Solving Related Rates Problems - AP Calculus AB

A circle expands with $\frac{dr}{dt}=4$ ; what is $\frac{dA}{dt}$ when $r=2$ , given $A=\pi r^2$ ?
Answer: $16\pi$ This related rates problem involves circle area $A = \pi r^2$ with $\frac{dr}{dt} = 4$ . Differentiate: $\frac{dA}{dt} = 2\pi r \frac{dr}{dt}$ . At $r=2$ , $\frac{dA}{dt} = 2\pi(2)(4) = 16\pi$ . A tempting misinterpretation is using the circumference derivative. The rate doubles with radius. A transferable strategy is to apply the power rule with chain rule for geometric rates.
Water fills a cone with $V=\frac{1}{3}\pi r^2h$ and $r=\frac{1}{2}h$ ; if $\frac{dh}{dt}=3$ , find $\frac{dV}{dt}$ at $h=4$ .
Answer: $12\pi$ units $^3$ /s This related rates problem concerns the volume of a cone where the radius r is half the height h, so r = h/2. Substitute into V = (1/3)πr²h to get V = (1/3)π(h/2)²h = (π/12)h³. Differentiate with respect to t: dV/dt = (π/4)h² dh/dt. At h = 4 and dh/dt = 3, dV/dt = (π/4)(16)(3) = 12π units³/s. A common mistake is to differentiate without substituting r in terms of h, leading to extra variables. The correct approach yields 12π by expressing everything in terms of one variable before differentiating. A transferable strategy is to express the quantity in terms of a single changing variable when possible, differentiate, and plug in values.
A kite string length $s$ satisfies $s^2=x^2+144$ ; if $\frac{dx}{dt}=2$ , find $\frac{ds}{dt}$ when $x=5$ .
Answer: $\frac{10}{13}$ Related rates in AP Calculus AB help analyze how rates interconnect in geometric contexts like distances. For kite string s² = x² + 144 with dx/dt = 2, differentiate to 2s ds/dt = 2x dx/dt, so ds/dt = (x/s) dx/dt. At x = 5, s = 13, ds/dt = (5/13)*2 = 10/13. A mistake might be using ds/dt = dx/dt without the ratio, getting 2. Always include the chain rule factors. The correct answer is 10/13. A transferable strategy is to differentiate squared relations carefully and solve for the target rate using current values.
A balloon's surface area is $S=4\pi r^2$ ; if $\frac{dr}{dt}=\frac{1}{2}$ , find $\frac{dS}{dt}$ when $r=6$ .
Answer: $24\pi \text{ units}^2 / \text{s}$ This problem applies related rates to the surface area of a balloon $S = 4\pi r^2$ . Differentiate: $\frac{dS}{dt} = 8\pi r \frac{dr}{dt}$ . At $r = 6$ and $\frac{dr}{dt} = \frac{1}{2}$ , $\frac{dS}{dt} = 8\pi(6)(\frac{1}{2}) = 24\pi$ units $^2$ /s. A common mistake is to use the volume derivative instead, leading to incorrect powers. The rate reflects faster area growth for larger radii. A transferable strategy is to differentiate surface or volume formulas, incorporate given rates, and evaluate at specified points.
For a rectangle with constant perimeter $P=60$ , length increases at $2$ cm/s; find $\frac{dA}{dt}$ when $L=20$ .
Answer: $-20$ cm $^2$ /s This related rates problem deals with a rectangle of constant perimeter P = 60 = 2L + 2W, so W = 30 - L. The area A = L W = L(30 - L), and dA/dt = (30 - 2L) dL/dt. At L = 20 and dL/dt = 2 cm/s, dA/dt = (30 - 40)(2) = -20 cm²/s. A common mistake is to assume area increases when length does, ignoring the decreasing width. The negative rate indicates area decrease past the maximum. A transferable strategy is to express dependent variables in terms of one, differentiate, and substitute rates and values carefully.
A triangle has area $A=\frac{1}{2}x y$ ; if $\frac{dx}{dt}=4$ and $y=10$ constant, find $\frac{dA}{dt}$ .
Answer: $20$ Related rates in AP Calculus AB apply to areas like triangle A=(1/2)xy with dx/dt=4 and y=10 constant. Then dA/dt = (1/2)(y dx/dt) since dy/dt=0, giving (1/2)*10*4=20. No x value needed as it's independent. A common mistake is unnecessarily including x or forgetting the 1/2. Verify which variables change. The correct answer is 20. A transferable strategy is to set rates of constants to zero and simplify the differentiated equation.
A sphere's radius increases at $2$ cm/s; what is $\dfrac{dV}{dt}$ when $r=3$ cm, given $V=\frac{4}{3}\pi r^3$ ?
