Solving Related Rate Problem: Find f(x,y) Change at T=3

[Eats Dirt]

Homework Statement

For a related rate problem I am only given the two rates say dx/dt=1 and dy/dt=2 and a the function which depends on both x and y say f(x,y)=4xy^2 find how it is changing at T=3

Homework Equations

Chain rule

The Attempt at a Solution

So I can take both partial derivatives with respect to the chain rule and sub in 1 and 2 for the dx/dt and dy/dt respectively I just have no idea how T comes into play and what to sub in for x and y. My thoughts are that possibly since dx/dt=1 and dy/dt=2 then y=2x and I can isolate for the one variable but then I still don't know what to do form there.

 

[aralbrec]

since dx/dt=1 and dy/dt=2

Those two tell you quite a bit about x and y as functions of time :)

 

[Eats Dirt]

Those two tell you quite a bit about x and y as functions of time :)

A bit more of a pointer would be much appreciated.

 

[HallsofIvy]

The chain rule says that
\frac{df}{dt}= \frac{\partial f}{\partial x}\frac{dx}{dt}+ \frac{\partial f}{\partial y}\frac{dy}{dt}

If f(x,y)= 4xy^2[/tex], what are \partial f/\partial x and \partial f/\partial y[/tex]?<br /> <br /> Now, if you are <b>only</b> told dx/dt= 1 and dy/dt= 2, then you do NOT have enough information to determine what x and y are at t= 3 and so could NOT evaluate the partial derivatives at t= 3.<br /> <br /> If you have a particular problem in mind then please tell us the exact statement of the problem.

 

[Eats Dirt]

The chain rule says that
\frac{df}{dt}= \frac{\partial f}{\partial x}\frac{dx}{dt}+ \frac{\partial f}{\partial y}\frac{dy}{dt}

If f(x,y)= 4xy^2[/tex], what are \partial f/\partial x and \partial f/\partial y[/tex]?<br /> <br /> Now, if you are <b>only</b> told dx/dt= 1 and dy/dt= 2, then you do NOT have enough information to determine what x and y are at t= 3 and so could NOT evaluate the partial derivatives at t= 3.<br /> <br /> If you have a particular problem in mind then please tell us the exact statement of the problem.

<br /> <br /> <br /> It states a cone grows in height starting from zero height dy/dt = 1 radius dx/dt = 2 how fast is the volume growing at t=3

 

[aralbrec]

It states a cone grows in height starting from zero height dy/dt = 1 radius dx/dt = 2 how fast is the volume growing at t=3

You need to find df/dt at t=3, so the first step is to perform the differentiation, as stated above. You'll find the result has x and y in it and to find a value for df/dt at a specific time, you will need to know x and y as functions of time.