A student solving a ladder problem notes that at the moment of interest x = 6 and y = 8, so she substitutes these into x² + y² = 100 first, obtaining 36 + 64 = 100. She then differentiates to find 0 = 0. What error did she make?
AShe used the Pythagorean theorem incorrectly for a moving triangle.
BShe substituted instantaneous values before differentiating, collapsing the variable relationship into a constant.
CShe should have differentiated each variable separately before writing the equation.
DThe equation x² + y² = 100 does not hold when the ladder is sliding.
Substituting before differentiating freezes the variables at a single instant, turning the equation into a numerical identity (36 + 64 = 100) whose derivative is zero. This destroys the rate information. The correct procedure: first differentiate x² + y² = 100 with respect to t (yielding 2x·dx/dt + 2y·dy/dt = 0), then substitute the known instantaneous values of x, y, and any given rates to solve for the unknown rate.
Question 2 Multiple Choice
The radius of a spherical balloon is increasing at 3 cm/s. What is dV/dt when the radius is 5 cm? (V = 4/3 πr³)
A12π cm³/s, because dV/dt = 4π(dr/dt)²
B100π cm³/s, because dV/dt = (4/3)π(dr/dt)³
C300π cm³/s, because dV/dt = 4πr² · dr/dt
D60π cm³/s, because dV/dt = 4πr · dr/dt
Differentiating V = (4/3)πr³ with respect to t via the chain rule gives dV/dt = 4πr² · dr/dt. At r = 5 and dr/dt = 3: dV/dt = 4π(25)(3) = 300π cm³/s. Option A mistakes dr/dt for r; option D drops the square on r; option B substitutes dr/dt in place of r before differentiating — the classic error of confusing a variable's current value with its rate of change.
Question 3 True / False
In a related-rates problem, all variables in the geometric equation are implicitly functions of time t, even when t does not appear explicitly in the equation.
TTrue
FFalse
Answer: True
This is the foundational conceptual shift in related rates. A balloon's radius r and volume V are not static numbers — they change as time passes, making them functions r(t) and V(t). The equation V = (4/3)πr³ holds at every instant, so it is an identity between two time-varying functions. Differentiating both sides with respect to t (and applying the chain rule to r) is only valid because r = r(t). If r were a fixed constant, dr/dt would be zero and the equation would carry no dynamic information.
Question 4 True / False
You can substitute the known instantaneous values of most position variables into the relating equation before differentiating, as long as you keep the rates (dx/dt, dy/dt) as unknowns.
TTrue
FFalse
Answer: False
This is the most common error in related rates. Substituting position values before differentiating turns variables into constants, making the chain rule yield zero. The rates dx/dt and dy/dt cannot be extracted once x and y have been replaced by numbers. The correct order is always: (1) write the geometric equation in terms of variables, (2) differentiate both sides with respect to t applying the chain rule, (3) then substitute known instantaneous values and known rates to solve for the unknown rate.
Question 5 Short Answer
Why must you differentiate the relating equation before substituting the known instantaneous values of the variables?
Think about your answer, then reveal below.
Model answer: Because substituting turns variables into constants, and the derivative of a constant is zero — destroying all information about rates. The chain rule can only extract dr/dt, dx/dt, etc., while r and x are still present as variables. Once you substitute x = 6 into x², you get 36, a number with no derivative relationship to time. Differentiation with respect to t must act while all quantities are still expressed as changing functions of t.
The relating equation (like x² + y² = L²) is true at every instant — it is an identity between functions of t, not a fact about one particular moment. Differentiating with respect to t converts this identity into a relationship between rates. Only after that step can you substitute specific values to find a specific rate at a specific moment. This ordering is what makes related rates a differentiation problem, not just an algebra problem.