# Homework Help: Am I doing it right?

1. Nov 19, 2005

### twoflower

Hi all,

we were given some recommended home excercises and since we hadn't been given the right results, I'm curous if I'm doing it right:

1. Find $\frac{dH}{dt}$, where

$$H(t) = sin (3x) - y$$

$$x = 2t^2 - 3$$

$$y = \frac{t^2}{2} - 5t + 1$$

This is what I did:

$$\frac{dH}{dt} = \frac{\partial H}{\partial x}\frac{du}{dt} + \frac{\partial H}{\partial y}\frac{dv}{dt} = 12t\cos (3x) - t + 5$$

Which seems kind of strange to me. Should I replace x with t? It would be this:

$$\frac{dH}{dt} = 12t\cos (6t^2 - 9) - t + 5$$

Is it ok?

Thank you.

Last edited: Nov 19, 2005
2. Nov 19, 2005

### Fermat

Yes, it's fine.

But you should have had dx/dt ,not du/dt, and dy/dt, not dv/dt, in the expression for dH/dt.

Why did it seem strange to you ?

3. Nov 19, 2005

### twoflower

I'm sorry, I forgot to mention that I defined

$$u(t) = 2t^2 - 3$$

$$v(t) = \frac{t^2}{2} - 5t + 1$$

So which form is ok? The first or the second?

I find the first one strange because of the mixing of x and t.

4. Nov 20, 2005

### Fermat

OK.
I'm not sure why you defined u and v like that but your mixing of symbols like that is a bit confusing and your differential eqn is wrong the way it is written.
By that I mean ...

$$\frac{dH}{dt} = \frac{\partial H}{\partial x}\cdot\frac{du}{dt} + \frac{\partial H}{\partial y}\cdot\frac{dv}{dt}$$

is just plain wrong where it has written,

$$\frac{\partial H}{\partial x}\cdot\frac{du}{dt}$$

You should have,

$$\frac{\partial H}{\partial x}\cdot\frac{dx}{dt}$$
or
$$\frac{\partial H}{\partial u}\cdot\frac{du}{dt}$$

The $$\partial u$$ should sort of "cancel" with $$du$$ to give$$\frac{\partial H}{dt}$$ so that you have $$\frac{dH}{dt}$$ on one side of the "=" sign and $$\frac{\partial H}{dt}$$ (twice) on the other side, making both sides consistent with each other.

So, the form of your differential eqn should be either,

$$\frac{dH}{dt} = \frac{\partial H}{\partial x}\cdot\frac{dx}{dt} + \frac{\partial H}{\partial y}\cdot\frac{dy}{dt}$$
or
$$\frac{dH}{dt} = \frac{\partial H}{\partial u}\cdot\frac{du}{dt} + \frac{\partial H}{\partial v}\cdot\frac{dv}{dt}$$

depending upon what definitions (u = f(t) or x = f(t)) you are using.

You are finding dH/dt, which is the rate at which H varies as t varies. i.e. dH/dt is a function of t, so should be expressed in terms of t.
You can work out things using x and y,or u and v (sometimes using a substitution can simplify working) but your final result shoud be converted to give an expresion that involves t only.

HTH

Last edited: Nov 20, 2005