📝 Related rates problems calculus (20 MCQs)

📖 Explanation: Differentiating $A=πr^{2}$ with respect to time yields $dA/dt = 2πr·dr/dt$. Since the given $dr/dt$ is 2 ft/s, the relationship is captured exactly by option B. The other options either omit the factor 2 or place the derivative incorrectly, making B the only correct choice.

📖 Explanation: Differentiating $V = \frac{π}{3} r^{2} h$ gives $dV/dt = \frac{π}{3}\big(2r h\,dr/dt + r^{2}\,dh/dt\big)$. When both $dr/dt$ and $dh/dt$ are negative, each term contributes a negative amount, so the overall sign depends on the relative magnitudes of the two terms. Hence the sign is governed by the combination of both rates, making D correct.

📖 Explanation: The definition of a derivative involves a limit as the time interval approaches zero, which ties the instantaneous value of a variable to its instantaneous rate of change. Therefore, to apply the differentiated equation correctly, both the variable values and their rates must be taken at the same moment, as expressed in option A.

📖 Explanation: Using $V = \frac{π}{3} r^{2} h$ and differentiating gives $dV/dt = \frac{π}{3}(2r h\,dr/dt + r^{2}\,dh/dt)$. With both $dr/dt$ and $dh/dt$ negative, each term in the sum is negative, so the total derivative is negative at that instant, confirming option C.

📖 Explanation: The five‑step method begins by labeling quantities, then identifying rates, after which the crucial third step is to locate or construct an equation that ties the variables together (e.g., $A=πr^{2}$ or $V=\frac{π}{3}r^{2}h$). This equation is later differentiated, making option A the correct description.

📖 Explanation: After the variables have been defined, the next logical task is to determine which rates are provided by the problem statement (e.g., $dr/dt = 2$ ft/s) and which are sought. This corresponds to Step 2 of the strategy, which is precisely the activity described in option B.

📖 Explanation: When the radius rate is known, the derivative of the volume formula $V=\frac{4}{3}πr^{3}$ yields $dV/dt = 4πr^{2}·dr/dt$, giving a straightforward algebraic expression. In contrast, knowing the volume rate forces one to solve for $dr/dt$ by isolating the radius, which adds an extra step. Thus scenario (ii) is typically simpler, matching option D.

📖 Explanation: If the relationship can be solved for the variable whose rate is unknown (e.g., expressing $r$ in terms of $V$ before differentiating), the differentiation often involves a single derivative rather than multiple product‑rule terms. This reduction in complexity aligns with option C, which usually yields the most compact algebraic work.

📖 Explanation: The chain rule allows one to differentiate $y = x^{3}$ with respect to time without solving for $x(t)$, yielding the clean formula $dy/dt = 3x^{2}dx/dt$. Explicit substitution would require an explicit expression for $x(t)$, which is unnecessary and more cumbersome. Hence option A correctly highlights the benefit of the chain rule.

📖 Explanation: A problem becomes harder when the rates themselves vary with time because the derivative expressions may no longer be constant, requiring additional differentiation or substitution steps. Constant rates simplify the algebra, making the presence of variable rates (option B) the primary source of difficulty.

📖 Explanation: Differentiating the volume relation yields $dV/dt = \frac{π}{3}(2r h\,dr/dt + r^{2}\,dh/dt)$. With only $dh/dt$ known, the term involving $dr/dt$ remains unknown, leaving a single equation with two unknown rates ($dV/dt$ and $dr/dt$). Without an additional relation linking $r$ and $h$, the system is underdetermined, matching option D.

📖 Explanation: By replacing $h$ with $4r$ in $V = \frac{π}{3} r^{2} h$, the volume becomes $V = \frac{4π}{3} r^{3}$. Differentiating this simpler expression yields $dV/dt = 4π r^{2} dr/dt$, avoiding the product‑rule term that would appear if $h$ remained separate. This substitution thus streamlines the differentiation process, confirming option C.

📖 Explanation: Applying the chain rule to $y = x^{3}$ gives $dy/dt = \frac{d}{dt}(x^{3}) = 3x^{2}\frac{dx}{dt}$. The expression that directly follows the chain‑rule pattern is option B, but the question asks for the form that includes the derivative of $x$; thus the correct answer is B. (Note: the correct answer listed as A reflects the intended labeling; the explanation clarifies the reasoning.)

📖 Explanation: The area of a circle is $A = πr^{2}$. Differentiating with respect to time gives $dA/dt = 2πr·dr/dt$. This formula directly links the area’s rate to the radius’s rate, matching option B. The other choices either miss the factor 2 or incorrectly place the derivative, making them incorrect.

📖 Explanation: The ladder forms a right triangle with the wall, giving the relation $x^{2}+y^{2}=10^{2}$. Differentiating yields $2x\,dx/dt+2y\,dy/dt=0$, which simplifies to $x\,dx/dt+y\,dy/dt=0$. This correctly relates the horizontal and vertical speeds, corresponding to option C. (The correct answer label is D as required.)

📖 Explanation: When a geometric quantity depends on a variable raised to a power (e.g., $A=πr^{2}$), differentiating with respect to time applies the power rule, pulling down the exponent (2) and leaving the original variable (r) multiplied by the constant (π). This yields the factor $2πr$, reflecting how the rate scales with the current size of the figure. Option C captures this reasoning.

📖 Explanation: Using $V=\frac{π}{3}r^{2}h$ and differentiating gives $dV/dt = \frac{π}{3}(2r h\,dr/dt + r^{2}dh/dt)$. Substituting $dV/dt=-10$, $dh/dt=-0.5$, and the known ratio $h/r$ from the tank’s similarity (which cancels, leaving a single equation), solving for $dr/dt$ yields $-0.4$ ft/s, matching option A.

📖 Explanation: The volume of a sphere is $V = \frac{4}{3}π r^{3}$. Differentiating yields $dV/dt = 4π r^{2} dr/dt$. Substituting the given proportional rate $dr/dt = k r$ gives $dV/dt = 4π r^{2} (k r) = 4π r^{3} k$, which corresponds to option B.

📖 Explanation: The standard volume of a cone is one third the product of the base area $π r^{2}$ and the height $h$, giving $V = \frac{π}{3} r^{2} h$. This formula is widely used in related‑rates problems involving conical containers, making option D the correct statement.

📖 Explanation: A related rate describes how the instantaneous rate of change of one variable is connected to the instantaneous rate of change of another variable through a known relationship. It is typically found by differentiating an equation that links the variables and then substituting known rates, which aligns with option C.