y − z

^{2}= − 6 at (1, − 3,2).

- To find the angle θ between two surfaces, it suffices to find the normal vectors N
_{1}and N_{2}to the surfaces at the point and apply cos θ = [(N_{1}×N_{2})/(|| N_{1}|||| N_{2}||)]. - Recall that ∇F is the normal vector to a surface at a point P. Let f(x,y,z) = xy
^{2}+ yz^{2}+ 3 and g(x,y,z) = x + y − z^{2}+ 6. - Then ∇f(x,y,z) = (y
^{2},2xy + z^{2},2yz) and ∇g(x,y,z) = (1,1, − 2z), so that N_{1}= ∇f(1, − 3,2) = (9, − 2, − 12) and N_{2}= ∇g(1, − 3,2) = (1,1, − 4).

_{1}×N

_{2})/(|| N

_{1}|||| N

_{2}||)] = [((9, − 2, − 12) ×(1,1, − 4))/(|| (9, − 2, − 12) |||| (1,1, − 4) ||)] = [55/(3√{458} )] and so θ = cos

^{ − 1}( [55/(3√{458} )] ) ≈ 31.1

^{o}

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

### Further Examples with Gradients & Tangents

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro
- Example 1: Parametric Equation for the Line Tangent to the Curve of Two Intersecting Surfaces
- Part 1: Question
- Part 2: When Two Surfaces in ℝ3 Intersect
- Part 3: Diagrams
- Part 4: Solution
- Part 5: Diagram of Final Answer
- Example 2: Gradients & Composite Functions
- Example 3: Cos of the Angle Between the Surfaces

- Intro 0:00
- Example 1: Parametric Equation for the Line Tangent to the Curve of Two Intersecting Surfaces 0:41
- Part 1: Question
- Part 2: When Two Surfaces in ℝ3 Intersect
- Part 3: Diagrams
- Part 4: Solution
- Part 5: Diagram of Final Answer
- Example 2: Gradients & Composite Functions 26:42
- Part 1: Question
- Part 2: Solution
- Example 3: Cos of the Angle Between the Surfaces 39:20
- Part 1: Question
- Part 2: Definition of Angle Between Two Surfaces
- Part 3: Solution

### Multivariable Calculus

### Transcription: Further Examples with Gradients & Tangents

*Hello and welcome back to educator.com and multi-variable calculus.*0000

*The last lesson we introduced the notion of the tangent plane. I thought it would be a good idea to go ahead to go ahead and do some further examples with the concepts we have been studying recently.*0004

*Just more practice with gradients, and tangents, and partial derivatives and things like that just to get a better sense of what is going on.*0012

*Just to get a sense of the different types of problems you are actually going to run across, and to use what you know about a particular situation to reason out a problem.*0022

*That is it, just for practice. Before we actually move on to the next topic, which is going to be directional derivatives.*0032

*Let us just go ahead and jump right on with the first example. Example 1*0038

*Now, our task is to find a parametric equation for the line to the curve of intersection of the 2 surfaces, x ^{2} + y^{2} + z^{2} = 49.*0054

*x ^{2} + y^{2} = 13 at the point (3,2,-6).*0112

*Let us see if that was the right point here, yes, the point (3,2,-6).*0134

*So, this is more than a little more complicated than what we have done before.*0150

*As far as the expression of the question is concerned, and that is going to happen sometimes.*0157

*Do not let the expression of the question lead you to believe that it is any more complicated than anything that you have done before.*0161

*That is the real issue. You just stop and think about it and understand what is going on.*0167

*As it is we only have a handful of terms at our disposal already, we have partial derivatives, which we know how to do, hopefully at this point.*0173

*We have the idea of the gradient vector and we have this idea of a tangent plane or a tangent line.*0180

*Now, let us just sort of pull back and see what it is they are actually asking us to do.*0185

*Find a parametric region for the line that is tangent to the curve of intersection of the 2 surfaces.*0190

*Here is what is happening. I will go ahead and describe it and I will write it out so that you have it for reference.*0199

*You know that any time 2 planes, a plane is just a surface in 3 space, it just happens to be a very simple type of surface, it is flat.*0205

*You know that when 1 plane meets another plane, where they intersect, it actually forms a line, right?*0213

