Question
The position of an object attached to a spring is given by \[y(t)=\frac{1}{6}cos(5t)-\frac{1}{4}sin(5t) \], where t is time in seconds. In the first 4 seconds, how many times is the velocity of the object equal to 0?
(A) Zero
(B) Three
(C) Five
(D) Six
(E) Seven
▶️ Answer/Explanation
Solution
1. Find the velocity function by differentiating the position function:
\( v(t) = y'(t) = -\frac{5}{6}\sin(5t) – \frac{5}{4}\cos(5t) \)
2. Set velocity equal to zero and solve:
\( -\frac{5}{6}\sin(5t) – \frac{5}{4}\cos(5t) = 0 \)
Simplify to: \( \frac{\sin(5t)}{\cos(5t)} = -\frac{3}{2} \) or \( \tan(5t) = -1.5 \)
3. Find solutions in the interval \( t \in [0,4] \):
\( 5t = \tan^{-1}(-1.5) + n\pi \) where n is an integer
Within 4 seconds, there are 6 solutions (n = 0 to 5)
✅ Answer: D) Six
▶️ Answer/Explanation
Solution
1. Given constraint: \( h + 2\pi r = 30 \) ⇒ \( h = 30 – 2\pi r \)
2. Volume formula: \( V = \pi r^2 h = \pi r^2 (30 – 2\pi r) = 30\pi r^2 – 2\pi^2 r^3 \)
3. Find critical points by taking derivative:
\( \frac{dV}{dr} = 60\pi r – 6\pi^2 r^2 \)
Set equal to zero: \( 60\pi r – 6\pi^2 r^2 = 0 \)
Factor: \( 6\pi r (10 – \pi r) = 0 \)
Solutions: \( r = 0 \) (minimum) or \( r = \frac{10}{\pi} \) (maximum)
4. Verify maximum using second derivative test:
\( \frac{d^2V}{dr^2} = 60\pi – 12\pi^2 r \)
At \( r = \frac{10}{\pi} \): \( \frac{d^2V}{dr^2} = 60\pi – 12\pi^2 (\frac{10}{\pi}) = -60\pi < 0 \) ⇒ maximum
✅ Answer: E) \(\frac{10}{\pi}\) cm