0
$\begingroup$

Suppose I have a function $h(\theta)$ measuring the height of a piston, with $\theta = \omega t$. I would like to know the vertical acceleration of this piston as $\omega$ changes at the point $\theta = \theta_0$. How would I differentiate $h$ to do this?

$\endgroup$
  • $\begingroup$ I have tried both partial derivation and the total derivative, I get meaningless results each time. E.g. $\frac{\partial h}{\partial \omega}(\omega t) = t \frac{\partial h}{\partial \theta}(\omega t)$. To me, this seems incorrect as you would expect the velocity to increase with $\omega$. Would you consider showing me what you mean? $\endgroup$ – Mikkel Rev Apr 7 at 10:15
1
$\begingroup$

If $\omega$ is not constant, I don't really see a reason to write $\theta = \omega t$. You can do it, I just don't see why it would be convenient. It's probably better to just think of $\theta(t)$.

So, I'm assuming that you have actual expressions for the functions $h(\theta)$ and $\theta(t)$ somewhere, even if you didn't put them in your question. If I misunderstood, please correct me.

We will use the chain rule:

$$\frac{d}{dx} f(g(x)) = \frac{df}{dg}\Big(g(x)\Big) \, \frac{dg}{dx}(x)$$

More specifically:

$$\frac{d}{dt}f(\theta(t)) = \frac{df}{d\theta}\Big( \theta(t) \Big) \, \frac{d\theta}{dt}(t)$$

You are interested in the vertical acceleration, so that is the second derivative of $h$ with respect to time:

$$\frac{d^2}{dt^2} h(\theta(t)) = \frac{d}{dt} \Bigg( \frac{dh}{d\theta} \bigg( \theta(t) \bigg) \, \frac{d\theta}{dt}(t) \Bigg) = \frac{d}{dt} \Bigg( \frac{dh}{d\theta} \bigg( \theta(t) \bigg) \Bigg) \, \frac{d\theta}{dt}(t) + \frac{dh}{d\theta} \bigg( \theta(t) \bigg) \, \frac{d^2\theta}{dt^2}(t) = \frac{d^2h}{d\theta^2} \bigg( \theta(t) \bigg) \bigg( \frac{d\theta}{dt} (t) \bigg)^2 + \frac{dh}{d\theta} \bigg( \theta(t) \bigg) \, \frac{d^2\theta}{dt^2}(t) $$

In your case, where you are interested in the acceleration at a specific angle $\theta = \theta_0$, you'll need to work out the values of all elements in the expression above at the moment that $\theta = \theta_0$ and use those to get your result.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.