Second derivative Assume a curve is given by the parametric eq...

Question

Second derivative Assume a curve is given by the parametric equations x=f(t) and y=g(t), where f and g are twice differentiable. Use the Chain Rule to show that y″x=(fʹ(t)g″(t)−gʹ(t)f″(t))/(fʹ(t))³.

Accepted Answer

Hello. In this video, we are given a pair of parametric equations on the boundary from 0 to pi, but what we want to do is we want to find the 2nd derivative with respect to t. So in order to find the 2nd derivative, we must first find the 1st derivative. Now the first derivative of parametric equations is defined as Dyd X, and this is equivalent to the derivative of y with respect to t divided by the derivative of x with respect to t. So let's go ahead and first find the respective derivatives for both of the parametric equations. Now X is defined as t minus sin of t, and y is defined as 1 minus cosine of t. Now the derivative of X with respect to T is going to equal to 1 minus cosine of T, and the derivative of Y with respect to T is going to equal to positive sin of T. So by plugging this into the derivative formula, we get the derivative DYDX is equal to sin of t. Divided by 1 minus cosine of t. So now that we have the 1st derivative, let's go ahead and define the 2nd derivative. Now the second derivative is defined as the following. D2y over dx2 is equivalent to the derivative with respect to t of the first derivative Dydx divided by the derivative of x with respect to t. So what we're going to do is we're going to take the derivative of sin oft divided by 1 minus cosinet first, and divide that by the derivative of X with respect to T. Now, in order to take the derivative of this function, we need to use the quotient rule. That is going to be 1 minus cosine of T multiplied by the derivative of sin of T, which is cosine of T. Minus sin of t. Multiplied by the derivative of 1 minus cosine of t, which is sin of t, and this will all be divided by the quantity 1 minus cosine of t quantity squared. This is going to simplify to be cosine of t. Minus cosine squad of t. Minus sin squared of t. All divided by 1 minus cosine of t quantity squared. Now, here, what we can go ahead and do is factor out a -1 from cosine 2 minus sin 2, and that's going to simplify to be cosine of T minus 1 divided by 1 minus cosine of T quantity squared. And by reversing the order, we get -1 divided by 1 minus cosine of t. Now, what we're going to do is we're going to take this quantity and divide it by the derivative with respect to X. So the second derivative is going to be 1 divided. By 1 minus cosine of t. And this would be divided by the derivative with respect to x, which is 1 minus cosine of t, and this is going to finally simplify to be 1-1 divided by 1 minus cosine of t quantity squared. This is going to be the 2nd derivative of the given parametric equations, and the overall solution to the problem is going to be option B. So I hope this video helps you in understanding how to approach this problem, and we will go ahead and see you all in the next video.

Second derivative Assume a curve is given by the parametric equations x=f(t) and y=g(t), where f and g are twice differentiable. Use the Chain Rule to show that y″x=(fʹ(t)g″(t)−gʹ(t)f″(t))/(fʹ(t))³.

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