Question

an object oscillates along a line, and its displacement in centimeters is given by y(t)=36(sin t - 1), where t≥0 is measured in seconds and y is positive in the upward direction. complete parts (a) through (f) below. a. graph the position function for 0≤t≤10. b. find the velocity of the oscillator, v(t)=y(t).

Explanation:

Step1: Recall derivative of sine - function

The position function is $y(t)=36(\sin t - 1)$. The derivative of $\sin t$ is $\cos t$ and the derivative of a constant is 0. Using the sum - difference rule of differentiation $(u - v)'=u'-v'$, where $u = 36\sin t$ and $v = 36$.

Step2: Differentiate $y(t)$

$v(t)=y'(t)=\frac{d}{dt}(36\sin t-36)$. By the constant multiple rule $\frac{d}{dt}(cf(t)) = c\frac{d}{dt}(f(t))$ and the sum - difference rule, we have $y'(t)=36\frac{d}{dt}(\sin t)-\frac{d}{dt}(36)$. Since $\frac{d}{dt}(\sin t)=\cos t$ and $\frac{d}{dt}(36) = 0$, then $v(t)=36\cos t$.

Answer:

$v(t)=36\cos t$