For more information, please see full course syllabus of College Calculus: Level II
For more information, please see full course syllabus of College Calculus: Level II
Parametric Curves
Main formula:
Slope m =
Arclength =
Hints and tips:
You can remember the formula for the slope of the tangent line by thinking that symbolically, the dt’s cancel, leaving you with dy ⁄ dx.
To find the equation of the tangent line, you also need a point. Use the given value of t into x(t) and y(t) to find it. Then you can use the pointslope formula from high school algebra (y − y_{0} = m(x − x_{0} )) to find the equation.
Sometimes you aren’t given a value of t, but the coordinates (x, y) instead. Then you must find which value of t gives you the correct (x(t), y(t)). Make sure you check that both x and y are correct for your value of t.
You can remember the arclength formula by recalling that it is derived from the distance formula between two points, which in turn comes from the Pythagorean Theorem.
Don’t make the common algebraic mistake of thinking that reduces to a + b! This is extremely wrong, and your teacher will likely be merciless if you do it
Many problems in Calculus II classes are “rigged” so that when you expand x′(t)² + y′(t)² , it becomes a perfect square that cancels nicely with the square root.
Often this perfect square is achieved by making one of x′(t)² and y′(t)² be something of the form (a − b)² = a² − 2ab + b². Then the other one changes it to a² + 2ab + b², which you can then factor as (a + b)².
Another common technique in arclength problems is to make a usubstitution for whatever is under the square root sign. Then (hopefully) you can manipulate the expression outside the square root into being the du. However, you might have to do several steps of algebraic manipulation, pulling factors in or out of the square root sign, before this works.
You may sometimes be able to use symmetry to find the arclength of part of a curve and then multiply by an appropriate factor to get the total arclength. This can be especially helpful if you just want to examine part of the curve where all the quantities involved are positive.
When it’s feasible, check that your answer makes sense. Unlike area integrals, which can be negative if a curve goes below the xaxis, arclength should always be positive! You might also be able to check that the curve looks about as long as your answer.
Parametric Curves
 Make a table

t 0 2 π [(3π)/2] 2π x(t) 0 1 0 − 1 0 y(t) 0 1.25 1.77 2.17 2.51
y(t) = t^{2}
x(t) = t + 2
What is its Cartesian equation?
 Solve for t with x
 x = t + 2
 t = x − 2
y(t) = t^{2}
y = (x − 2)^{2}
y(t) = t − 9
x(t) = √t
for t ≥ 0
What is its Cartesian equation?
 Solve for t, we start with x
 x = √t
 t = x^{2}
y(t) = t − 9
y = x^{2} − 9
x(t) = [cos(t)/3]
y(t) = [sin(t)/4]
for t ≥ 0
 Isolate x and y
 x = [cos(t)/3]
 3x = cos(t)
 y = [sin(t)/4]
 4y = sin(t)
cos^{2}(t) + sin^{2}(t) = 1
(3x)^{2} + (4y)^{2} = 1
9x^{2} + 16y^{2} = 1
x(t) = √2 sin(t)
y(t) = 2cos(t) + 3
for t ≥ 0
 Isolate x
 x = √2 sin(t)
 [x/(√2 )] = sin(t)
 Isolate y
 y = 2cos(t) + 3
 y − 3 = 2cos(t)
 [(y − 3)/2] = cos(t)
([x/(√2 )])^{2} + ([(y − 3)/2])^{2} = 1
[(x^{2})/2] + [((y − 3)^{2})/4] = 1
x(t) = t + 5
y(t) = t^{3} + 5
