Find d 2 y d x 2 \frac{d^2y}{dx^2} for the parametric curve at the given point. x = 8 cos t x=8\cos t , y = 6 sin t y=6\sin t , t = π 4 t=\frac{\pi}{4}

So in this problem, we're asked to find the second derivative for the parametric curve given at the point. So you can see that we have our x of t, which is eight cosine of t, our y of t, which is six sine of t, then I can see we also have a t value of pi over four. So how do we go about solving this? Well, a first step here since we're looking for the second derivative in this problem is going to be to first find the first derivative. And to find the first derivative, I'm looking for dy over dx, which is the same thing as x or, excuse me, y prime of t over x prime of t. So let's find the derivatives for y and x with respect to t. And we'll go ahead and start with y. So I want to find y' Well, that's just going to be the derivative of six sin. Six is going to stay the same, and the derivative of sin is cos. So six cos is going to be y' Now next, we're going to find x'. Well, x' is just going to be the derivative of eight cos. Eight will stay the same, and the derivative of cos with respect to t is going to be negative sine. So we're going to get negative eight sin of t, and that's our x' of t. So to find dy over dx, we need to take our y', which is six cosine of t, and then we need to divide that by x', which I can see here is negative eight sine of t. Now there's a couple of things I can do from here to go ahead and simplify this. Six over eight, that's going to reduce to threefour, and then we have this negative sign. So we're going to get this negative three fourths out front. Now, what is cosine over sine? Well, cosine over sine is an identity that's actually cotangent. And the way that you can remember this is sine over cosine gives us tangent. So so it makes sense that cosine over sine, which would be the reciprocal of tangent, would be cotangent. So that's how we can go ahead and get this, and that's going to be our derivative dy over dx. So we've gone ahead and found dy over dx, but what we're looking for is d squared y over dx squared. We're actually looking for the second derivative. How do I go about doing that? Well recall, if you want to find the second derivative, d squared y over d x squared, what you want to do is you want to take the derivative with respect to t of your d y over d x that you found. Then once you do this, you want to divide that by the derivative of x with respect to t, and that's going to give you the second derivative that you're looking for. So let's go ahead and do this. Well, what I'm looking for is the derivative of this now. So I'm looking for the derivative with respect to t of dy over dx. So I'm going to go ahead and start with this piece right here in the numerator. Well, doing this, I'm going to be taking the derivative of negative threefour times the cotangent of t. How can I take this derivative? Well, the negative three fourths, that's a constant, so that's just going to stay the same. But what is the derivative of cotangent? There should be a cotangent right here. What's the derivative of cotangent? Well, the thing that I know is the derivative of cotangent is actually negative cosecant squared. So what that means is if we take the derivative of this, we're going to get negative three fourths times, and that's going to be negative cosecant squared t. So this right here is going to be what happens when we take the derivative of cotangent. And notice the negative signs here are going to cancel. So we're going to end up with positive three fourths cosecant squared t, and that's going to be our result for the derivative of dy over dx with respect to t. So this is what we get for the derivative of dy over dx. Now next, what I need is my dx over dt right here, which I can go ahead and see that that's just this negative eight sine t. So we can start to put this all together. The derivative, the second derivative of y with respects to x is going to be the derivative of dy over dx with respect to t, which we figured out right here is three fourths cosecant squared t. And that's going to be divided by dx over dt, which I can see right here is negative eight sine of t. And the reciprocal of sine so when we have one over sine, that actually is cosecant. So we can actually write this whole thing as three over that's gonna be negative negative three over four times eight, four times eight right there, which is gonna give us 32. And then we're going to have cosecant squared times cosecant, which is gonna be cosecant cubed t. Again, one over sine t, that's just gonna be the cosecant. And then we went ahead and combined this eight into the denominator there. And that's how we got to this next step here. So now we have our second derivative with respect to x. Now what I'm asked to do is evaluate this at the point t is equal to pi over four. So to do this as a final step, I just need to take pi over four and plug it in for t. So if we're looking for the second derivative at the given point, which I can see here is our t is equal to pi over four, but t equals pi over four, well, that's just going to be equal to negative three over 32 cosecant cubed of pi over four. Now what you can do at this point is you have a few options. You can check this on a calculator or verify with the unit circle, but you can also use this identity that we know about. Where we know that one over the sine of t is equal to cosecant of t. And the reason why this is gonna help us is because we're usually a bit more familiar with the sine than we are with the cosecant. So what we can recognize is what's gonna happen if we have the sine of pi over four. And I know the sine of pi over four gives me square root two over two on a unit circle. Now in this case, I can see that we're dealing with something cubed. So if we have sine cubed pi over four, that's just going to be the square root of two over two cubed. And so what you can do is you can plug this in and then take the reciprocal and put it in there. So what this is really gonna come up to is negative three over 32. That's going to be multiplied by, in this case, the cosecant is one over sine. So we could say that same thing as one over, and then since it's cosecant cubed pi over four, we know sine cubed pi over four is this. So I'm going to put that in the denominator. It's going to be square root two over two cubed. Now from here, I want to simplify this further. So we have negative three over 32, and then what is going to be multiplied by one over the square root of two over two cubed. What exactly is this? Well, square root of two cubed, that's square root of two times square root of two times square root of two. Two of these square roots will cancel each other when we combine these, so we're just gonna get two square root two. And then on bottom, we're gonna have two cubed, which is the same thing as eight. Now I can go ahead and simplify this harder because two over eight, that's the same thing as one fourth. So you can go ahead and say this is the same thing as square root two over four. Now continuing to simplify this, well, we have negative three over 32. And then what I can go ahead and do is flip this. Because if I flip this and bring it to the numerator, that's going to give me four over the square root of two. And notice what we have here. We have three times four over 32 times square root of two. But I can actually reduce this because notice that we have a four and we have 32. Four goes into thirty two eight times. So what that means is it will become a one and that will become an eight. So what we end up getting is negative three on top over, and it's going to be eight square root of two. Now the last step I'm gonna take here is to rationalize our denominator. Because I don't want this radical in the bottom here, so I'm going to multiply the top and bottom of this fraction by the square root of two. That'll cancel the square roots right there, and it's going to leave me with two times eight, which is going to give me 16 in the denominator. And then on top, we'll have three times the square root of two, and that right there is our final solution. So this will be the solution to the problem, negative three times the square root of two over 16. That is how you can find the second derivative of a parametric curve if we're looking for it at a certain t value or at a certain point. Hope you found this helpful, and let's move on and get some more practice. See you in the next one.

