Applications of Derivatives Question - None

Question

Match the following:

(A) If the radius of a circle is 6 cm and it increases at the rate of 0.06 cm/s, then the rate of increase of area is ___

(B) If $V = x^3$ and $\frac{\delta x}{x} \times 100 = 2\%$, then $\frac{\delta V}{V} \times 100$ = ___

(C) If $(x-2)\frac{dx}{dt} = 3\frac{dx}{dt}$, then $x$ = ___

(D) If $A = \frac{\sqrt{3}}{4}x^2$ and $\frac{dx}{dt} = 30$, then $\frac{dA}{dt}$ at $x=1$ is ___

Accepted Answer

Problem A: Rate of increase of area

Given: r = 6 cm, dr/dt = 0.06 cm/s

A = πr²

dA/dt = 2πr(dr/dt) = 2π(6)(0.06) = 0.72π cm²/s ≈ 2.26 cm²/s → q

Problem B: Percentage error in volume

V = x³, so dV/dx = 3x²

For small changes: (ΔV/V) × 100 ≈ 3(Δx/x) × 100

Given (δx/x) × 100 = 2%, therefore (δV/V) × 100 = 3 × 2 = 6% → r

Problem C: Solving for x

(x-2)(dx/dt) = 3(dx/dt)

Assuming dx/dt ≠ 0, divide both sides by dx/dt:

x - 2 = 3

x = 5 → p

Problem D: Rate of change of area at x=1

A = (√3/4)x²

dA/dt = (√3/4) · 2x · (dx/dt) = (√3/2)x(dx/dt)

At x = 1 with dx/dt = 30:

dA/dt = (√3/2)(1)(30) = 15√3 → s

∴ Answer: A→q; B→r; C→p; D→s

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