 Isolate t with x
 x = t + 5
 x − 5 = t
y = x^{3} − 15x^{2} + 75x − 125 + 125
y = x^{3} − 15x^{2} + 75x
x(t) = 1 + t
y(t) = t^{2} − 9
 Isolate t with x
 x = 1 + t
 − t = 1 − x
 t = x − 1
y = (x − 1)^{2} − 9
y = x^{2} − 2x + 1 − 9
y = x^{2} − 2x − 8
x(t) = t^{2} − 3
y(t) = 1 − t
for t − 3
 Isolate t with x
 x = t^{2} − 3
 x + 3 = t^{2}
 √{x + 3} = t
 Substitute into y(t)
 y = 1 − √{x + 3}
 Graph the equation using shifts
 y = 1 − √{x + 3}
x(t) = arccos([t/4]) − [(π)/2]
y(t) = [t/8] − 3 for − 2π ≤ t ≤ 2π
 Isolate t with x
 x = arccos([t/4]) − [(π)/2]
 x + [(π)/2] = arccos([t/4])
 cos(x + [(π)/2]) = [t/4]
 4cos(x + [(π)/2]) = t
 Observe half  angle trig identity
 cos(u + [(π)/2]) = − sinu
 4cos(x + [(π)/2]) = t
 − 4sinx = t
 Substitute into y(t)
 y = [( − 4sin(x))/8] − 3
 y = − [sinx/2] − 3
 Graph using shifts, reflections, and stretching
 y = − [sinx/2] − 3
 The graph y = sinx has been shifted 3 units down to y = − 3, and reflected. It also has been compressed by a factor of 2.
 Remember to graph within the bounds − 2π ≤ t ≤ 2π.
 Find y′
 y′ = [3/2](x − 8)^{1/2}dx
 Use Arc Length Formula
 Arc Length = ∫_{a}^{b} √{1 + ( [dy/dx] )^{2}} dx
 = ∫_{10}^{17} √{1 + ( [3/2](x − 8)^{1/2} )^{2}} d x
 = ∫_{10}^{17} √{1 + ( [9/4](x − 8) )} dx
 = ∫_{10}^{17}√{1 + [9/4]x − 18} dx
 = ∫_{10}^{17} √{[9/4]x − 17} dx
 Use u substitution
 u = [9/4]x − 17
 du = [9/4]dx
 ∫_{10}^{17} √{[9/4]x − 17} dx = ∫_{10}^{17} [4/9]√u du
 = [4/9]∫_{10}^{17} √u du
 = [4/9][ [2/3]u^{[3/2]} ]_{10}^{17}
 = [4/9][ [2/3]( [9/4]x − 17 )^{[3/2]} ]_{10}^{17}
 Use log properties and trig identities to alter
 − lncosx = ln( cosx )^{ − 1}
 = ln[1/cosx]
 = lnsecx
[dy/dx] = [secxtanx/secx] = tanx dx
 Find the derivative of − lncosx
 y′ = tanx
 Apply Arc Length equation
 ∫_{a}^{b} √{1 + ( [dy/dx] )^{2}} dx = ∫_{1}^{3} √{1 + ( tanx )^{2}} dx
 = ∫_{1}^{3} √{1 + tan^{2}x} dx
 Use Pythagorean identity
 ∫_{1}^{3} √{1 + tan^{2}x} dx = ∫_{1}^{3} √{sec^{2}x} dx
 = ∫_{1}^{3} secxdx
 Use trig integral identity
 ∫_{}^{} secxdx = lnsecx + tanx + C
 ∫_{1}^{3} secxdx = [ lnsecx + tanx ]_{1}^{3}
 = lnsec(3) + tan(3) − lnsec(1) + tan(1)
x = [2/3]t^{[3/2]}, y = 2t + 7
 Find the deratives using product rule
 x′ = √t
 y′ = 2
 Apply the parametric Arc Length Formula
 Arc Length = ∫_{a}^{b} √{( [dx/dt] )^{2} + ( [dy/dt] )^{2}} dt
 = ∫_{ − 1}^{4} √{( √t )^{2} + ( 2 )^{2}} dt
 = ∫_{ − 1}^{4} √{t + 4} dt
 Use substitution with u = t + 4
 du = dt
 ∫_{ − 1}^{4} √{t + 4} dt = ∫_{ − 1}^{4} √u du
 = [3/2][ t + 4 ]_{ − 1}^{4}
 = [3/2]( 4 + 4 − ( − 1 + 4) )
 = [3/2]( 8 + 3 )
 x = e^{t}cost, y = e^{t}sint
 Find the deratives using product rule
 x′ = e^{t}cost − e^{t}sint
 y′ = e^{t}sint + e^{t}cost
 Apply the parametric Arc Length Formula
 Arc Length = ∫_{a}^{b} √{( [dx/dt] )^{2} + ( [dy/dt] )^{2}} dt
 = ∫_{0}^{3} √{( e^{t}cost − e^{t}sint )^{2} + ( e^{t}sint + e^{t}cost )^{2}} dt
 = ∫_{0}^{3} √{2e^{2t}(cos^{2}t + sin^{2}t)} dt
 = ∫_{0}^{3} √{2e^{2t}} dt
 = √2 ∫_{0}^{3} e^{t}dt
 = √2 [ e^{t} ]_{0}^{3}
 x = lnsint, y = t
 Find the deratives using trig integral identity
 x′ = cott
 y′ = 1
 Apply the parametric Arc Length Formula
 Arc Length = ∫_{a}^{b} √{( [dx/dt] )^{2} + ( [dy/dt] )^{2}} dt
 = ∫_{1}^{[(3π)/4]} √{( cott )^{2} + ( 1 )^{2}} dt
