Question
The sides of the rectangle above increase in such a way that \(\frac{dz}{dt}=1 \) = and \(\frac{ dx}{dy}= 3\frac{dy}{dt}\) . At the instant when x = 4 and 3 y = , what is the value of \(\frac{dx}{dt}\)?
(A) \(\frac{1}{3}\) (B) 1 (C) 2 (D)\(\sqrt{5}\) (E) 5
▶️Answer/Explanation
Ans:B
Question
What is the instantaneous rate of change at x = 2 of the function f given by \(f(x)=\frac{x^2-2}{x-1}\) ?
(A) 2 (B) \(\frac{1}{6}\) (C)\(\frac{1}{2}\) (D) 2 (E) 6
▶️Answer/Explanation
Ans:D
\(f'(x)=\frac{(x-1)(2x)-(x^2-2)(1)}{(x-1)^2}; f'(2)\frac{(2-1)(4)-(4-2)(1)}{(2-1)^2}=2\)
▶️ Answer/Explanation
Solution
Step 1: Define Rates
Area: \( A = \pi r^2 \)
Area rate: \( \frac{dA}{dt} = 2\pi r \cdot \frac{dr}{dt} \)
Circumference: \( C = 2\pi r \)
Circumference rate: \( \frac{dC}{dt} = 2\pi \cdot \frac{dr}{dt} \)
Given: \( \frac{dA}{dt} = 2 \cdot \frac{dC}{dt} \)
Step 2: Solve for Radius
Substitute: \( 2\pi r \cdot \frac{dr}{dt} = 2 \cdot (2\pi \cdot \frac{dr}{dt}) \)
\( 2\pi r \cdot \frac{dr}{dt} = 4\pi \cdot \frac{dr}{dt} \)
Divide by \( \frac{dr}{dt} \) (non-zero): \( 2\pi r = 4\pi \)
\( r = 2 \)
Answer: D
▶️ Answer/Explanation
Solution
Step 1: Volume Rate
Volume: \( V = \frac{4}{3}\pi r^3 \)
Derivative: \( \frac{dV}{dt} = 4\pi r^2 \cdot \frac{dr}{dt} \)
Given: \( \frac{dV}{dt} = -2\pi \), \( r = 5 \)
\( -2\pi = 4\pi (5)^2 \cdot \frac{dr}{dt} \)
\( -2\pi = 100\pi \cdot \frac{dr}{dt} \)
\( \frac{dr}{dt} = -\frac{2\pi}{100\pi} = -\frac{1}{50} \)
Step 2: Surface Area Rate
Surface area: \( S = 4\pi r^2 \)
Derivative: \( \frac{dS}{dt} = 8\pi r \cdot \frac{dr}{dt} \)
Substitute: \( r = 5 \), \( \frac{dr}{dt} = -\frac{1}{50} \)
\( \frac{dS}{dt} = 8\pi (5) \cdot \left(-\frac{1}{50}\right) = 40\pi \cdot \left(-\frac{1}{50}\right) = -\frac{40\pi}{50} = -\frac{4\pi}{5} \)
Rate of decrease: \( \frac{4\pi}{5} \)
Answer: A