← Back to learning objective

Velocity from a damped oscillation position function

A particle moves along a straight line so that its position at time t≥0t \geq 0 is given by s(t)=e−tcos⁡(t)s(t) = e^{-t}\cos(t), representing a damped oscillation.

Find the velocity of the particle at t=π2t = \dfrac{\pi}{2}.

Hint

Differentiate s(t)=e−tcos⁡(t)s(t) = e^{-t}\cos(t) using the product rule to get v(t)=s′(t)v(t) = s'(t), then substitute t=π/2t = \pi/2, recalling that cos⁡(π/2)=0\cos(\pi/2) = 0 and sin⁡(π/2)=1\sin(\pi/2) = 1.

Step-by-step solution

Apply the product rule to s(t)=e−tcos⁡(t)s(t) = e^{-t}\cos(t): v(t)=s′(t)=ddt[e−t]cos⁡(t)+e−tddt[cos⁡(t)]=−e−tcos⁡(t)−e−tsin⁡(t)v(t) = s'(t) = \frac{d}{dt}[e^{-t}]\cos(t) + e^{-t}\frac{d}{dt}[\cos(t)] = -e^{-t}\cos(t) - e^{-t}\sin(t). This simplifies to v(t)=−e−t(cos⁡(t)+sin⁡(t))v(t) = -e^{-t}(\cos(t) + \sin(t)).

Substitute t=π/2t = \pi/2: since cos⁡(π/2)=0\cos(\pi/2) = 0 and sin⁡(π/2)=1\sin(\pi/2) = 1, we get v(π/2)=−e−π/2(0+1)=−e−π/2v(\pi/2) = -e^{-\pi/2}(0 + 1) = -e^{-\pi/2}. This is a negative value, meaning the particle is moving in the negative direction at that instant.

Related problems

Practice similar Calculus problems interactively.

Start Closing the Gap