How Do You Calculate the Derivative of a Function with Parametric Equations?

[Cpt Qwark]

Homework Statement

Given f(x,y)=ycosx,x(t)=t^2,y(t)=sint

Calculate \frac{df((x(t),y(t))}{dt}
t∈ℝ

Homework Equations

For parametric equations I know
\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}

The Attempt at a Solution

So far I have found
\frac{∂f}{∂x}=-ysinx
\frac{∂f}{∂y}=cosx
\frac{dx}{dt}=2t
\frac{dy}{dt}=sint

but I'm not too sure where to go afterwards.

 

[Noctisdark]

Why do you just write f(x(t),y(t)) just change x but x(t) and y(t) and you'll get f as a WHOLE function of time, all you need there is takeout the derivative :)

 

[pasmith]

Homework Statement

Given f(x,y)=ycosx,x(t)=t^2,y(t)=sint

Calculate \frac{df((x(t),y(t))}{dt}
t∈ℝ

Homework Equations

For parametric equations I know
\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}

The Attempt at a Solution

So far I have found
\frac{∂f}{∂x}=-ysinx
\frac{∂f}{∂y}=cosx
\frac{dx}{dt}=2t
\frac{dy}{dt}=sint

but I'm not too sure where to go afterwards.

Apply the chain rule.

 

[Cpt Qwark]

Why do you just write f(x(t),y(t)) just change x but x(t) and y(t) and you'll get f as a WHOLE function of time, all you need there is takeout the derivative :)

oh so just f(x,y)=sintcost^2

 

[Noctisdark]

oh so just f(x,y)=sintcost^2

Yes, But write f(t) = sin(t)cos(t²) and pull out the derivative :3

 

[HallsofIvy]

If f(x,y)= y cos(x) with x= t^2 and y= sin(t), then, yes, f(x)= sin(t)cos(t^2) and you can differentiate that.

But I think it is more likely that you expected to use the chain rule:
\frac{df}{dt}= \frac{\partial f}{\partial x}\frac{dx}{dt}+ \frac{\partial f}{\partial y}\frac{dy}{dt}

 

[vela]

Homework Statement

Given f(x,y)=y \cos x,x(t)=t^2,y(t)=\sin t, calculate \frac{df((x(t),y(t))}{dt}, t∈\mathbb{R}.

Homework Equations

For parametric equations I know
\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}

The Attempt at a Solution

So far I have found
\begin{align*}\frac{\partial f}{\partial x}&=-y\sin x \\
\frac{\partial f}{\partial y}&=\cos x \\
\frac{dx}{dt}&=2t \\
\frac{dy}{dt}&=\sin t
\end{align*} but I'm not too sure where to go afterwards.

It's curious why you would calculate these derivatives if you didn't have some idea already in mind about how to solve the problem. Perhaps you just weren't clear on the details. When you reached this point, a good idea would have been to consult your notes or textbook for an example like this problem to see how to put it all together.