*Now if I take, instead of 2 planes that are intersecting, if I actually take 2 surfaces, that are intersecting, there point of intersection actually ends up being a curve.*0223

*Of course we will draw all of this out in just a minute.*0234

*They are asking, they are saying, I have a surface, x ^{2} + y^{2} = 49, and I have another x^{2} + y^{2} = 13.*0239

*They are going to meet in some sort of curve in 3-space. Well, at that curve, they want you to find the equation of the tangent line that hits the point of intersection and is actually tangent to the curve itself.*0245

*That is what they are asking. So, let us do a little bit of discussion and then we will draw some pictures and then we will actually start the problem because this is really, really important.*0263

*So, I will write out everything here. When 2 surfaces in R3 intersect, their points of intersection form a curve in R3.*0273

*Again, we will be drawing it out in just a minute so that you can actually see it.*0307

*Now, here is what is important. When we form the tangent plane to surface 1, and the tangent plane to which we can do, which we already know how to do.*0313

*The tangent plane to surface 2 at a point where they intersect, the line of intersection of these 2 planes, of these two tangent planes is, it is the line we are looking for.*0342

*It is the line tangent to the curve of intersection.*0394

*So, before I draw it out, let me just tell you what is going on once again.*0408

*I have a curve, I have another curve. Where they intersect is going to be some sort of another curve.*0412

*Well, any point along that curve where they intersect, I can actually form a plane that is tangent to one of these surfaces, and I can form the plane that is tangent to the other surface.*0417

*Now I have a plane, and another plane. Those 2 planes, when they intersect, they intersect in a line. Well, because they have that point p in common, that one point.*0427

*The line where the 2 tangent planes intersect, that is the line that happens to be tangent to the curve of intersection. That is what we are doing.*0442

*We are going from the surface to the tangent plane, the intersection of the 2, that is going to be our line. This is what it looks like.*0448

*So, I am going to do that on the next page here.*0456

*Okay, so, drawings, here we go. Let me see if I can get this right.*0461

*So I have one surface going this way, let us just go like this, like that, like that. How is that?*0467

*Now I have got another surface here, so I am going to go... let us go down, let us go up like this. Let us go up here, and we will just go like that.*0472

*So, let me draw, let me make a little bit of an erasure here so that I can see.*0489

*Okay. That is what we mean. If we have one surface, and another surface, where they intersect, where they actually intersect is a curve because surfaces are curved.*0502

*They actually intersect at a curve. That is what is happening here.*0513

*Now, what I can do is I can actually form the plane that is perpendicular... the plane that is tangent... so, let us say that we have some point on there.*0521

*So, what we want is we want the line that is going to be tangent to that curve. That is what we are looking for.*0532

*Well, if I form the plane that is... so we will call this surface 2, and we will call this surface 1.*0538

*If I form the tangent plane at that point, just surface 2, and if I form the tangent plane to that point at surface 1, I am going to end up with 2 planes.*0546

*Now, let me go ahead and draw those. So, I am not going to draw on top of it, I am going to draw it over here.*0553

*So, we have 1 plane, we will call this the tangent through 1, and now I will go ahead and draw this one.*0562

*Let me make this again... little dots here, so that we know we are actually talking -- uh, it is not very good, we want to be clear. Something like that.*0581

*This is t2, this is the plane that is tangent to the surface along this to this surface at that point that is that.*0597

*This t1 right here, that is the -- let me go ahead and draw this one in 2 -- this plane is the one that is tangent to this surface at that point. It is on top of that one.*0607

*They intersect in a line -- let me go back to this.*0621

*There is a gradient vector that is going to be perpendicular to surface 2, that is going to be pointing that way.*0629

*There is a gradient vector that is going to be pointing this way that is perpendicular to surface 1, right?*0635

*The is a velocity. This is a curve. The velocity vector is this way. The gradient vectors are perpendicular to the surface, that is the whole idea.*0643

*Well, same idea here, we have used the gradient vector to find these planes, so there is some vector that is going that way that is perpendicular to this plane.*0652

*There is some vector that is going out that way, that is perpendicular to this plane.*0661

*Now, let me go ahead and take this line, and this curve together. I am going to put these 2 together.*0668

*What I have is the curve of intersection, I have the line of intersection, this is the point where they meet, so this line is this line, this curve is this curve.*0678

