Calculus BC · Unit 4
Relate two changing quantities, differentiate with respect to time, then substitute.
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An equation between the quantities comes first
Name the lengths, areas, or volumes. Write the formula that connects them at every moment. That formula does not yet contain the rates.
Differentiate with respect to time
Every changing quantity gets a \(d/dt\). Then substitute the numbers from the instant you care about, including the rate you were given.
Find a missing length before you substitute
If the ladder problem gives one leg, the other leg comes from the Pythagorean theorem. Put that length in after you differentiate.
A negative result is a decreasing quantity
If \(dy/dt\) comes out negative, that length is getting shorter at that instant. Report the rate and the direction.
One formula, one known rate. Area, volume, or circumference.
\(A = \pi r^2\)
\(\dfrac{dA}{dt} = 2\pi r\,\dfrac{dr}{dt}\)
\(r = 6\) and \(\dfrac{dr}{dt} = 2\)
\(2\pi(6)(2)\)
\(24\pi\)
Answer: \(24\pi\) cm²/s
\(A = s^2\)
\(\dfrac{dA}{dt} = 2s\,\dfrac{ds}{dt}\)
\(2(4)(3)\)
\(24\)
Answer: 24 cm²/s
\(V = s^3\)
\(\dfrac{dV}{dt} = 3s^2\,\dfrac{ds}{dt}\)
\(3(16)(0.5)\)
\(24\)
Answer: 24 cm³/s
\(\dfrac{dS}{dt} = 8\pi r\,\dfrac{dr}{dt}\)
\(\dfrac{dr}{dt} = -1\)
\(8\pi(5)(-1)\)
\(-40\pi\)
Answer: \(-40\pi\) cm²/min
\(\dfrac{dV}{dt} = 4\pi r^2\,\dfrac{dr}{dt}\)
\(4\pi(100)(3)\)
\(1200\pi\)
Answer: \(1200\pi\) cm³/s
\(C = 2\pi r\)
\(\dfrac{dC}{dt} = 2\pi\,\dfrac{dr}{dt}\)
\(2\pi\cdot 4\)
\(8\pi\)
The rate does not depend on the current radius.
Answer: \(8\pi\) cm/s
Pythagoras, a cone, or two cars. Find the missing length, then substitute.
\(x^2 + y^2 = 100\)
When \(x = 6\), \(y = 8\).
\(2x\,\dfrac{dx}{dt} + 2y\,\dfrac{dy}{dt} = 0\)
\(\dfrac{dy}{dt} = -\dfrac{x}{y}\,\dfrac{dx}{dt}\)
\(-\dfrac{6}{8}\cdot 2\)
\(-\dfrac{3}{2}\)
The top is falling at \(\dfrac{3}{2}\) ft/s.
Answer: \(\dfrac{3}{2}\) ft/s downward
Similar triangles: \(\dfrac{r}{h} = \dfrac{4}{12} = \dfrac{1}{3}\), so \(r = \dfrac{h}{3}\).
\(V = \dfrac{1}{3}\pi r^2 h = \dfrac{\pi h^3}{27}\)
\(\dfrac{dV}{dt} = \dfrac{\pi h^2}{9}\,\dfrac{dh}{dt}\)
Substitute \(\dfrac{dV}{dt} = 2\) and \(h = 6\).
\(2 = \dfrac{\pi\cdot 36}{9}\,\dfrac{dh}{dt}\)
\(2 = 4\pi\,\dfrac{dh}{dt}\)
\(\dfrac{dh}{dt} = \dfrac{1}{2\pi}\)
Answer: \(\dfrac{1}{2\pi}\) m/min
After 1 hour, \(x = 30\), \(y = 40\), and \(z = 50\).
\(z^2 = x^2 + y^2\)
\(2z\,\dfrac{dz}{dt} = 2x\,\dfrac{dx}{dt} + 2y\,\dfrac{dy}{dt}\)
\(50\,\dfrac{dz}{dt} = 30\cdot 30 + 40\cdot 40\)
\(50\,\dfrac{dz}{dt} = 2500\)
\(\dfrac{dz}{dt} = 50\)
Answer: 50 mph
Similar triangles: \(\dfrac{15}{x + s} = \dfrac{6}{s}\).
