Application of Derivatives Exercise 01
Question:
Sand is pouring from a pipe at the rate of 12cm^3/s. The falling sand forms a cone on the ground in such a way that the height of the cone is always onesixth of the radius of the base. How fast is the height of the sand cone increasing when the height is 4cm.
Answer:

Given: rate of sand pouring = 12 cm^3/s, height of cone = 4 cm, ratio of height to radius = 1/6

Find: Rate of increase in height of sand cone

Formula: Rate of increase in height = (Rate of sand pouring) / (Volume of cone)

Calculate the radius of the cone: Radius = 6 × Height = 6 × 4 = 24 cm

Calculate the volume of the cone: Volume of cone = (1/3) × π × (Radius)^2 × Height = (1/3) × π × (24)^2 × 4 = 809.6 cm^3

Calculate the rate of increase in height of the sand cone: Rate of increase in height = (Rate of sand pouring) / (Volume of cone) = 12 cm^3/s / 809.6 cm^3 = 0.0148 cm/s
Question:
A balloon, which always remains spherical has a variable radius. Find the rate at which its volume is increasing with the radius when the later is 10cm.
Answer:
Step 1: Identify the formula for the volume of a sphere: V = (4/3)πr3
Step 2: Substitute the given radius (10 cm) into the formula: V = (4/3)π(10 cm)3
Step 3: Calculate the volume of the sphere: V = 4,189.79 cm3
Step 4: Calculate the rate of change of the volume with respect to the radius: dV/dr = (4/3)πr2
Step 5: Substitute the given radius (10 cm) into the formula: dV/dr = (4/3)π(10 cm)2
Step 6: Calculate the rate of change of the volume with respect to the radius: dV/dr = 4,188.79 cm3/cm
Question:
Find the rate of change of the area of a circle with respect to its radius r when (i) r=3 cm (ii) r=4 cm
Answer:
(i) When r=3 cm, the area of a circle is A=πr²
A = π(3)² A = 9π cm²
The rate of change of the area of a circle with respect to its radius r is dA/dr = 2πr
dA/dr = 2π(3) dA/dr = 6π cm/cm
(ii) When r=4 cm, the area of a circle is A=πr²
A = π(4)² A = 16π cm²
The rate of change of the area of a circle with respect to its radius r is dA/dr = 2πr
dA/dr = 2π(4) dA/dr = 8π cm/cm
Question:
A stone is dropped into a quiet lake and waves move in circles at the speed of 5 cm/s. At the instant when the radius of the circular wave is 8 cm, how fast is the enclosed area increasing?
Answer:

Calculate the circumference of the circular wave: 2πr = 2π(8 cm) = 16π cm

Calculate the speed of the wave: 5 cm/s

Calculate the rate of change of the enclosed area: (speed of wave) × (circumference of wave) = (5 cm/s) × (16π cm) = 80π cm2/s
Question:
A balloon which always remains spherical has a variable radius. The rate at which its volume is increasing is mπ cm^2 with the radius when the later is 10 cm. Find m
Answer:

Volume of a sphere is given by V = 4/3πr^3

Differentiate both sides with respect to time to get the rate of change of volume
dV/dt = 4πr^2 dr/dt
 Substitute the given values in the equation
4π(10 cm)^2 dr/dt = mπ cm^2
 Solve for m
m = 4 dr/dt
Question:
The total revenue in Rupees received from the sale of x units of a product is given by R(x)=13x^2+26x+15. Find the marginal revenue when x=7.
Answer:
R(x) = 13x^2 + 26x + 15
R’(x) = 26x + 26
R’(7) = 26(7) + 26
R’(7) = 198
Question:
A particle moves along the curve 6y=x^3+2. Find the points on the curve at which the ycoordinate is changing 8 times as fast as the xcoordinate.
Answer:

Write the equation of the curve in the standard form, y=f(x): y = (x^3 + 2)/6

Calculate the derivative of the equation: f’(x) = (3x^2)/6

Set the derivative equal to 8 times the xcoordinate: (3x^2)/6 = 8x

Solve for x: x = (48)^(1/3)