*There is that vector, and there is that vector. This line is perpendicular to both vectors. It is perpendicular to that vector, and it is perpendicular to that vector.*0691

*Because it is perpendicular, well, we know perpendicularity means that any vector along here dotted with this is equal to 0, orthogonality.*0703

*This vector dotted with this is equal to 0, orthogonality.*0714

*I can form 2 equations, and 2 unknowns, and I can find this vector. That is what I am doing here. Hopefully that made sense.*0720

*Now let us go ahead and actually do it. That was our motivation.*0726

*So, I guess I can start at the bottom down here.*0731

*So, let us take f1, very carefully f1 is x ^{2} + y^{2} + z^{2}.*0735

*The gradient of f1 = well, derivative with respect to x is 2x, derivative with respect to y is 2y, derivative with respect to z is 2z.*0744

*Now the gradient of f1, evaluated at p is equal to (2x,2y,2z), evaluated at the point (3,2,-6).*0757

*This is a common notation for evaluation of a certain something at a given point. You just draw a little line and at the bottom you put whatever it is.*0775

*If it is a vector you just put a vector, if it is a point you just put 5.*0783

*That is going to equal, when I put 3 in for x I get 6, when I put 2 in there I get 4, and when I put -6 in here -- whoops, let me make this a little clearer, this is not zz, this is 2z.*0787

*Okay. I get -12, if I am not mistaken.*0801

*So, the vector (6,4,-12) is perpendicular to surface 1.*0809

*That is what we found when we take the gradient and we evaluate it, the gradient vector is going to be perpendicular to the surface. Here is the surface.*0821

*The gradient vector is going to be perpendicular to it. That is what we found.*0827

*Now, let us move on to function 2.*0833

*f(2) = x ^{2} + y^{2}... okay, so now I am going to stop and just remind you of something if you do not know... well, if you know it is a reminder, if you do not, just something to be vigilant about. I am going to write this out.*0837

*Just because z is not in this equation does mean it is not a surface.*0857

*You can treat any function... so this can be f(x,y,z) = x ^{2} + y^{2}, z does not have to show up.*0864

*If you are treating this like a surface in 3-space, all it means is that z can take any value that you want. It is an infinite number of values.*0875

*What this actually is, is a cylinder in 3-space. The x ^{2} + y^{2}, so in the x,y plane you have a circle, but since z can be anything it is just a cylinder going up and down along the z-axis.*0884

*Because, z can be anything, we are not specifying a constraint on z.*0898

*So, know that you can do that, you can have a function of 3 variables, but all of the variables do not have to be there, as long as you interpret it properly.*0903

*Z can be anything. What you do is you get a surface, it just depends on what the problem is asking and what is going on.*0913

*So you have to be very, very vigilant about what is happening here. Let me actually write that down.*0920

*Note that just because z is not in the equation, this can still be a function of 3 variables f(x,y,z).*0929

*As a surface, absence of z means z takes on whole values.*0968

*This is a cylinder of radius sqrt(13), centered along the z-axis, I will just write that.*0998

*Okay, so now let us go ahead and continue on with the mathematics. Again, it is the algebra that matters.*1026

*So, f(2) = x ^{2} + y^{2}. The gradient of f(2) = (2x,2y,0), there is no z.*1033

*The gradient of f(2) evaluated at the point p, is equal to the (2x,2y,0), evaluated at (3,2,-6).*1046

*What we get is 2 × 3 is 6, 2 × 2 is 4, we get (6,4,0).*1060

*So, the vector (6,4,0) is perpendicular to surface 2.*1069

*There we go. We found 2 vectors, one perpendicular to surface 1, one perpendicular to surface 2. They both meet at that point. Now we can do it.*1075

*Here we go. A vector in the direction of the line of intersection is perpendicular to both gradient vectors, right?*1086

*Go back to that picture... to both gradient vectors.*1113

*Now, we will let v = this vector. We will let v be this vector. In other words, where we had that curve, this line, we had 1 vector going this way, 1 vector going this way, this line is perpendicular to both vectors.*1123

*So, if we... what we want is we want to find a vector in this direction or in this direction so that we can write a parametric equation because they want the parametric equation.*1153

*So, we will let v be this vector, and we will call v... we will give it components of (v1,v2,v3).*1162