16. Parametric Equations & Polar Coordinates

Calculus with Parametric Curves

Calculus

16. Parametric Equations & Polar Coordinates

Calculus with Parametric Curves

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Multiple Choice

Find $\frac{d^{2} y}{d x^{2}}$ for the parametric curve at the given point.
$x equals 8 cosine t$ , $y equals 6 sine t$ , $t = \frac{π}{4}$

$\frac{3}{2}$

$- \frac{3}{2}$

$- \frac{3 \sqrt{2}}{16}$

$\frac{3 \sqrt{2}}{16}$

Verified step by step guidance

Step 1: Recall that for parametric equations x = f(t) and y = g(t), the second derivative of y with respect to x, denoted as d²y/dx², can be computed using the formula:

\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}

. This involves finding dy/dx, differentiating it with respect to t, and dividing by dx/dt.

Step 2: Compute dx/dt and dy/dt from the given parametric equations. For x = 8cos(t), differentiate with respect to t to get dx/dt = -8sin(t). For y = 6sin(t), differentiate with respect to t to get dy/dt = 6cos(t).

Step 3: Find dy/dx using the chain rule. Recall that dy/dx = (dy/dt) / (dx/dt). Substitute dy/dt = 6cos(t) and dx/dt = -8sin(t) into the formula to get dy/dx = (6cos(t)) / (-8sin(t)). Simplify this expression.

Step 4: Differentiate dy/dx with respect to t to find d/dt(dy/dx). Use the quotient rule for differentiation: if u(t) = 6cos(t) and v(t) = -8sin(t), then d/dt(dy/dx) = [v(t)u'(t) - u(t)v'(t)] / [v(t)]². Compute u'(t) and v'(t), substitute them into the formula, and simplify.

Step 5: Divide d/dt(dy/dx) by dx/dt to find d²y/dx². Substitute dx/dt = -8sin(t) and the expression for d/dt(dy/dx) into the formula for d²y/dx². Evaluate the result at t = π/4 by substituting t = π/4 into the expressions for sin(t) and cos(t). Simplify the final expression.