# Thread: Differentiation of Multivariable Functions

1. ## Differentiation of Multivariable Functions

I need help with finding the derivative of z, with respect to t, in the following questions:

a) $\displaystyle z\; =\; x^{2}\; -\; y^{2}$, with $\displaystyle x\; =\; t$ and $\displaystyle y\; =\; t^{2}$

b) $\displaystyle z\; =\; \frac{x}{y}$, with $\displaystyle x\; =\; \sin \left( t \right)\;$ and $\displaystyle y\; =\; \cos \left( t \right)$

I believe the method for doing both questions is:

$\displaystyle \frac{dz}{dt}=\frac{\partial z}{\partial x}\frac{dx}{dt}\; +\; \frac{\partial z}{\partial y}\frac{dy}{dt}$

Is this correct? Because no matter how many times I try to do the above questions using that method, I get answers that don't match the ones in the text book.

My answer for a) is $\displaystyle \frac{dz}{dt}=2t-t^{4}-2t^{3}$ and my answer for b) is $\displaystyle \frac{dz}{dt}=1\; +\; \frac{\sin ^{2}t}{\cos ^{2}t}$

2. Originally Posted by DJ Hobo
I need help with finding the derivative of z, with respect to t, in the following questions:

a) $\displaystyle z\; =\; x^{2}\; -\; y^{2}$, with $\displaystyle x\; =\; t$ and $\displaystyle y\; =\; t^{2}$

b) $\displaystyle z\; =\; \frac{x}{y}$, with $\displaystyle x\; =\; \sin \left( t \right)\;$ and $\displaystyle y\; =\; \cos \left( t \right)$

I believe the method for doing both questions is:

$\displaystyle \frac{dz}{dt}=\frac{\partial z}{\partial x}\frac{dx}{dt}\; +\; \frac{\partial z}{\partial y}\frac{dy}{dt}$

Is this correct? Because no matter how many times I try to do the above questions using that method, I get answers that don't match the ones in the text book.

My answer for a) is $\displaystyle \frac{dz}{dt}=2t-t^{4}-2t^{3}$ and my answer for b) is $\displaystyle \frac{dz}{dt}=1\; +\; \frac{\sin ^{2}t}{\cos ^{2}t}$
Your formula is correct but I don't see how you could have gotten that answer for (a) from that formula!

For (a) $\displaystyle \frac{\partial z}{\partial x}= 2x$, $\displaystyle \frac{\partial z}{\partial y}= -2y$, $\displaystyle \frac{dx}{dt}= 1$, and [tex]\frac{dy}{dt}= 2t.

$\displaystyle \frac{dz}{dt}= \frac{\partial z}{\partial x}\frac{dx}{dt}+ \frac{\partial z}{\partial y}\frac{dy}{dt}$
$\displaystyle = (2x)(1)+ (-2y)(2t)= (2t)(1)+ (-2t^2)(2t)$

Your answer for (b) is correct. Is it possible that the answer in the book is $\displaystyle 1+ tan^2(t()$?

3. Originally Posted by HallsofIvy
Your formula is correct but I don't see how you could have gotten that answer for (a) from that formula!

For (a) $\displaystyle \frac{\partial z}{\partial x}= 2x$, $\displaystyle \frac{\partial z}{\partial y}= -2y$, $\displaystyle \frac{dx}{dt}= 1$, and [tex]\frac{dy}{dt}= 2t.

$\displaystyle \frac{dz}{dt}= \frac{\partial z}{\partial x}\frac{dx}{dt}+ \frac{\partial z}{\partial y}\frac{dy}{dt}$
$\displaystyle = (2x)(1)+ (-2y)(2t)= (2t)(1)+ (-2t^2)(2t)$

Your answer for (b) is correct. Is it possible that the answer in the book is $\displaystyle 1+ tan^2(t()$?
Heh I'm gonna have to keep trying with a), then. And no, the answer in the book isn't $\displaystyle 1+ tan^{2}t$. It's $\displaystyle \frac{dz}{dt}=\sec ^{2}t$.

EDIT - I understand why the answer for a) is so, now. Thanks for your help