*Now, well, the vector v is perpendicular to the first gradient, so the grad of f1 evaluated at p · v = 0, and the gradient of f2 evaluated at p · v = 0.*1173

*So, we have (6,4,-12) · (v1,v2,v3) = 0, and (6,4,0) · (v1,v2,v3) = 0.*1199

*When we do this we get 6v1 + 4v2 - 12v3 = 0, and 6v1 + 4v2 + 0v3 = 0.*1220

*We have 2 equations, well, we have 2 equations and 3 unknowns but fortunately one of them is... so 6v1 + 4v2 = 0, because that is 0, right?*1247

*Essentially we have, let us go ahead and we will let v3 = 0, because again it is... this one does not have a v3, this one has a v3, so it does not matter what it is.*1263

*We can go ahead and set equal to anything that we want. In this case, I am going to go ahead and set it to 0, that is the easiest number.*1277

*So, I have got v3 = 0. Then I have 6v1 + v2 = 0, because this part and this part are the same.*1285

*So, I am going to get 6v1 = -4v2, that means v1 is -2/3v2.*1301

*Now, I can pick v2 as anything that I want, and then v1 will be contingent upon v2, and v3 is always going ot be 0.*1314

*Let us go ahead and take v2, so let us come up here, let v2 = 3. I think it is going to be the easiest if I let v2 = 3, the v's cancel.*1321

*Therefore, then v1 = -2.*1340

*So our vector v = v1, which is -2, v2 which is 3, v3 which is 0. Now we can write out parametric equation.*1348

*The line is equal to... you remember, it is the point p + some vector v × t.*1362

*A line passing through p in the direction of the vector p. Well, we know what p is. P is equal to (3,2,-6).*1370

*You notice I have decided to write it as a column vector instead of a row vector + t × v which is (-2,3,0). There you go.*1379

*We have 2 surfaces. Where they met was a curve. I wanted to find the equation, the parametric equation of the line tangent to that curve.*1395

*I found both gradients. Both gradients define tangent planes. Well, it is the intersection of those tangent planes, because they both pass through p, is going to be the line that actually passes through p and is going to be tangent to the point at p, which happens to be tangent to the surfaces.*1406

*That is all we did here, very, very nice. Let us go ahead and draw that this looks like, so that you can actually see it.*1429

*What you have is the following. Let me see if I can do this right. So, I have got... let me go ahead and let me just draw a couple of... let me draw that one, let me draw that one, and I will go ahead and... so, so this is a sphere.*1437

*Now, I am going to go ahead and draw the cylinder.*1466

*Now, let me go ahead and make this... yeah.*1479

*Something like that. So this was our f1, this was x ^{2} + y^{2} + z^{2}, and it was equal to 49.*1498

*So, this was a circle centered at the origin. I have not yet put the axes in but this is where the origin is, right there.*1511

*It is centered at the origin and it has a radius of 7.*1517

*This is f2 right here, f2, which is equal to x ^{2} + y^{2} = 13. It is a cylinder of radius sqrt(13) that passes, that goes along the axis.*1523

*The points of intersection are these points right here. Let us just say it is like that. That is one particular point.*1538

*In this particular case it was what, 3,6 something like that, -12, what was our point? Oh, (3,2,-6).*1546

*(3,2,-6) wherever it happens to be... well, okay, you know what, let us go ahead and... if you go along the x axis 3, and you go along the y axis 2, -6, so the point is going to be probably somewhere around there.*1558

*That is what we have done. We have basically found the line. We found the equation of that line. That is what we have done.*1573

*Where they meet is a curve. If we happen to find tangent planes to this, we can get a tangent plane that is perpendicular to the surface here, we can get a tangent that is perpendicular to the surface there.*1583

*Where those -- or here -- where those tangent planes actually meet, they form a line. That is this line.*1597

*now, let us do another example. So, this example will go back to our idea of gradient and composite function forming f(c(t)).*1603

*So, let f(x,y) = e ^{9x} + 2y.*1620

*g(x,y) = sin(4x+y).*1632

*Do not let the statement of the problem intimidate you. Often times the problem is a lot easier than you think it is, but kids, they look at these equations and all of a sudden they freak out.*1642

*That sort of just stops them in their tracks. Read the question, it is probably a lot easier than you think. It is just equations.*1652