 = ∫_{1}^{[(3π)/4]} √{cot^{2}t + 1} dt
 Use Pythagoream Identity
 ∫_{1}^{[(3π)/4]} √{cot^{2}x + 1} dt = ∫_{1}^{[(3π)/4]} √{csc^{2}x} dt
 = ∫_{1}^{[(3π)/4]} csctdt
 = [ − lncsct + cott ]_{1}^{[(3π)/4]}
 Find [dr/(dθ)]
 r′ = − 2sinθ
 Apply the polar Arc Length Formula
 Arc Length = ∫_{a}^{b} √{r^{2} + ( [dr/(dθ)] )^{2}} dθ
 = ∫_{0}^{5π} √{( 2cosθ )^{2} + ( − 2sinθ )^{2}} dθ
 = ∫_{0}^{5π} √{4cos^{2}θ+ 4sin^{2}θ} dθ
 = ∫_{0}^{5π} √4 √{cos^{2}θ+ sin^{2}θ} dθ
 = 2∫_{0}^{5π} √{cos^{2}θ+ sin^{2}θ} dθ
 = 2∫_{0}^{5π} √1 dθ
 = 2[ θ]_{0}^{5π}
 Convert using Polar definitions
 x = rcosθ
 = cosθ(cosθ)
 = cos^{2}θ
 y = rsinθ
 = cosθsin θ
 Find [dx/(dθ)] and [dy/(dθ)]
 x′ = − 2sinθcosθ
 y′ = cos( 2x )
 Apply parametric Arc Length Formula (but not necessarily solve)
 Arc Length = ∫_{a}^{b} √{( [dx/dt] )^{2} + ( [dy/dt] )^{2}} dt
 = ∫_{a}^{b} √{( [dx/(dθ)] )^{2} + ( [dy/(dθ)] )^{2}} dθ
 Find [dr/(dθ)]
 r′ = e^{2θ}
 Arc Length = ∫_{a}^{b} √{r^{2} + ( [dr/(dθ)] )^{2}} dθ
 Apply previous equation with known bounds
 √5 ∫_{0}^{1} e^{2θ}dθ = [(√5 )/2][ e^{2θ} ]_{0}^{1}
*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
Parametric Curves
Lecture Slides are screencaptured 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 0:00
 Important Equations 0:05
 Slope of Tangent Line
 Arc length
 Lecture Example 1 1:40
 Lecture Example 2 4:23
 Lecture Example 3 8:38
 Additional Example 4
 Additional Example 5
College Calculus 2 Online Course
I. Advanced Integration Techniques  

Integration by Parts  24:52  
Integration of Trigonometric Functions  25:30  
Trigonometric Substitutions  30:09  
Partial Fractions  41:22  
Integration Tables  20:00  
Trapezoidal Rule, Midpoint Rule, Left/Right Endpoint Rule  22:36  
Simpson's Rule  21:08  
Improper Integration  44:18  
II. Applications of Integrals, part 2  
Arclength  23:20  
Surface Area of Revolution  28:53  
Hydrostatic Pressure  24:37  
Center of Mass  25:39  
III. Parametric Functions  
Parametric Curves  22:26  
Polar Coordinates  30:59  
IV. Sequences and Series  
Sequences  31:13  
Series  31:46  
Integral Test  23:26  
Comparison Test  22:44  
Alternating Series  25:26  
Ratio Test and Root Test  33:27  
Power Series  38:36  
V. Taylor and Maclaurin Series  
Taylor Series and Maclaurin Series  30:18  
Taylor Polynomial Applications  50:50 
Transcription: Parametric Curves
Welcome back and we are trying some more examples of arc lengths.0000
And other problems involving parametric equations.0005
We have here the length of a curve given by x(t) = cos^{2}(t) and y(t) = sin^{2}(t) as t goes from 0 to π/2.0010
Remember our arc length formula is x'^{2} + y'^{2}, take the square root of that and integrate it.0023
Let us calculate x'.0029
X is cos^{2}(t), so that would be 2 × cos(t) × sin(t) using the chain rule there,0034
The  sign comes from the derivative of cosine.0044
y'(t) is 2 × sin(t) × derivative of sin which is cos(t).0049
If we square each one of those, x'(t)^{2} is 4cos^{2}(t) sin^{2}(t).0059
y'(t)^{2} is 4sin^{2}(t) cos^{2}(t).0070
If we add those up, x'^{2} + y'^{2},0082
Well those are the same thing.0091
We just get the sqrt(8sin^{2}(t)cos^{2}(t)).0094
The sqrt(8) is 2×sqrt(2).0102
The square root of sin^{2} and cos^{2}, since sin and cos are positive when t is between 0 and π/2,0105
