Differentiating $v= \frac{d \theta}{dt}-r \sin \theta~i + \frac{d \theta}{dt} r \cos \theta~j$, with respect to time using the product rule

Question

This is what I'm trying to differentiate with respect to time:

The answer is supposed to be found using the product rule:

However, I can't see how the product rule would be used here and what the products actually are so that I can at least try to differentiate.

Answer 1

the product is $$r(d\theta/dt) \cdot (-\sin \theta \mathbf{i} + \cos \theta \mathbf{j})$$ thus by using the product rule,

$$r(\frac{d\theta}{dt})' \cdot (-\sin \theta \mathbf{i} + \cos \theta \mathbf{j})+r(\frac{d\theta}{dt}) \cdot (-\sin \theta \mathbf{i} + \cos \theta \mathbf{j})'$$ $$=r \frac{d^2\theta}{dt^2}(-\sin \theta \mathbf{i} + \cos \theta \mathbf{j})+r(\frac{d\theta}{dt})(-\cos \theta \mathbf{i} - \sin \theta \mathbf{j})(\frac{d\theta}{dt})$$ $$=r \frac{d^2\theta}{dt^2}(-\sin \theta \mathbf{i} + \cos \theta \mathbf{j})-r(\frac{d\theta}{dt})^2(\cos \theta \mathbf{i} + \sin \theta \mathbf{j})$$

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The chain rule was used when differentiating the vector, since $\theta$ is also a function of $t$.

Answer 2

Assuming you're differentiating with respect to $t$ then \begin{align} \frac{d}{dt}\left(-r \sin \theta \frac{d \theta}{dt}\hat{i}+r \cos \theta \frac{d \theta}{dt}\hat{j} \right) &= -\frac{dr}{dt}\sin \theta \frac{d \theta}{dt}\hat{i}-r \cos \theta \left(\frac{d \theta}{dt}\right)^{2}\hat{i}-r \sin \theta \frac{d^{2} \theta}{dt^{2}}\hat{i} \\ &+\frac{dr}{dt} \cos \theta \frac{d \theta}{dt}\hat{j} -r \sin \theta \left(\frac{d \theta}{dt}\right)^{2}\hat{j}+r \cos \theta \frac{d^{2} \theta}{dt^{2}}\hat{j} \end{align} And then collect like terms and simplify.