Solving Home Exercises: Am I Doing It Right?

[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

<br /> H(t) = sin (3x) - y<br />

<br /> x = 2t^2 - 3<br />

<br /> y = \frac{t^2}{2} - 5t + 1<br />This is what I did:

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

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

<br /> \frac{dH}{dt} = 12t\cos (6t^2 - 9) - t + 5<br />

Is it ok?

Thank you.

 

[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 ?

 

[twoflower]

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 ?

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

<br /> u(t) = 2t^2 - 3<br />

<br /> v(t) = \frac{t^2}{2} - 5t + 1<br />

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

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

 

[Fermat]

I'm sorry, I forgot to mention that I defined
<br /> u(t) = 2t^2 - 3<br />
<br /> v(t) = \frac{t^2}{2} - 5t + 1<br />
So which form is ok? The first or the second?
I find the first one strange because of the mixing of x and t.

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