Implicit differentiation; related rates; optimization practice questions.

The curve x³ + y³ = 9 passes through the point (1, 2). Determine the value of dy/dx at this point.

• 1/4
• −1/4
• −1/2
• −4

Answer: −1/4

Implicit differentiation gives 3x² + 3y²(dy/dx) = 0, so dy/dx = −x²/y² = −1/4 at (1, 2). Dropping the minus sign or inverting the ratio are the usual errors.

Air is pumped into a spherical balloon so that its volume increases at 32π cm³ s⁻¹. Calculate the rate of increase of the radius, in cm s⁻¹, at the instant when the radius is 4 cm.

• 2
• 8
• 0.5
• 1/(8π)

Answer: 0.5

From V = (4/3)πr³ we get dV/dt = 4πr²(dr/dt); substituting 32π = 4π(16)(dr/dt) gives dr/dt = 0.5 cm s⁻¹.

A 13 m ladder leans against a vertical wall. The foot of the ladder slides away from the wall at 1.2 m s⁻¹. Calculate the speed, in m s⁻¹, at which the top slides down the wall when the foot is 5 m from the wall.

• 0.5
• 1.2
• 2.88
• 0.25

Answer: 0.5

With x² + y² = 169, differentiating gives x(dx/dt) + y(dy/dt) = 0; at x = 5, y = 12, so dy/dt = −(5)(1.2)/12 = −0.5, a downward speed of 0.5 m s⁻¹.

Given that sin y = x, where −π/2 < y < π/2, determine dy/dx in terms of y.

• cos y
• −1/cos y
• sin y
• 1/cos y

Answer: 1/cos y

Differentiating implicitly: cos y (dy/dx) = 1, so dy/dx = 1/cos y, which equals 1/√(1 − x²) — the derivative of arcsin x.

Two positive real numbers x and y satisfy xy = 64. Determine the minimum possible value of x + y.

• 8
• 32
• 16
• 64

Answer: 16

With S = x + 64/x, S′ = 1 − 64/x² = 0 gives x = 8, so y = 8 and the minimum sum is 16; S″ = 128/x³ > 0 confirms a minimum.

An open-topped box has a square base of side x cm and volume 32 cm³. Determine the value of x that minimises the total external surface area of the box.

• 2
• 4
• 8
• 16

Answer: 4

With h = 32/x², the surface area is S = x² + 4xh = x² + 128/x; S′ = 2x − 128/x² = 0 gives x³ = 64, so x = 4.