Question

suppose z=(y^4)(x^3). at the instant when x=1 and y=3, x is decreasing at 2 units/sec and y is increasing at 2 units/sec. how fast is z changing at this instant? is z increasing or decreasing?

Answer 1

Based on the chain rule:\[
\frac{dz}{dt}=\frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt}
\]

Answer 2

"x is decreasing at 2 units/sec"\[
\frac{dx}{dt}=-2
\]"y is increasing at 2 units/sec"\[
\frac{dy}{dt}=2
\]

Answer 3

@chels.408 Do you get it?

Answer 4

not really..

Answer 5

Do you know partial derivatives?

Answer 6

no

Answer 7

You must know multivariable calculus to do this problem, so you need to know partial derivatives.

Answer 8

\[
z=(y^4)(x^3)
\]You can do partial derivative with respect to \(x\)?

Answer 9

\[
z=f(x,y)=(y^4)(x^3)
\]\[
\frac{\partial z}{\partial x}=f_x(x,y)=3(y^4)(x^2)
\]

Answer 10

I just treat \(y\) as a constant and differentiate with respect to \(x\).
Do you understand?

Answer 11

oooh I see. thanks

Answer 12

So find \(f_x(1,3)\) and \(f_y(1,3)\)

Answer 13

\[
z'=f_x(1,3)x'+f_y(1,3)y'
\]
This should make it a bit more clear.

Answer 14

"x is decreasing at 2 units/sec"\[
x'=-2
\]"y is increasing at 2 units/sec"\[
y'=2
\]

Answer 15

@chels.408 Can you do it now?

Answer 16

yes thank you