Second-Order Partial Derivative of a Parametric Function

[molybdenum42]

The problem is from an online homework assignment. I know it's probably fairly simple, but my brain isn't grasping it right now for some reason.[The Problem]

We know:

r(t) = <3t² - 8t + 3, -9t² + 2t + 7>

And we are asked to find d²y/dx².[Background Information]

My understanding of d²y/dx² is that it is the second derivative with respect to x (the first derivative of the function having been with respect to both x and y).

In other words, it's broken down like this:

(d/dx)(dy/dx) = d²y/dx²
The derivative (with respect to x) of the first derivative of the parametric function, r(t), is equal to that mess on the right hand side of the equation.

[Attempt at a Solution]

So we know that:

(dy/dx) = (dy/dt) / (dx/dt) (From the textbook.)

And the above function, r(t), can be broken down into two parts:

x(t) = -9t² + 2t + 7
y(t) = 3t² - 8t + 3

Therefore:

(dx/dt) = 6t - 8
(dy/dt) = -18t + 2

And

(dy/dx) = (-18t + 2) / (6t - 8)So, now, here's where I feel like I'm guessing a little bit. Following that logic, would d²y/dx² = (6) * [(-18t + 2)/(6t - 8)]?
Heh, this is probably a silly question, but thanks very much in advance for any help!

 

[ehild]

x(t) = -9t² + 2t + 7
y(t) = 3t² - 8t + 3

Therefore:

(dx/dt) = 6t - 8
(dy/dt) = -18t + 2

And

(dy/dx) = (-18t + 2) / (6t - 8)


So, now, here's where I feel like I'm guessing a little bit. Following that logic, would d²y/dx² = (6) * [(-18t + 2)/(6t - 8)]?

dy/dx is correct, but I can not follow your logic afterwards. d²y/dx² =d(dy/dx)/dx = (d(dy/dx)/dt)/(dx/dt). Are you sure you did the derivative of the fraction correctly?



ehild