# Application Of Derivatives: Math Trivia Quiz

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| By Tanmay Shankar
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Do you know the applications of derivatives? Many people have a negative feeling about math and think they won’t understand how to solve fundamental problems. Problems involving derivatives are easy to solve, as all one has to follow specific procedures. The quiz below is a perfect way to test out your solving skills. Do give it a shot as you refer to your notes and keep a lookout for more quizzes like it!

• 1.

### Find the interval in which the function f given by f(x) = log sin x is strictly increasing on and strictly decreasing on.

• A.
• B.
• C.

Both A & B

• D.

None of these

C. Both A & B
Explanation
The function f(x) = log sin x is strictly increasing on interval A and strictly decreasing on interval B. This means that as x increases within interval A, the value of f(x) also increases. Similarly, as x increases within interval B, the value of f(x) decreases. Therefore, the correct answer is Both A & B.

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• 2.

### Find the rate of change of the area of a circle with respect to its radius r when r = 4 cm

• A.

• B.

• C.

• D.

C. 8π
Explanation
The rate of change of the area of a circle with respect to its radius can be found using the formula for the area of a circle, A = πr^2. To find the derivative of A with respect to r, we can use the power rule of differentiation. Taking the derivative of A = πr^2, we get dA/dr = 2πr. When r = 4 cm, the rate of change of the area is 2π(4) = 8π.

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• 3.

### Find the points on the curve x2 + y2- 2x-3=0 at which the tangents are parallel to the x-axis.

• A.

(1, 2) and (1, −2)

• B.

(1, -2) and (1, 2)

• C.

(-1, 2) and (1, −2)

• D.

(-1, 2) and (1, 2)

A. (1, 2) and (1, −2)
Explanation
The equation of the given curve is x^2 + y^2 - 2x - 3 = 0. To find the points on the curve at which the tangents are parallel to the x-axis, we need to find the points where the derivative dy/dx is equal to 0. Taking the derivative of the equation with respect to x, we get 2x + 2y(dy/dx) - 2 = 0. Since we want the tangents to be parallel to the x-axis, the slope dy/dx should be 0. Substituting dy/dx = 0 into the derivative equation, we get 2x - 2 = 0, which simplifies to x = 1. Plugging x = 1 into the original equation, we get y^2 - 2y - 2 = 0. Solving this quadratic equation, we find y = 2 and y = -2. Therefore, the points on the curve where the tangents are parallel to the x-axis are (1, 2) and (1, -2).

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• 4.

### It is given that at x = 1, the function x4− 62x2 + ax + 9 attains its maximum value, on the interval [0, 2]. Find the value of a.

• A.

120

• B.

110

• C.

-110

• D.

-120

A. 120
Explanation
At x = 1, the function x^4 - 62x^2 + ax + 9 attains its maximum value on the interval [0, 2]. This means that the derivative of the function at x = 1 is equal to 0. Taking the derivative of the function, we get 4x^3 - 124x + a. Setting this equal to 0 and solving for x, we find that x = ±√(31/a). Since x = 1 is in the interval [0, 2], we can conclude that √(31/a) = 1. Squaring both sides, we get 31/a = 1, which implies a = 31. Therefore, the value of a is 31.

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• 5.

### For the curve y = 4x3 − 2x5, find all the points at which the tangents passes through the origin.

• A.

(0, 0), (1, 2), and (1, −2)

• B.

(0, 0), (1, 2), and (−1, −2)

• C.

(0, 0), (1, 2), and (−1, 2)

• D.

(0, 0), (1, -2), and (−1, −2)

B. (0, 0), (1, 2), and (−1, −2)
Explanation
The given curve is y = 4x^3 - 2x^5. To find the points at which the tangent passes through the origin, we need to find the points where the slope of the tangent is equal to zero. The slope of the tangent can be found by taking the derivative of the curve with respect to x. Differentiating y = 4x^3 - 2x^5, we get dy/dx = 12x^2 - 10x^4. Setting this derivative equal to zero and solving for x, we find x = 0, x = 1, and x = -1. Plugging these values back into the original equation, we get the points (0, 0), (1, 2), and (-1, -2), which are the points at which the tangents pass through the origin.

