Exercise 6.1 answers: Application of Derivatives

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(a) r = 3 cm

1. Area of a circle: A = πr².
2. Rate of change of A with respect to r means dA/dr.
3. Differentiate: dA/dr = 2πr.
4. Put r = 3: dA/dr = 2π×3 = 6π.

Answer6π cm² per cm

(b) r = 4 cm

1. Use the same rate, dA/dr = 2πr.
2. Put r = 4: dA/dr = 2π×4 = 8π.

Answer8π cm² per cm

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1. Let the edge of the cube be a. Volume V = a³ and surface area S = 6a².
2. Differentiate V with respect to time: dV/dt = 3a² × da/dt.
3. We are told dV/dt = 8, and a = 12, so 8 = 3×144×da/dt, giving da/dt = 1/54 cm/s.
4. Differentiate S with respect to time: dS/dt = 12a × da/dt.
5. Put a = 12 and da/dt = 1/54: dS/dt = 12×12×(1/54) = 144/54 = 8/3.

Answer8/3 cm² per second

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1. Area of a circle: A = πr².
2. Differentiate with respect to time: dA/dt = 2πr × dr/dt.
3. We are given dr/dt = 3 cm/s and r = 10 cm.
4. dA/dt = 2π×10×3 = 60π.

Answer60π cm² per second

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1. Let the edge of the cube be a. Volume V = a³.
2. Differentiate with respect to time: dV/dt = 3a² × da/dt.
3. We are given da/dt = 3 cm/s and a = 10 cm.
4. dV/dt = 3×100×3 = 900.

Answer900 cm³ per second

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1. Let r be the radius of the circular wave. Enclosed area A = πr².
2. Differentiate with respect to time: dA/dt = 2πr × dr/dt.
3. We are given dr/dt = 5 cm/s and r = 8 cm.
4. dA/dt = 2π×8×5 = 80π.

Answer80π cm² per second

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1. Circumference of a circle: C = 2πr.
2. Differentiate with respect to time: dC/dt = 2π × dr/dt.
3. We are given dr/dt = 0.7 cm/s.
4. dC/dt = 2π×0.7 = 1.4π = 7π/5.

Answer7π/5 cm per second (that is, 1.4π cm/s)

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(a) the perimeter

1. The length x is decreasing, so dx/dt = −5 cm/min. The width y is increasing, so dy/dt = 4 cm/min.
2. Perimeter P = 2(x + y).
3. Differentiate with respect to time: dP/dt = 2(dx/dt + dy/dt) = 2(−5 + 4) = −2.

Answer−2 cm/min: the perimeter is decreasing at 2 cm per minute

(b) the area of the rectangle

1. Area A = xy. Both x and y change with time, so use the product rule.
2. dA/dt = x × dy/dt + y × dx/dt.
3. Put x = 8, y = 6, dx/dt = −5, dy/dt = 4: dA/dt = 8×4 + 6×(−5) = 32 − 30 = 2.

Answer2 cm²/min: the area is increasing at 2 cm² per minute

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1. Let r be the radius of the balloon. Volume of a sphere: V = (4/3)πr³.
2. Differentiate with respect to time: dV/dt = 4πr² × dr/dt.
3. We are given dV/dt = 900 cm³/s and r = 15 cm.
4. 900 = 4π×225×dr/dt = 900π×dr/dt, so dr/dt = 1/π.

Answer1/π cm per second

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1. Volume of a sphere: V = (4/3)πr³.
2. "Rate at which the volume is increasing with the radius" means dV/dr, not a rate with time.
3. Differentiate: dV/dr = 4πr².
4. Put r = 10: dV/dr = 4π×100 = 400π.