\(15s = 6(x + s)\)
\(9s = 6x\)
\(s = \dfrac{2}{3}x\)
\(\dfrac{ds}{dt} = \dfrac{2}{3}\,\dfrac{dx}{dt}\)
\(\dfrac{2}{3}\cdot 4\)
\(\dfrac{8}{3}\)
Answer: \(\dfrac{8}{3}\) ft/s
The tip is at distance \(x + s\) from the post.
\(x + s = x + \dfrac{2}{3}x = \dfrac{5}{3}x\)
\(\dfrac{d}{dt}(x + s) = \dfrac{5}{3}\cdot 4\)
\(\dfrac{20}{3}\)
Answer: \(\dfrac{20}{3}\) ft/s
\(A = LW\)
\(\dfrac{dA}{dt} = \dfrac{dL}{dt} W + L\,\dfrac{dW}{dt}\)
\(2\cdot 5 + 8\cdot(-1)\)
\(2\)
Answer: 2 cm²/s
An angle, a diagonal, a cylinder, or a rope. Keep the known rate on its own line.
\(\tan\theta = \dfrac{100}{x}\)
\(\sec^2\theta\,\dfrac{d\theta}{dt} = -\dfrac{100}{x^2}\,\dfrac{dx}{dt}\)
When \(x = 100\), \(\tan\theta = 1\), so \(\theta = \dfrac{\pi}{4}\) and \(\sec^2\theta = 2\).
\(2\,\dfrac{d\theta}{dt} = -\dfrac{100}{10000}\cdot 8\)
\(2\,\dfrac{d\theta}{dt} = -\dfrac{2}{25}\)
\(\dfrac{d\theta}{dt} = -\dfrac{1}{25}\)
The angle is decreasing at \(\dfrac{1}{25}\) radian per second.
Answer: \(-\dfrac{1}{25}\) rad/s
\(z^2 = L^2 + W^2\)
\(z = \sqrt{64 + 25} = \sqrt{89}\)
\(2z\,\dfrac{dz}{dt} = 2L\,\dfrac{dL}{dt} + 2W\,\dfrac{dW}{dt}\)
\(2\sqrt{89}\,\dfrac{dz}{dt} = 2(8)(2) + 2(5)(-1)\)
\(2\sqrt{89}\,\dfrac{dz}{dt} = 22\)
\(\dfrac{dz}{dt} = \dfrac{11}{\sqrt{89}}\)
Answer: \(\dfrac{11}{\sqrt{89}}\) cm/s
\(V = \pi r^2 h\)
\(\dfrac{dV}{dt} = \pi\left(2r\,\dfrac{dr}{dt}\, h + r^2\,\dfrac{dh}{dt}\right)\)
\(\pi\bigl(2\cdot 4\cdot 0.2\cdot 10 + 16\cdot(-0.5)\bigr)\)
\(\pi(16 - 8)\)
\(8\pi\)
Answer: \(8\pi\) cm³/s
\(x^2 + 144 = z^2\)
When \(z = 13\), \(x = 5\).
\(2x\,\dfrac{dx}{dt} = 2z\,\dfrac{dz}{dt}\)
\(\dfrac{dz}{dt} = -2\)
\(5\,\dfrac{dx}{dt} = 13(-2)\)
\(\dfrac{dx}{dt} = -\dfrac{26}{5}\)
The boat approaches at \(\dfrac{26}{5}\) ft/s.
Answer: \(\dfrac{26}{5}\) ft/s toward the dock
\(\dfrac{dy}{dt} = -\dfrac{x}{y}\cdot 2\)
When \(x = 6\), \(y = 8\), and the top falls at \(\dfrac{3}{2}\) ft/s.
When \(x = 8\), \(y = 6\), and the top falls at \(\dfrac{8}{3}\) ft/s.
\(\dfrac{8}{3} > \dfrac{3}{2}\)
The top is falling faster when the bottom is 8 ft from the wall.
Answer: faster when the bottom is 8 ft out: \(\dfrac{8}{3}\) ft/s
Differentiate both sides with respect to \(t\).
Each variable produces its own rate.
\(2x\,\dfrac{dx}{dt} + 2y\,\dfrac{dy}{dt} = 0\)
Answer: A. \(2x\,\dfrac{dx}{dt} + 2y\,\dfrac{dy}{dt} = 0\)
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