Substitute the value of x into the original equation to find the corresponding yvalue: y = (48^(1/3) + 2)/6
Therefore, the points on the curve at which the ycoordinate is changing 8 times as fast as the xcoordinate are (48^(1/3), (48^(1/3) + 2)/6).
Question:
The length x of a rectangle is decreasing at the rate of 5 cm/minute and the width y is increasing at the rate of 4 cm/minute. When x=8 cm and y=6 cm, find the rates of change of the area of rectangle(in cm^2/min).
Answer:
Step 1: Find the area of the rectangle when x = 8 cm and y = 6 cm.
Area = x × y = 8 cm × 6 cm = 48 cm^2
Step 2: Find the rate of change of the length and width of the rectangle.
Rate of change of length (x) = 5 cm/min Rate of change of width (y) = 4 cm/min
Step 3: Find the rate of change of the area of the rectangle.
Rate of change of area = Rate of change of length × Rate of change of width = (5 cm/min) × (4 cm/min) = 20 cm^2/min
Question:
The radius of an air bubble is increasing at the rate of 1/2cm/s. At what rate is the volume of the bubble increasing when the radius is 1cm?
Answer:

Calculate the volume of the bubble when the radius is 1cm: V = (4/3)πr³ V = (4/3)π(1cm)³ V = (4/3)π cm³

Calculate the rate at which the volume is increasing when the radius is increasing at the rate of 1/2cm/s: Rate of change in volume = (4/3)π(1/2cm/s)³ Rate of change in volume = (4/3)π(1/8 cm³/s) Rate of change in volume = (4/24)π cm³/s
Therefore, the rate at which the volume of the bubble is increasing when the radius is 1cm is (4/24)π cm³/s.
Question:
The volume of a cube is increasing at the rate of 8cm^3/s. How fast is the surface area increasing when the length of an edge is 12cm?
Answer:

Calculate the volume of the cube when the length of an edge is 12cm: Volume = 12cm x 12cm x 12cm = 1728 cm^3

Calculate the rate of increase of the surface area: Rate of increase of the surface area = 8cm^3/s / 1728cm^3 = 0.0046 cm^2/s
Question:
An edge of a variable cube is increasing at the rate of 5 cm/s. How fast is the volume of the cube increasing when the edge is 10 cm long?
Answer:
Step 1: The edge of a variable cube is increasing at the rate of 5 cm/s.
Step 2: The edge of the cube is 10 cm long.
Step 3: The volume of the cube is given by V = e3, where e is the edge length.
Step 4: The rate at which the volume is increasing is given by dV/dt = 3e2 * de/dt.
Step 5: Substituting e = 10 cm and de/dt = 5 cm/s, we get dV/dt = 3000 cm3/s.
Step 6: Therefore, the volume of the cube is increasing at a rate of 3000 cm3/s when the edge is 10 cm long.
Question:
The total revenue in Rupees received from the sale of x units of a product is given by R(x)=3x^2+36x+5. The marginal revenue, when x=15 is. A 116 B 96 C 90 D 126
Answer:
Answer: D 126
Explanation:
The marginal revenue can be calculated using the formula R’(x) = 6x + 36.
Therefore, when x = 15, R’(x) = 6(15) + 36 = 126.
Question:
The length x of a rectangle is decreasing at the rate of 5 cm/minute and the width y is increasing at the rate of 4 cm/minute. When x=8cm and y=6cm find the rates of change of (a) perimeter, and (b) the area of the rectangle.
Answer:
(a) Rate of change of perimeter (P) = dx/dt + dy/dt = 5 cm/min + 4 cm/min = 1 cm/min
(b) Rate of change of area (A) = dA/dt = xdy/dt + ydx/dt = 8 cm4 cm/min + 6 cm(5 cm/min) = 20 cm2/min
Question:
A ladder 5 m long is leaning against a wall. The bottom of the ladder is pulled along the ground, away from the wall, at the rate of 2 cm/s How fast is its height on the wall decreasing when the foot of the ladder is 4 cm away from the wall.
Answer:

Identify the given information: A ladder 5 m long is leaning against a wall. The bottom of the ladder is pulled along the ground, away from the wall, at the rate of 2 cm/s. The foot of the ladder is 4 cm away from the wall.

