Question

chain rule: problem 3 (1 point) the displacement of a particle on a vibrating string is given by the equation s(t)=11 + \frac{1}{4}sin(4pi t), where s is measured in centimeters and t in seconds. find the velocity of the particle after t seconds.

Explanation:

Step1: Recall velocity - displacement relation

Velocity $v(t)$ is the derivative of displacement $s(t)$ with respect to time $t$, i.e., $v(t)=\frac{ds(t)}{dt}$. Given $s(t)=11 + \frac{1}{4}\sin(4\pi t)$.

Step2: Differentiate the constant term

The derivative of a constant $C$ with respect to $t$ is 0. The derivative of 11 with respect to $t$ is 0, since $\frac{d}{dt}(11) = 0$.

Step3: Differentiate the sine - term using chain rule

Let $u = 4\pi t$. Then $\frac{du}{dt}=4\pi$. The derivative of $\sin(u)$ with respect to $u$ is $\cos(u)$. By the chain - rule $\frac{d}{dt}(\frac{1}{4}\sin(4\pi t))=\frac{1}{4}\cos(4\pi t)\cdot\frac{d}{dt}(4\pi t)$.

Step4: Calculate the derivative of the sine - term

Since $\frac{d}{dt}(4\pi t)=4\pi$, then $\frac{1}{4}\cos(4\pi t)\cdot\frac{d}{dt}(4\pi t)=\frac{1}{4}\cos(4\pi t)\cdot4\pi=\pi\cos(4\pi t)$.

Step5: Find the velocity function

$v(t)=\frac{ds(t)}{dt}=0+\pi\cos(4\pi t)=\pi\cos(4\pi t)$.

Answer:

$v(t)=\pi\cos(4\pi t)$