Question

For r(t)=⟨tsin(t),e^t,t+π⟩, then r′(t) is equal to
Select one:
a. ⟨sin(t)−tcos(t),e^t,1⟩+c where c is an arbitrary constant vector

b. ⟨sin(t)+tcos(t),e^t,t⟩


c. ⟨cos(t),e^t,π⟩


d. ⟨sin(t)−tcos(t),e^t⟩


e. ⟨sin(t)+tcos(t),e^t,1⟩

Answer 1

Consider the vector function $r(t)=⟨ t \sin(t),e^{t},t+\pi⟩$

Differentiate with respect to $t$ as,

$r'(t)=⟨ \frac{d}{dt}[t \sin(t)],\frac{d}{dt}[e^{t}],\frac{d}{dt}[t+\pi]⟩$

Here, differentiate $t \sin(t)$ using product rule as shown below:

$=⟨ t\frac{d}{dt}[ \sin(t)]+\sin(t)\frac{d}{dt}(t),e^{t},1⟩$

$=⟨ t\cos(t)+\sin(t),e^{t},1⟩$

Therefore, the derivative of vector function is $r'(t)=⟨ t\cos(t)+\sin(t),e^{t},1⟩$.

Hence, option (e) is correct.