This is sin(t) × cos(t).0117
So, that is what we want to integrate.0122
The integral from t=0, to t=π/2.0125
I will write the 2×sqrt(2) on the outside.0130
sin(t) cos(t) dt.0138
This integral is not too bad.0143
We can use the substitution u = sin(t).0145
The reason that works so well is because du = cos(t) dt.0149
So, what we have here is the integral of u du.0156
That gives us u^{2}/2.0160
Now to convert things back into t.0169
2×sqrt(2) × sin^{2}(t)/2.0172
Evaluated from t=0 to t=π/2.0180
Those 2's cancel each other so we get sqrt(2) × sin(π/2)^{2},0188
That is just one.0195
Minus sin(0)^{2}, which is just 0.0197
So our arc length is just sqrt(2).0200
Again there, the calculus worked nicely.0206
What we did was find x' and y' and plugged them into this Pythagorean Formula.0208
Then we integrated to get the answer.0215
There is actually another way to see how this problem works.0218
We can see through this problem geometrically.0220
If you actually try graph the path that these equations are describing, 0226
Notice that cos^{2}(t) + sin^{2}(t) = 1.0228
So, cos^{2}(t) = x, sin^{2}(t) = y.0240
So this path is actually taking place on the line y+x = 1.0245
So there is that line,0255
And if you plug in t=0, then x=1 and y=0.0257
There is t=0.0264
At the point (1,0).0266
If you plug in t  π/2, then y = 1 and x=0.0269
So, there is t=π/2 at the point (0,1).0277
So, what this path is really doing is just following a straight line from (1,0) to (0,1).0286
Of course the length of that line is sqrt(2).0293
That is kind of a geometric check on the calculus we just did.0297
Let us try one more example.0000
We want to find the length of the curve given by x(t) = 7 + 2t,0002
And y(t) = e^{t} + e^{t}, where t goes from 0 to 1.0006
Again we want to find x' and y', square each one, add them up and take the square root.0013
x'(t) = 2, y'(t), y = e^{t} + e^{t}, so the derivative of e^{t} is just e^{t}.0020
The derivative of e^{t} is e^{t} × the derivative of t, which is 1.0037
So we get e^{t}.0047
So x'(t)^{2} + y'(t)^{2}.0050
Well, 2^{2} is just 4. 0058
Now, if we square y'(t), then we are going to follow the formula (a + b)^{2},0060
Is a^{2} + 2ab + b^{2}.0074
Here a is e^{t}, so (e^{t})^{2} is e^{2t}, b is e^{t},0080
So this is 2ab, well ab is e^{t} × e^{t},0090
Then + b^{2} is e^{t}, so that is e^{2t},0102
But look at this e^{t} × e^{t}, is e^{0} which is 1.0111
This is just 2.0116
We also have a 4 here, so what we get is e^{2t}  2 + 4, is just + 2.0120
+ e^{2t}.0131
The clever thing to do here is to write that 2 as 2e^{t} × e^{t} again + e^{2t}.0134
This is (e^{t} + e^{t})^{2}.0145
The square root of all that, x'(t)^{2} + y'(t)^{2}.0155
The square root cancels that for a fixed square,0164
So we get e^{t} + e^{t}.0168
That is what we want to integrate to find the arc length.0173
We take the integral from t=0 to t=1.0175
We are getting those bounds from the original problem of e^{t} + e^{t} dt.0181
That integral is not too bad.0192
The integral of e^{t}, is just e^{t} itself.0194
The integral of e^{t}, is e^{t}/the derivative of t, which is 1.0197
That is the same as multiplying by 1.0207
So this is e^{t}  e^{t}.0210
Then we want to evaluate that from t=0 to t=1.0213