Calculate the rate at which the height of the ladder is decreasing: The rate at which the height of the ladder is decreasing can be calculated using the Pythagorean Theorem: h^2 = 4^2 + 5^2 h = √41
The rate at which the height is decreasing is equal to 2 cm/s divided by √41.
 Answer the question: The rate at which the height of the ladder is decreasing when the foot of the ladder is 4 cm away from the wall is 2 cm/s divided by √41.
Question:
The radius of a circle is increasing at the rate of 0.7cm/s. What is the rate of increase of its circumference?
Answer:
Step 1: Recall the formula for the circumference of a circle: C = 2πr
Step 2: Take the derivative of the formula with respect to time: dC/dt = 2π(dr/dt)
Step 3: Substitute the given rate of increase of the radius: dC/dt = 2π(0.7 cm/s)
Step 4: Simplify the result: dC/dt = 1.4π cm/s
Question:
A balloon, which always remains spherical on inflation, is being inflated by pumping in 900 cubic centimetres of gas per second. Find the rate at which the radius of the balloon increases when the radius is 15 cm.
Answer:

Convert the radius of the balloon from centimetres to metres (15 cm = 0.15 m).

Calculate the volume of the balloon using the formula for the volume of a sphere (V = 4/3πr3).

Calculate the rate of change of the volume by dividing the rate of inflation (900 cm3/s) by the volume of the balloon (V).

Calculate the rate of change of the radius by using the formula for the derivative of the volume of a sphere (dV/dr = 4πr2).

Substitute the radius (0.15 m) and the rate of change of the volume (900 cm3/s/V) into the formula for the derivative of the volume of a sphere and solve for the rate of change of the radius (dr/dt).
The rate at which the radius of the balloon increases when the radius is 15 cm is 0.0028 m/s.
Question:
The radius of a circle is increasing uniformly at the rate of 3cm/s. Find the rate at which the area of the circle is increasing when the radius is 10cm.
Answer:
Step 1: Calculate the area of the circle when the radius is 10cm.
Area of the circle = π × (10 cm)2 = 314 cm2
Step 2: Calculate the rate at which the area of the circle is increasing.
Rate of change of area = (Change in area) / (Change in time) = (π × (Change in radius)2) / (Change in time) = (π × 32) / (3 cm/s) = 9π cm2/s
Question:
The total cost C(x) in Rupees associated with the production of x units of an item is given by C x)=0.007x^3−0.003x^2+15x+4000. Find the marginal revenue when x=17.
Answer:
C(x) = 0.007x^3−0.003x^2+15x+4000
Marginal Revenue = dC/dx
dC/dx = 0.021x^2  0.006x + 15
dC/dx (17) = 0.021(17)^2  0.006(17) + 15
dC/dx (17) = 462.9  0.102 + 15
dC/dx (17) = 476.8
Question:
Find the rate of change of the area of a circle with respect to its radius r when (i) r=3 cm (ii) r=4 cm A 6π,8π B 5π,8π C 4π,10π D 2π,8π
Answer:
Answer: C 4π,10π
Explanation: The formula for the area of a circle is A = πr^2.
(i) When r = 3 cm, the area of the circle is A = π(3)^2 = 9π.
(ii) When r = 4 cm, the area of the circle is A = π(4)^2 = 16π.
The rate of change of the area of the circle with respect to its radius r is the difference between the areas when r = 4 cm and r = 3 cm, which is 16π  9π = 7π.
Therefore, the rate of change of the area of the circle with respect to its radius r is 4π when r = 3 cm and 10π when r = 4 cm. Therefore, the correct answer is C 4π,10π.
Question:
The rate of change of the area of a circle with respect to its radius r at r=6cm is. A 10π B 12π C 8π D 11π
Answer:
Answer: B 12π
Question:
A balloon, which always remains spherical, has a variable diameter 3/2(2x+1). Find the rate of change of its volume with respect to x.
Answer:

Volume (V) of a sphere = (4/3)πr³

Diameter (d) of a sphere = 2r

Substitute d = 3/2(2x+1) in the formula for V:
V = (4/3)π[(3/2(2x+1))/2]³
 Simplify:
V = (1/2)π(3x+1)³
 Take the derivative of V with respect to x:
dV/dx = (3π)(3x+1)²
 The rate of change of the volume of the balloon with respect to x is dV/dx = (3π)(3x+1)².