I don't know how to begin solving this problem. If anyone could help me get the equations that I need and small hints, I would hugely appreciate it!
The voltage across a resistor is given by $V(t) = e^{-0.3t}\sin t$
a. What is the instantaneous rate of change of the voltage at $t=3$ and $t=7$.
b. What are the linear approximations (or linearizations) of $V(t)$ at $t=3$ and $t=7$? Sketch the voltage and linearizations.
On
Hints.
a. One interpretation of the derivative of a function at a point of its domain is as the rate of change of the function at that point. Can you now find the rate of change of $V$ with respect to $t$? In short, differentiate, then substitute the given values of $t$ to find the needed rates.
b. Since $dV=V'dt,$ then for small values of $\Delta t,$ we have that $dV\approx\Delta V,$ so that we have the approximation $$\Delta V=V'\Delta t,$$ or $$V-V_0=V'_0(t-t_0),$$ where $V_0,\,V'_0$ are the values of the voltage and its rate of change at the point $t_0.$ Can you now continue?
Finally, to sketch the voltage, note that it is a decaying sinusoid. Decaying because the exponential term decreases as $t$ increases, and $\sin$ is bounded. At the indicated points especially, the linear approximations will be the tangent lines to the curve.
For part a, the "instantaneous rate of change of a function" at a given point is the derivative of the function at that point. So you need to take the derivative of the voltage and then plug in the given values of $t$. You will want to use the product rule.
b.) The linearization of a function $f$ at a given point $a$ is the function $g$ given by$$g(x) = f'(a)(x-a) + f(a)$$ So for instance, one linearization would be $$V'(3)(t-3) + f(3)$$
So you just need to plug in what you got for $V'(3)$ in part a.