*You can do derivatives of basic questions, it is not a problem. Again, we only have a handful of things at our disposal at this point. It is not like we have a toolbox full of 150 techniques.*1658

*We only have the gradient, the partial derivative and equations for lines and dot products, and the equations for line and planes.*1666

*Not too much to work with, so we should be able to hammer this out.*1677

*sin(4x+y). Okay, let c be a curve, such that c(0) = (0,0). Oh this is kind of nice.*1680

*C(0,0), okay, we are dealing with something reasonably straightforward here.*1697

*Given, df/dt is equal to, given df/dt at t=0, is equal to 2 and dg/dt... no, let me actually, this is not quite correct, let me give you a little bit more... d(f(c))/dt at t=0 equals 2.*1704

*d(g(c)), because we are forming the composite function g(c) and d(g)/dt evaluated at t=0 equals 1, our task is to find c'(t).*1743

*Okay, let us do it. So, we have a function. We have another function. We have a curve, and we know that the value of the curve at t = 0 is (0,0).*1760

*We know that the derivative, of the composite function f(c(0)) = 2, we know that the derivative of g, the composite function g(c) at t, that is equal to 1. How can we find c'(t).*1772

*Okay, let us just sort of see what is going on. C gradf, okay, so let us go ahead and sort of... again, you just sort of start with what you know.*1787

*They tell me that the d(f(c(t)))dt, so let us expand this derivative, we know that it is the gradient of f evaluated at that point, dotted with c'(t). Let us write that out.*1802

*So, the d(f(c(0)))dt, that is equal to the gradient of f at c(0) · c'(t).*1814

*We know that that equals 2, because they gave us that. Okay, well that is nice.*1838

*Now, let me write this a little... I want to do these in parallel so I am going to make a little more room here. I am actually going to write this down below. I am going to go like this...*1843

*That is equal to the gradient of f evaluated at c(0) · c'(t), and we know that that equals 2, because I want to go in parallel.*1857

*I know that the d(g) evaluated at c(0) dt, I know that that is equal to the gradient of g at c(0) · c'(t).*1874

*They tell me that that is equal to 1. Oh, okay, well this is really, really great. We have the gradient of something dotted with c'(t) = 2, and we have the gradient of something else dotted with c' = 1.*1893

*This suggests two equations and two unknowns, so let us just follow this through and see where this is actually going to take us.*1905

*So, this suggests 2 equations and 2 unknowns. The unknowns are c(t), but that is a vector and so a vector has components, x and y, that is why we have two unknowns.*1911

*Again, do not expect that you are just supposed to look at a problem and know where you are supposed to go. A lot of times you just have to start, and see where you go.*1928

*If you hit a wall, it is not a big deal, people hit walls all the time. Professional mathematicians do, scientists do all the time. You just try another path, that is the learning process.*1937

*Do not think that this is supposed to be like 6th, 7th grade math where you are supposed to look at a problem and know exactly what you are supposed to do and it is going to fall out.*1948

*You just have to start going somewhere, based on what you know, and hopefully something will make itself clear. That is the way to do it, no worries.*1956

*Now, let us go ahead and actually find that and that, because we have the functions.*1965

*Well, we know that df/dx... let me see, what do we get, 9, 0, 2, so df/dx of the gradient of f is going to be -- I do not know if I should write it... that will be fine... so let us go ahead and do df/dx.*1971

*So df/dx is equal to, well when we differentiate with respect to x, the e ^{9x+2y}, we get 9 ×, let me rewrite the functions so that we have them on this page. I think it is a good idea. We want to be as clear as possible.*2001

*So f = e ^{9x+2y}, and let me go ahead and put g over here, so g = sin(4x+y). There we go. Now that we have them as reference.*2021

*So df/dx is going to equal 9 × e ^{9x+2y}, right? The derivative is this × the derivative of this, and the derivative of this with respect to x is just 9.*2038

*Then when we evaluate it at t = 0, we end up with 0,0 = 9.*2057

*So we have that number, now df/dy that is going to equal 2 × e ^{9x+2y}, right? because the derivative of this is e^{9x+2y} × the derivative of that.*2066

*The derivative of this with respect to y, this is a constant, is 2.*2081

*So when we evaluate that at t = 0, we get 2, so now we have those numbers, and we do the same thing for g.*2084