Answer400π cm³ per cm of radius

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1. Let x be the distance of the foot from the wall and y be the height on the wall, both in cm.
2. The ladder length is fixed: x² + y² = 500² (since 5 m = 500 cm).
3. When x = 400 cm (4 m), y = √(500² − 400²) = √90000 = 300 cm, by the Pythagoras rule.
4. Differentiate x² + y² = 500² with respect to time: 2x(dx/dt) + 2y(dy/dt) = 0.
5. We are given dx/dt = 2 cm/s. Put x = 400, y = 300: 2(400)(2) + 2(300)(dy/dt) = 0.
6. 1600 + 600(dy/dt) = 0, so dy/dt = −1600/600 = −8/3.
7. The negative sign means the height is decreasing, at a rate of 8/3 cm/s.

Answer8/3 cm per second

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1. The curve is 6y = x³ + 2. Differentiate both sides with respect to time: 6(dy/dt) = 3x²(dx/dt).
2. The condition "y-coordinate changing 8 times as fast as x-coordinate" means dy/dt = 8(dx/dt).
3. Substitute: 6×8×(dx/dt) = 3x²×(dx/dt). Since dx/dt is not zero, cancel it: 48 = 3x².
4. So x² = 16, giving x = 4 or x = −4.
5. Find y from the curve equation 6y = x³ + 2.
6. At x = 4: 6y = 64 + 2 = 66, so y = 11. Point: (4, 11).
7. At x = −4: 6y = −64 + 2 = −62, so y = −31/3. Point: (−4, −31/3).

Answer(4, 11) and (−4, −31/3)

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1. The bubble is spherical, so its volume is V = (4/3)πr³.
2. Differentiate with respect to time: dV/dt = 4πr² × dr/dt.
3. We are given dr/dt = 1/2 cm/s and r = 1 cm.
4. dV/dt = 4π×1×(1/2) = 2π.

Answer2π cm³ per second

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1. The diameter is (3/2)(2x + 1), so the radius is half of that: r = (3/4)(2x + 1).
2. Volume of a sphere: V = (4/3)πr³.
3. Substitute r: V = (4/3)π × [(3/4)(2x + 1)]³ = (9/16)π(2x + 1)³.
4. Differentiate with respect to x: dV/dx = (9/16)π × 3(2x + 1)² × 2 = (27/8)π(2x + 1)².

Answer(27π/8)(2x + 1)²

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1. Let h be the height and r the radius of the sand cone, with h = r/6, so r = 6h.
2. Volume of a cone: V = (1/3)πr²h. Substitute r = 6h: V = (1/3)π(6h)²h = 12πh³.
3. Differentiate with respect to time: dV/dt = 36πh² × dh/dt.
4. We are given dV/dt = 12 cm³/s and h = 4 cm.
5. 12 = 36π×16×dh/dt = 576π×dh/dt, so dh/dt = 1/(48π).

Answer1/(48π) cm per second

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1. Marginal cost is the rate at which the total cost changes with output, that is, dC/dx.
2. C(x) = 0.007x³ − 0.003x² + 15x + 4000.
3. Differentiate: dC/dx = 0.021x² − 0.006x + 15.
4. Put x = 17: dC/dx = 0.021×289 − 0.006×17 + 15 = 6.069 − 0.102 + 15 = 20.967.

Answer₹20.967 (about ₹20.97)

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1. Marginal revenue is the rate at which the total revenue changes with the number sold, that is, dR/dx.
2. R(x) = 13x² + 26x + 15.
3. Differentiate: dR/dx = 26x + 26.
4. Put x = 7: dR/dx = 26×7 + 26 = 182 + 26 = 208.

Answer₹208

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1. Area of a circle: A = πr².
2. Differentiate: dA/dr = 2πr.
3. Put r = 6: dA/dr = 2π×6 = 12π.
4. This matches option (B).

Answer(B) 12π

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1. Marginal revenue is dR/dx.
2. R(x) = 3x² + 36x + 5. Differentiate: dR/dx = 6x + 36.
3. Put x = 15: dR/dx = 6×15 + 36 = 90 + 36 = 126.
4. This matches option (D).

Answer(D) 126

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