That is e^{1}  e^{1}  e^{0} + e^{0}.0221
The e^{0}'s are both 1 so those cancel each other out.0232
This is e  e^{1}.0235
I will write this as 1/e, and that is our answer for the arc length.0240
Again, what we did there, we looked at the x and y that we were given,0246
We took the derivative of each one, squared them, added them up, took their square root.0252
And integrated to get our answer for the arc length.0257
Thanks for watching, this has been educator.com.0261
Hi this is educator.com and we are here to talk about parametric curves.0000
The idea about parametric curves is that you are given the equation x(t) and y(t) and those define how a point is moving around in the plane.0007
The x(t) gives you the x coordinate at a particular time and the y(t) gives you the y coordinate at a particular time.0019
There are basically two calculus problems associated with parametric equations.0026
One is to find the tangent line to occur at a particular point.0030
The way you are going to do that is we will fine the slope of the tangent line by looking at d(y) dt.0039
So, just looking at the derivative of the y equation, and dividing by d(x) d(t), which is the derivative of the x equation.0048
That will give us the slope of the tangent line and we will also know one point on the tangent line.0052
We can use the point slope equation to find the slope of the tangent line. 0059
The other equation that you use with parametric equations a lot is 0063
You find the length of a curve and essentially the formula from that comes from the Pythagorean distance formula.0075
You look at x'(t^{2} + y'(t^{2}) and then you find the square root of that.0076
That is a unit of arc length representing the length travelled in a very small amount of time.0084
Then you integrate that from your starting time to your ending time and that formula represents the total length of the curve.0089
Let us try this out with some examples0097
The first examples is the equations are x(t) = t+1, y(t) = t^{2}.0100
Maybe I will just graph a couple of points there.0109
If t = 0 then x = 1 and y = 0.0111
If t = 1 then x = 2 and y = 1.0118
If t = 2 then x = 3 and y = 4.0125
This point is travelling along a parabolic path here.0133
What we are asked to do is find the tangent line at t=1.0138
At t=1, remember, x = 2 and y = 1.0143
We are trying to find the tangent line at that point right there.0149
We need to find the slope but our slope is d(y) dt/d(x) dt.0156
Now dy dt, since y = t^{2} is 2t.0170
Dx dt, since x = t+1 is just 1.0174
That is 2t, and when we plug in t = 1, that gives us the slope of 2.0180
Now we have the slope and we have a point and it is just an algebra problem to find the equation of a line.0189
We use the point slope formula, y  y_{0}, which is 1 here, is equal to the slope × x  x_{0} which is 2.0195
This is 2x  4.0207
We get the equation of the tangent line, y = 2x3.0213
To recap there, what we are trying to do is to find the tangent line to a curve that is defined by parametric equations.0220
What we did was we plugged in the time value we were given, t =1 to get a point on the tangent line.0230
Then we used our equation for the slope dy dt/dx dt.0240
We figured those out using the equations for x and y that we were given.0245