*dg/dx, that is equal to... the cos = 4 × cos(4x+y), right?*2094

*When we evaluate that at t=0, in other words when t=0, they are saying that c(0) is (0,0). That is why I put 0 in for y and 0 in for x. Does that make sense?*2121

*Let me make sure that we know what is going on. Remember that we had specified that one of the things they gave us was c(0) = (0,0).*2135

*So over here, when I say evaluated at t=0, I am actually doing (0,0). X is 0, y is 0, because of this thing that they gave us. That is how I am evaluating it at t=0.*2144

*In fact, let me make it a little more clear then. Instead of doing t=0, let us just say at (0,0), because we are forming f(c(t)), right?*2160

*Because of the gradient right here, this gradient of f(c) at 0. c(0) is (0,0), so it is the gradient of f at (0,0).*2173

*Hopefully that will make it maybe a little bit more clear, sorry about that.*2181

*So when we do that, we end up with 4, right? When we put (0,0) in for here. cos(0) is 1. 1 × 4, okay?*2185

*Then, let us go back to black here, so dg/dy, that is going to equal the cos(4x+y), because the derivative of y is 1, so when we evaluate that at (0,0), that is equal to 1.*2195

*Now, let us let c(t), c'(t) = uv, that is our components, now we can form the... we have the gradient of f evaluated at c(0) · c'(t).*2218

*We had the gradient of g evaluated at c(0) · c'(t).*2248

*The gradient of f was (9,2) · uv and we said that that was equal to 2, and this was equal to 1.*2256

*This one, we have (4,1) · uv, and this is equal to 1,*2270

*So, we get 9u + 2v = 2, and we get 4u + v = 1.*2280

*When we solve this, I will not go through the solution... solving, we get u = 0, v = 1, so, c'(t), actually it is c'(0) which is what we are doing.*2292

*We are finding c'(0). c'(0) is equal to 0 and 1.*2317

*There we go. Standard usage, we just followed what it is that they gave us again. The only real issue with these problems is keeping track of everything on the page.*2324

*All it means is you have to do a little looking, a little going back and forth, just trying to keep it all straight.*2335

*It is expected, I mean this is multi variable calculus, the problems are going to get a little bit more complicated.*2340

*There is going to be a lot more going on, it is not just 1 derivative, you have partial derivatives, you just need to keep it all straight. So, the answer is just go slowly, that is it, just go slowly until you are comfortable with what is what and where is where.*2346

*Let us go ahead and do one more example. This time you may be using the, well, let us see what the example says. Example number 3.*2361

*This says find the cosine of the angle between the surfaces x ^{2} + y^{2} + z^{2} = 38, and x^{2} - y^{2} = -38, at the point (-1... let us make sure we write everything clearly... (-1,1,-6).*2372

*We are starting to get these stray lines again so I am going to move on to the next page after writing this.*2435

*So, find the cosine of the angle between the surfaces x ^{2} + y^{2} + z^{2} = 38, and x - z^{2} - y^{2} = -38 at the point (-1,1,-6).*2441

*At that point, the two curves hit each other. We want to find the angle that the two curves make with each other, the two surfaces make with each other.*2454

*So, this is a good time to give a quick definition of what we mean by the angle between 2 surfaces.*2467

*The angle between 2 surfaces... nope, the angle between 2 surfaces, is defined as the angle between the normal vectors, the normal vectors to the surfaces at that point.*2475

*We know how to find the vector that is normal to the surface at that point. It is the gradient vector, evaluated at that point, that is the whole idea. You have a surface, the gradient vector is going to be perpendicular to that surface at that point. Very, very, nice.*2523

*So, the angle between 2 surfaces, where the 2 surfaces is defined as the angle between the normal vectors to the surfaces. You have a surface, you have a normal vector; you have another surface, you have another normal vector.*2541

*You have 2 normal vectors, that is the angle between the surfaces, nice and easy.*2552

*So, f(1) = x ^{2} + y^{2} + z^{2}.*2561

*Now, the gradient of f(1) = (2x,2y,2z).*2570

*When we evaluate at the particular point, we end up with (-2,2,-12). That is one of them.*2583

*Now the gradient of f(2).*2594

*So, this vector is normal to the surface at the particular point we are discussing. So the gradient of f(2) is going to be 1 - 2y - 2z, right? df/dx, df/dy, df/dz, it is a vector.*2598

