AP Calculus AB Related Rates Quiz: 10 Questions with Explanations

A \(2\pi r\)B \(2\pi r\frac{dr}{dt}\)C \(\pi r^{2}\frac{dr}{dt}\)D \(2\pi\frac{dr}{dt}\)

Differentiate with respect to t using the chain rule: 2 pi r times dr/dt. Forgetting dr/dt gives only 2 pi r.

A \(\frac{3}{4}\text{ ft/s}\)B \(-\frac{4}{3}\text{ ft/s}\)C \(-1\text{ ft/s}\)D \(-\frac{3}{4}\text{ ft/s}\)

x dx/dt + y dy/dt = 0 gives 3(1) + 4 dy/dt = 0, so dy/dt = -3/4. The negative sign shows the top is sliding down.

A after differentiatingB before differentiatingC only at the end of the graphD never

If you substitute first, a changing quantity becomes a constant and its derivative disappears. Differentiate first.

A \(V=\frac{1}{3}\pi h^{3}\)B \(V=\frac{\pi}{4}h^{2}\)C \(V=\frac{1}{3}\pi r^{2}h\)D \(V=\frac{\pi}{12}h^{3}\)

Similar triangles give r = h/2, so V = (1/3) pi (h/2)^2 h = (pi/12) h^3. The last choice still has two variables.

A \(36\text{ cm}^{3}\text{/s}\)B \(12\text{ cm}^{3}\text{/s}\)C \(8\text{ cm}^{3}\text{/s}\)D \(24\text{ cm}^{3}\text{/s}\)

dV/dt = 3x^2 dx/dt = 3(4)(3) = 36.

A \(12x\)B \(12x\frac{dx}{dt}\)C \(6x^{2}\frac{dx}{dt}\)D \(12\frac{dx}{dt}\)

Differentiate with respect to t and use the chain rule on x: 12x dx/dt. Forgetting dx/dt leaves only 12x.

A \(7\)B \(3.5\)C \(4.8\)D \(5\)

z dz/dt = x dx/dt + y dy/dt gives 10 dz/dt = 6(4) + 8(3) = 48, so dz/dt = 4.8.

A \(\sec^{2}\theta\cdot\frac{1}{100}\frac{dh}{dt}\)B \(\cos^{2}\theta\cdot\frac{1}{100}\frac{dh}{dt}\)C \(\frac{1}{100}\frac{dh}{dt}\)D \(\sin\theta\cdot\frac{dh}{dt}\)

Differentiate: sec^2(theta) dtheta/dt = (1/100) dh/dt, so dtheta/dt = cos^2(theta) (1/100) dh/dt. Forgetting the chain rule factor on theta is the usual slip.

A negativeB positiveC zeroD it depends on the units

A decreasing quantity has a negative derivative with respect to time.

A \(\frac{dL}{dt}\frac{dW}{dt}\)B \(L\frac{dW}{dt}\)C \(\frac{dL}{dt}W+L\frac{dW}{dt}\)D \(\frac{dL}{dt}+\frac{dW}{dt}\)

Use the product rule, because both L and W change with time.