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• 6.

### Sand is pouring from a pipe at the rate of 12 cm3 /s. The falling sand forms a cone on the ground in such a way that the height of the cone is always one-sixth of the radius of the base. How fast is the height of the sand cone increasing when the height is 4cm?

A.
Explanation
To find how fast the height of the sand cone is increasing, we can use the related rates formula. Let's denote the radius of the base of the cone as r and the height of the cone as h. We are given that h = (1/6)r and we want to find dh/dt when h = 4 cm.

First, we need to find an equation that relates r and h. Since the cone is formed by the falling sand, we know that the volume of the cone is changing with time. The volume of a cone is given by V = (1/3)πr^2h.

We can differentiate this equation with respect to time (t) to find how the volume is changing with time: dV/dt = (1/3)π(2rh(dr/dt) + r^2(dh/dt)).

Since we are given that sand is pouring from the pipe at a constant rate of 12 cm^3/s, we know that dV/dt = 12 cm^3/s.

Plugging in the known values, we have 12 = (1/3)π(2r(12) + r^2(dh/dt)).

We also know that h = (1/6)r, so we can substitute this into the equation: 12 = (1/3)π(2r(12) + r^2(dh/dt)).

Now we can solve for dh/dt when h = 4 cm. We can substitute r = 6h into the equation and solve for dh/dt.

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• 7.

### Find the points on the curve y = x3 at which the slope of the tangent is equal to the y-coordinate of the point.

• A.

(0, 0) and (2, 27)

• B.

(0, 0) and (3, 27)

• C.

(0, 0) and (3, 25)

• D.

(0, 0) and (3, 25)

B. (0, 0) and (3, 27)
Explanation
The slope of the tangent to the curve y = x^3 is equal to the derivative of the function y = x^3. Taking the derivative, we get dy/dx = 3x^2. To find the points on the curve where the slope of the tangent is equal to the y-coordinate of the point, we need to find the values of x for which 3x^2 = y. Substituting y = x^3, we get 3x^2 = x^3. Solving this equation, we find x = 0 and x = 3. Evaluating y for these values of x, we get y = 0 and y = 27. Therefore, the points on the curve where the slope of the tangent is equal to the y-coordinate of the point are (0, 0) and (3, 27).

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• 8.

### 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, which of the following is true?

• A.

Perimeter is decreasing at the rate of 2 cm/min

• B.

Area of the rectangle is increasing at the rate of 2 cm2/min

• C.

Both A & B

• D.

None of these

C. Both A & B
Explanation
As the length of the rectangle is decreasing and the width is increasing, the perimeter of the rectangle is affected by both changes. The decrease in length contributes to a decrease in the perimeter, while the increase in width contributes to an increase in the perimeter. Therefore, the perimeter is decreasing at a rate of 5 cm/minute - 4 cm/minute = 1 cm/minute.

Similarly, the area of the rectangle is affected by both changes. The decrease in length results in a decrease in the area, while the increase in width results in an increase in the area. Therefore, the area is increasing at a rate of 5 cm/minute * 6 cm - 4 cm/minute * 8 cm = 2 cm^2/minute.

Hence, both statement A and statement B are true.

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• 9.

### The volume of a cube is increasing at the rate of 8 cm3 /s. How fast is the surface area increasing when the length of an edge is 12 cm?

D.
Explanation
The volume of a cube is given by V = s^3, where s is the length of one edge. To find how fast the surface area is increasing, we need to differentiate the surface area formula with respect to time. The surface area of a cube is given by A = 6s^2. Differentiating both sides with respect to time, we get dA/dt = 12s * ds/dt. Plugging in the given values, we have dA/dt = 12(12)(8) = 1152 cm^2/s. Therefore, the surface area is increasing at a rate of 1152 cm^2/s.

