Question

The distance $s$ in meters travelled by a particle in $t$ seconds is given by $s=e^{t}(4\cos 3t+5\sin 3t)$. Then the velocity of the particle at time $t$ is given by

• A. $e^{t}(19\cos 3t-7\sin 3t)$
• B. $e^{t}(12\cos 3t-7\sin 3t)$
• C. $e^{t}(16\cos 3t+9\sin 3t)$
• D. $e^{t}(14\cos 3t-9\sin 3t)$
• E. $e^{t}(19\cos 3t+7\sin 3t)$

Hint:

Logic Tip: A massive time-saving rule: The derivative of $e^{at} \cdot f(t)$ is always $e^{at}[a \cdot f(t) + f'(t)]$. Here $a=1$, so you simply add the original function to its derivative: $(4\cos 3t + 5\sin 3t) + (-12\sin 3t + 15\cos 3t)$, getting $19\cos 3t - 7\sin 3t$ instantly.

Answer 1

The Correct Option is A

Concept:
Velocity $v$ is the first derivative of the position function $s$ with respect to time $t$. To differentiate the given expression, we must use the Product Rule: $\frac{d}{dt}(uv) = u'v + uv'$, along with the Chain Rule for the trigonometric terms.
Step 1: Identify the terms for the Product Rule.
Let $u = e^t$ and $v = (4\cos 3t + 5\sin 3t)$. The derivatives are: $u' = e^t$ $v' = 4(-3\sin 3t) + 5(3\cos 3t) = -12\sin 3t + 15\cos 3t$
Step 2: Apply the Product Rule.
$$v = \frac{ds}{dt} = u'v + uv'$$ $$v = e^t(4\cos 3t + 5\sin 3t) + e^t(-12\sin 3t + 15\cos 3t)$$
Step 3: Factor out $e^t$ and simplify.
$$v = e^t[ (4\cos 3t + 5\sin 3t) + (-12\sin 3t + 15\cos 3t) ]$$ Group the sine and cosine terms together: $$v = e^t[ (4\cos 3t + 15\cos 3t) + (5\sin 3t - 12\sin 3t) ]$$ $$v = e^t(19\cos 3t - 7\sin 3t)$$