Question

hw14 the chain rule (target c4; §3.6)
score: 10/11 answered: 10/11
question 11
a mass attached to a vertical spring has position function given by (s(t)=2sin(2t)) where (t) is measured in seconds and (s) in inches.
find the velocity at time (t = 4).
find the acceleration at time (t = 4).
question help: message instructor

Explanation:

Step1: Recall velocity - position relationship

Velocity $v(t)$ is the derivative of position function $s(t)$. Using the chain - rule, if $s(t)=2\sin(2t)$, then $v(t)=s^\prime(t)$. The derivative of $\sin(u)$ with respect to $x$ is $\cos(u)\cdot u^\prime$, where $u = 2t$ and $u^\prime=2$. So $v(t)=2\cos(2t)\cdot2 = 4\cos(2t)$.

Step2: Calculate velocity at $t = 4$

Substitute $t = 4$ into $v(t)$: $v(4)=4\cos(2\times4)=4\cos(8)$.

Step3: Recall acceleration - velocity relationship

Acceleration $a(t)$ is the derivative of velocity function $v(t)$. Since $v(t)=4\cos(2t)$, using the chain - rule, the derivative of $\cos(u)$ with respect to $x$ is $-\sin(u)\cdot u^\prime$, where $u = 2t$ and $u^\prime = 2$. So $a(t)=v^\prime(t)=4(-\sin(2t))\cdot2=-8\sin(2t)$.

Step4: Calculate acceleration at $t = 4$

Substitute $t = 4$ into $a(t)$: $a(4)=-8\sin(2\times4)=-8\sin(8)$.

Answer:

Velocity at $t = 4$: $4\cos(8)$ inches per second
Acceleration at $t = 4$: $-8\sin(8)$ inches per second²