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• 10.

### Find both the maximum value and the minimum value of 3x4 − 8x3 + 12x2 − 48x + 25 on the interval [0, 3]

• A.

The absolute maximum value is 25

• B.

The absolute minimum value is − 39

• C.

Both A & B

• D.

None of these

C. Both A & B
Explanation
The given polynomial is a fourth-degree polynomial, which means it is a continuous function. To find the maximum and minimum values of the polynomial on the interval [0, 3], we can take the derivative of the polynomial and set it equal to zero to find critical points. By finding the critical points and evaluating the polynomial at the endpoints of the interval, we can determine the maximum and minimum values. In this case, the maximum value of 25 occurs at x = 0, and the minimum value of -39 occurs at x = 3. Therefore, the correct answer is both A & B.

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• 11.

A.
• 12.

### Find the equation of the tangent to the curve  which is parallel to the line 4x − 2y + 5 = 0.

• A.

48x + 24y = 23

• B.

48x – 24y = 25

• C.

48x – 24y = 23

• D.

48x – 24y = 28

C. 48x – 24y = 23
Explanation
The equation of a line parallel to 4x - 2y + 5 = 0 will have the same slope as the given line. The given line can be rearranged to the form 2y = 4x + 5, which gives a slope of 2. Therefore, any line parallel to this line will also have a slope of 2. The equation 48x - 24y = 23 has a slope of 2, making it parallel to the given line.

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• 13.

### A wire of length 28 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a circle. What should be the length of the two pieces so that the combined area of the square and the circle is minimum?

B.
Explanation
To find the length of the two pieces that will result in the minimum combined area of the square and the circle, we can use the concept of calculus optimization. Let's assume that the length of the wire used for the square is x, then the length of the wire used for the circle will be 28 - x. The area of the square is x^2, and the area of the circle is π((28 - x)/(2π))^2 = (28 - x)^2/(4π). To minimize the combined area, we need to find the value of x that minimizes the sum of the areas. By taking the derivative of the sum of the areas with respect to x and setting it equal to zero, we can find the critical point. Solving for x will give us the length of the wire that should be used for the square, and 28 - x will give us the length for the circle.

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• 14.

### Find two positive numbers x and y such that their sum is 35 and the product x2y5 is a maximum

• A.

10 & 25

• B.

15 & 20

• C.

5 & 30

• D.

None of these

A. 10 & 25
Explanation
To find two positive numbers x and y such that their sum is 35 and the product x^2y^5 is a maximum, we can use the concept of maximizing a function subject to a constraint. In this case, we want to maximize the function f(x, y) = x^2y^5, subject to the constraint x + y = 35. We can rewrite the constraint as y = 35 - x and substitute it into the function. This gives us f(x) = x^2(35 - x)^5. To find the maximum, we can take the derivative of f(x) with respect to x, set it equal to zero, and solve for x. By solving this equation, we find that x = 10. Substituting this value back into the constraint equation, we find that y = 25. Therefore, the two positive numbers that satisfy the given conditions are 10 and 25.

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• 15.

### Find the semi-vertical angle of the cone of the maximum volume and is of given slant height.

D.
Explanation
The semi-vertical angle of a cone affects its volume. To find the maximum volume, we need to determine the optimal semi-vertical angle. By using calculus, we can differentiate the volume formula with respect to the semi-vertical angle and set it equal to zero to find the critical point. Solving this equation will give us the value of the semi-vertical angle that maximizes the volume. However, without the specific formula for the volume of the cone or the given slant height, it is not possible to provide a precise explanation.

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• Current Version
• Mar 22, 2023
Quiz Edited by
ProProfs Editorial Team
• Dec 04, 2013
Quiz Created by
Tanmay Shankar

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