*When we evaluate it at the particular point in question, we end up with (1,-2,12), and again I sure hope that you checked my arithmetic here and arithmetic is... well, arithmetic is arithmetic.*2619

*It is important for the sake of computation, but it is the map that is important. We want to understand what is going on underneath.*2634

*Now, we know that the cos(θ) happens to be, remember? vector a, the cos between 2 vectors a and b is the dot product of those vectors divided by the norms of the vectors, the product of the norms of the vectors.*2642

*So, that is our standard. So what we have is, the gradient of f(1) dotted with the gradient of f(2)/norm(gradf(f1)) × norm(gradf(f2)). Wow, look at all of that crazy notation.*2660

*So, the gradient of f1 dotted with gradient of f2 is just this dotted with that, so we get -2, 2 × -2 is -4, 12 × -12 is -144, okay?*2683

*Then, the norm of this thing, if you do it, it is going to equal 4 + 4 + 144, so it is going to be sqrt(152), and here we are going to have 1 + 4 + 144, this is sqrt(149).*2700

*So, we end up with cos(θ) = -150 over... and when I actually do that, it is going to be -150.5 + 0.9967, which is what we wanted.*2720

*If you want θ itself, it is going to be 4.67 degrees. There you go.*2740

*Again, hopefully I did my arithmetic correct, but ultimately, arithmetic is not the real issue, we want to understand the mathematics.*2746

*In this particular case we have two surfaces that are intersecting. We want to find the angle of that intersection where they happen to intersect at a given point.*2753

*Well, we find the vector that is normal to one surface, which is the gradient vector at that point. We find the normal to the other surface, which is the gradient vector to that surface at that point, and then we just go ahead and treat those vectors.*2761

*We do our normal a · b/norm(a)norm(b), that gives us the cosine of the angle between them.*2776

*Hopefully these examples have helped to again, just develop a little more of the sense of the kind of problems that you are looking at.*2783

*There is only a handful of techniques at your disposal at this point, but we want to get good and comfortable with this notion of gradients, and tangent planes, tangent lines, surfaces, things like that, because these are our basic tools.*2793

*The fundamental tools are the actual tools that keep showing up over and over and over again.*2807

*You have done enough mathematics at this point to realize that some of the higher end techniques do not always show up.*2813

*It is the things that you learn at the beginning that keep showing up over and over and over again.*2821

*Thank you for joining us here at educator.com, and we will see you next time here for a discussion of directional derivatives. Take care, bye-bye.*2825

1 answer

Last reply by: Professor Hovasapian

Mon Oct 13, 2014 6:32 PM

Post by sam sohirad on October 11, 2014

hi raffi for example 1, can we use the cross product to find a vector in the direction of the space curve (intersection curve of the planes)?

I dont know if my thought process is correct but as i understood the two planes intersect and then we have a curve of intersection. we are looking for the tangent line at a point on this curve. in order to find the equation of a line we need 2 things; 1 a point on the line (3,2,-6) and a vector pointing in the direction of this line. since the 2 surfaces meet and create a curve, we can find the gradient vector at a certain point on the line. this will give us 2 normal vectors at a certain point of intersection. we can then go on to find the cross product of these 2 normal vectors at a mutual point to find a perpendicular vector at this point, thereby giving us the direction of the line. we then plug into our equation X=tv+P and have the vector equation of a line at the point.

if i use this method will i get the equation of a line that touches the point (3,2,-6)making it an equation of a tangent line?

i also understand what you said about the planes and when they intersect we get a line and that is the equation of our tangent line at that point, but i dont see how we used the planes in example 1? to me it seems we found the gradients of the surfaces.

5 answers

Last reply by: Josh Winfield

Sat May 11, 2013 8:07 AM

Post by Josh Winfield on February 12, 2013

Example 2:You ask to find c'(t) but found c'(0) instead. Which one where we supposed to find?

Example 3: âˆš(152) x âˆš(149) = 150.5 not -150.5

So the angle you get arccos(-150/150.5) = 175.33 (degrees)

1 answer

Last reply by: Professor Hovasapian

Thu Sep 6, 2012 2:43 PM

Post by Ian Vaagenes on September 6, 2012

Great course Raffi, you have a real gift for making things understandable. Any chance you'll do a few lectures on measure theory/probability theory?