Answer: $72\pi$ cm $^3$ /s This problem involves related rates, where the volume of a sphere changes as its radius increases. The volume is given by V = (4/3)πr³, and we differentiate both sides with respect to time t to get dV/dt = 4πr² dr/dt. Given dr/dt = 2 cm/s and r = 3 cm, we substitute to find dV/dt = 4π(3)²(2) = 72π cm³/s. A tempting misinterpretation is to plug in the values before differentiating, which would incorrectly ignore the chain rule. Instead, always differentiate first and then substitute the known values. The correct reasoning confirms the rate of volume increase at that instant. A transferable strategy is to identify the relating equation, differentiate implicitly with respect to time, solve for the desired rate, and evaluate at the given point.
A rectangle has constant area $A=100$ with $A=LW$ ; if $\frac{dL}{dt}=2$ , find $\frac{dW}{dt}$ when $L=20$ .
Answer: $-\frac{1}{2}$ Related rates in AP Calculus AB for constant areas like A=100=LW with dL/dt=2, differentiate L dW/dt + W dL/dt=0, dW/dt=-(W/L) dL/dt. At L=20, W=5, dW/dt=-(5/20)*2=-1/2. Shows inverse change. A mistake might be using dW/dt = -dL/dt, getting -2. Include the ratio. The correct answer is -1/2. A transferable strategy is to set dA/dt=0 for constants and solve for compensatory rates.
A circle's radius decreases at $1$ cm/s; what is $\dfrac{dA}{dt}$ when $r=7$ , given $A=\pi r^2$ ?
Answer: $-14\pi$ cm $^2$ /s Related rates in AP Calculus AB for areas like circle $A=\pi r^2$ with $dr/dt=-1$ , $dA/dt=2\pi r \cdot dr/dt$ . At r=7, $dA/dt=2\pi \cdot 7 \cdot (-1)=-14\pi$ . Indicates decreasing area. A common error is using $\pi r \cdot dr/dt$ , getting $-7\pi$ . Include the 2 from power rule. The correct answer is $-14\pi$ cm $^2$ /s. A transferable strategy is to differentiate power functions carefully and interpret negative rates as decreases.
A point moves on $x^2 - y^2 = 15$ ; if $\frac{dx}{dt} = 2$ , find $\frac{dy}{dt}$ at $(4,1)$ .
Answer: $8$ For hyperbolas in AP Calculus AB related rates, $x^2 - y^2 = 15$ with $\frac{dx}{dt} = 2$ , differentiate to $2x \frac{dx}{dt} - 2y \frac{dy}{dt} = 0$ , so $\frac{dy}{dt} = \frac{x}{y} \frac{dx}{dt}$ . At (4,1), $\frac{dy}{dt} = (4/1)*2 = 8$ . This shows the relation. A mistake might be switching signs, getting -8 for $y^2 - x^2$ . Check the equation form. The correct answer is 8. A transferable strategy is to differentiate conic sections implicitly and isolate the desired rate using point values.
A right triangle has legs $x$ and $y$ with hypotenuse $13$ ; if $\frac{dx}{dt}=5$ , find $\frac{dy}{dt}$ when $x=5$ .
Answer: $-\frac{25}{12}$ In this related rates problem, a right triangle has legs x, y and hypotenuse 13, so x² + y² = 169. Differentiate: 2x dx/dt + 2y dy/dt = 0, dy/dt = -(x/y) dx/dt. At x = 5, y = 12, dx/dt = 5, dy/dt = -(5/12)(5) = -25/12. A common mistake is to use the hypotenuse in the rate formula incorrectly. The negative indicates y decreases as x increases. A transferable strategy is to use the Pythagorean theorem, differentiate, and plug in after solving for the rate.
A rectangle has diagonal $10$ with sides $x,y$ so $x^2+y^2=100$ ; if $\frac{dx}{dt}=1$ , find $\frac{dy}{dt}$ at $x=8$ .
Answer: $-\frac{4}{3}$ Related rates in AP Calculus AB connect changing quantities through implicit differentiation with respect to time. For a rectangle with diagonal 10, so $x^2 + y^2 = 100$ and $\frac{dx}{dt} = 1$ , differentiate to get $2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0$ , solving for $\frac{dy}{dt} = -\left(\frac{x}{y}\right) \frac{dx}{dt}$ . At $x = 8$ , $y = 6$ , $\frac{dy}{dt} = -\frac{8}{6} \times 1 = -\frac{4}{3}$ . A tempting error is using $y = \sqrt{100 - 64} = 6$ but forgetting the negative sign, getting $\frac{4}{3}$ . Remember rates can be negative indicating decrease. The correct answer is $-\frac{4}{3}$ . A transferable strategy is to solve for the unknown rate after differentiating and consider signs based on context.
A cube's surface area is $S=6s^2$ ; if $\frac{ds}{dt}=2$ , find $\frac{dS}{dt}$ when $s=3$ .
Answer: $72$ This related rates problem finds the surface area change for a cube S=6s². Differentiate: dS/dt=12s ds/dt. At s=3, ds/dt=2, dS/dt=12(3)(2)=72. A tempting misinterpretation is using volume instead. The rate increases linearly with s. A transferable strategy is to differentiate polynomial expressions and evaluate step-by-step.

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