WEBVTT mathematics/calculus-ii/murray
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Hi this is educator.com and we are here to talk about parametric curves.
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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.
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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.
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There are basically two calculus problems associated with parametric equations.
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One is to find the tangent line to occur at a particular point.
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The way you are going to do that is we will fine the slope of the tangent line by looking at d(y) dt.
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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.
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That will give us the slope of the tangent line and we will also know one point on the tangent line.
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We can use the point slope equation to find the slope of the tangent line.
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The other equation that you use with parametric equations a lot is
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You find the length of a curve and essentially the formula from that comes from the Pythagorean distance formula.
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You look at x'(t² + y'(t²) and then you find the square root of that.
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That is a unit of arc length representing the length travelled in a very small amount of time.
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Then you integrate that from your starting time to your ending time and that formula represents the total length of the curve.
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Let us try this out with some examples
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The first examples is the equations are x(t) = t+1, y(t) = t².
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Maybe I will just graph a couple of points there.
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If t = 0 then x = 1 and y = 0.
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If t = 1 then x = 2 and y = 1.
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If t = 2 then x = 3 and y = 4.
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This point is travelling along a parabolic path here.
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What we are asked to do is find the tangent line at t=1.
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At t=1, remember, x = 2 and y = 1.
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We are trying to find the tangent line at that point right there.
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We need to find the slope but our slope is d(y) dt/d(x) dt.
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Now dy dt, since y = t² is 2t.
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Dx dt, since x = t+1 is just 1.
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That is 2t, and when we plug in t = 1, that gives us the slope of 2.
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Now we have the slope and we have a point and it is just an algebra problem to find the equation of a line.
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We use the point slope formula, y - y₀, which is 1 here, is equal to the slope × x - x₀ which is 2.
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This is 2x - 4.
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We get the equation of the tangent line, y = 2x-3.
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To recap there, what we are trying to do is to find the tangent line to a curve that is defined by parametric equations.
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What we did was we plugged in the time value we were given, t =1 to get a point on the tangent line.
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Then we used our equation for the slope dy dt/dx dt.
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We figured those out using the equations for x and y that we were given.
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We plugged in the same value of t and we got our slope and then we had a point on our slope,
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And we could use the old point-slope formula to find the equation for our tangent line.
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Let us find another tangent line.
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This time the curve is x(t) = cos(t) y(t) = sin(t).
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You will hopefully recognize that as the equations for a circle because cos²(t) + sin²(t) = 1.
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Those are the equations that define a point moving around in a circle.
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We want to find the tangent line at the point, (sqrt(3/2), 1/2).
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That is about right there on the circle
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The difference between this one and the previous one is we have not been given a t value.
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We have to figure out what value of t gives us the point (sqrt(3/2), 1/2).
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What value of t if you plug it in to cos(t) sin(t) gives us the sqrt(3/2)?
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The answer is π/6 because the cos(π/6) is sqrt(3/2) and the sin(π/6) is 1/2.
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So, we know that t is π/6.
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Again, to find the slope, we use dy dt/dx dt.
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Now y = sin(t) so the derivative there is cos(t).
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X = cos(t), the derivative of that is -sin(t).
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Then we plug in the value t = π/6 to get a number for the slope so the cos(π/6) = sqrt(3/2)
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The sin(π/6) = 1/2, and we still have our negative there.
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The two's cancel so we get our slope is -sqrt(3).
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That gives us the slope and we also have a point, so I will plug those into the point slope formula.
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y - 1/2 = -sqrt(3) x - sqrt(3/2).
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Those values are coming from plugging the points in there.
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We can simplify this a bit.
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This is = sqrt(3)x.
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The minuses cancel each other out plus sqrt(3) × sqrt(3) = 3.
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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.
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We get our tangent line to be y = -sqrt(3)x + 2.
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The way that this problem was different from the previous one was that we were not given a value of t.
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We had to look at the point we were given and we had to figure out what the value of t should be.
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From then on we used the same formulas to find out the slope and the equation of the tangent line.
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By the way, this problem you can also check the answer geometrically if you draw that tangent line.
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It forms a 30-60-90 triangle with the x and y axis.
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That is 30 degrees and that is 60 degrees.
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We know what those angles are because the tangent line is perpendicular to the radius of the circle there,
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We also know that the radius of the circle is 1 so we have a smaller 30-60-90 triangle in there.
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With a short side of length 1, so the long side has length 2.
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That confirms that the y intercept of this tangent line = 2.
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That is a little check on our work using some trigonometry and no calculus at all
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Let us try an example of arc length now.
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We are given the curve x(t) = 6 - 2t³, y(t) = 8 + 3t².
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We want to find the length of that curve from t = 0 to t = 2.
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Remember our arc length formula says you want to take the integral of sqrt of x'(t)² + y'(t)² dt.
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x'(t), the derivative of x if x is 6 - 2t³,
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You take its derivative and the 6 just goes away, so the derivative is -6t².
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y'(t), if y' is 8 + 3t², again the 8 does not have any effect.
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The 3t² you take the derivative and you get 6t.
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What we want then is sqrt(x')², well that is (-6t)², so that is 36t⁴
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+ y'² is 36t².
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That can simplify because we can pull a 36 out and just get 6,
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we can pull a t² out and get t.
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Under the radical, what we have left is just t² + 1.
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The arc length = integral from t=0 to t=2 of 6t × sqrt(t²+1) dt.
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Now, this integral is not too bad because what we can do is make a little substitution.
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u = t² + 1.
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Then, du = 2t, dt.
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The reason that works so nicely is that we already have the t and the dt, so we basically have du.
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In fact, 6t dt, is just 3 du.
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Then, we still have the integral from t = 0 to t = 2 of now the sqrt(u) du.
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You can think of the square root as u^1/2.
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To integrate that we get u^3/2/3/2 which is the same as multiplying by 2/3.
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Then we still have that 3 on the outside.
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This is evaluated from t=0 to t=2.
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I have to convert the u's back into t's
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These 3's cancel, so we get 2 × u was (t² + 1)^3/2 evaluated from t=0 to t=2.
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This gives us 2 now if we plug in t=2, we get 2² + 1, so that is 5^3/2.
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- 0² + 1, so that is 1^3/2.
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We can simplify that to a little bit.
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5^3/2 is the same as 5 × 5^1/2.
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1^3/2 is just 1.
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We get our final answer for the arc length there.
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What made this problem work is having this formula that came from the pythagorean theorem.
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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,
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Simplifying a little bit if we can, then we integrate and we get out answer.
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We will try a couple more examples later.