We plugged in the same value of t and we got our slope and then we had a point on our slope,0250
And we could use the old pointslope formula to find the equation for our tangent line.0259
Let us find another tangent line.0263
This time the curve is x(t) = cos(t) y(t) = sin(t).0265
You will hopefully recognize that as the equations for a circle because cos^{2}(t) + sin^{2}(t) = 1.0272
Those are the equations that define a point moving around in a circle.0282
We want to find the tangent line at the point, (sqrt(3/2), 1/2).0288
That is about right there on the circle0294
The difference between this one and the previous one is we have not been given a t value.0297
We have to figure out what value of t gives us the point (sqrt(3/2), 1/2).0302
What value of t if you plug it in to cos(t) sin(t) gives us the sqrt(3/2)?0308
The answer is pi/6 because the cos(pi/6) is sqrt(3/2) and the sin(pi/6) is 1/2.0318
So, we know that t is pi/6.0328
Again, to find the slope, we use dy dt/dx dt.0330
Now y = sin(t) so the derivative there is cos(t).0340
X = cos(t), the derivative of that is sin(t).0345
Then we plug in the value t = pi/6 to get a number for the slope so the cos(pi/6) = sqrt(3/2)0354
The sin(pi/6) = 1/2, and we still have our negative there.0364
The two's cancel so we get our slope is sqrt(3).0370
That gives us the slope and we also have a point, so I will plug those into the point slope formula.0376
y  1/2 = sqrt(3) x  sqrt(3/2).0380
Those values are coming from plugging the points in there.0394
We can simplify this a bit.0396
This is = sqrt(3)x.0398
The minuses cancel each other out plus sqrt(3) × sqrt(3) = 3.0400
Then we can bring this half over to the other side so we get y = sqrt(3)x + 3/2 + 1/2, is just 2.0410
We get our tangent line to be y = sqrt(3)x + 2.0429
The way that this problem was different from the previous one was that we were not given a value of t.0435
We had to look at the point we were given and we had to figure out what the value of t should be.0440
From then on we used the same formulas to find out the slope and the equation of the tangent line.0450
By the way, this problem you can also check the answer geometrically if you draw that tangent line.0454
It forms a 306090 triangle with the x and y axis.0464
That is 30 degrees and that is 60 degrees.0470
We know what those angles are because the tangent line is perpendicular to the radius of the circle there,0476
We also know that the radius of the circle is 1 so we have a smaller 306090 triangle in there.0484
With a short side of length 1, so the long side has length 2.0501
That confirms that the y intercept of this tangent line = 2.0506
That is a little check on our work using some trigonometry and no calculus at all0511
Let us try an example of arc length now.0519
We are given the curve x(t) = 6  2t^{3}, y(t) = 8 + 3t^{2}.0520
We want to find the length of that curve from t = 0 to t = 2.0530
Remember our arc length formula says you want to take the integral of sqrt of x'(t)^{2} + y'(t)^{2} dt. 0536
x'(t), the derivative of x if x is 6  2t^{3},0549
You take its derivative and the 6 just goes away, so the derivative is 6t^{2}.0556
y'(t), if y' is 8 + 3t^{2}, again the 8 does not have any effect.0568
The 3t^{2} you take the derivative and you get 6t.0575
What we want then is sqrt(x')^{2}, well that is (6t)^{2}, so that is 36t^{4}0584
+ y'^{2} is 36t^{2}.0596
That can simplify because we can pull a 36 out and just get 6, 0600
we can pull a t^{2} out and get t.0610
Under the radical, what we have left is just t^{2} + 1.0615
The arc length = integral from t=0 to t=2 of 6t × sqrt(t^{2}+1) dt.0620
Now, this integral is not too bad because what we can do is make a little substitution.0645
u = t^{2} + 1.0650
Then, du = 2t, dt.0654
The reason that works so nicely is that we already have the t and the dt, so we basically have du.0658
In fact, 6t dt, is just 3 du.0666
Then, we still have the integral from t = 0 to t = 2 of now the sqrt(u) du.0674
You can think of the square root as u^{1/2}.0683
To integrate that we get u^{3/2}/3/2 which is the same as multiplying by 2/3.0687
Then we still have that 3 on the outside.0698
This is evaluated from t=0 to t=2.0701
I have to convert the u's back into t's0706
These 3's cancel, so we get 2 × u was (t^{2} + 1)^{3/2} evaluated from t=0 to t=2.0709
This gives us 2 now if we plug in t=2, we get 2^{2} + 1, so that is 5^{3/2}.0727
 0^{2} + 1, so that is 1^{3/2}.0737
We can simplify that to a little bit.0744
5^{3/2} is the same as 5 × 5^{1/2}.0746
1^{3/2} is just 1.0749
We get our final answer for the arc length there.0758
What made this problem work is having this formula that came from the pythagorean theorem.0762
We just take the x(t) and the y(t) that we are given and we plug them into this formula which involves computing a couple derivatives,0768
Simplifying a little bit if we can, then we integrate and we get out answer.0775
We will try a couple more examples later.0782
1 answer
Last reply by: Dr. William Murray
Thu Nov 20, 2014 3:20 PM
Post by David Llewellyn on November 19, 2014
How do the table and the graph in the answer to the practice question relate to the actual question? Shouldn't the graph show the relationship between x and y and not the two relationships between x (albeit a different function than in the question) and y and t?
1 answer
Last reply by: Dr. William Murray
Fri Aug 16, 2013 6:21 PM
Post by Timothy Davis on August 14, 2013
Hi Dr. Murray,
Thanks for answering my last question. I thoroughly enjoy your videos. You mentioned that once we find the tangent line, we can use the point slope formula to find the equation of the tangent line. My question is, if we have the slope of the tangent line already defined by dy/dt / dx /dt, why do we need to find the equation of the tangent line? What is the difference between the slope of the tangent line and the equation of the tangent line? You go over this around 1:15 into your lecture.
1 answer
Last reply by: Dr. William Murray
Wed Apr 3, 2013 11:52 AM
Post by omatseye ugen on March 12, 2012
good job . but i thought this video will involve mostly multivar. calculus
1 answer
Last reply by: Dr. William Murray
Wed Apr 3, 2013 11:48 AM
Post by Allison Walsh on April 6, 2011
Awesome!
6 answers
Last reply by: Dr. William Murray
Wed Apr 3, 2013 11:46 AM
Post by Wen Geng on March 7, 2011
Dr. Murray's lectures are clear, quick and concise. Very good, unlike the Chemistry ones.