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Sample Calculus 1 Exam Questions

Updated on October 23, 2015
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TR Smith is a product designer and former teacher who uses math in her work every day.

Using the first and second derivatives of a function to find its minimum or maximum value is the technique you learn in Calculus I to solve optimization problems; it is the key concept tested on Calc I exams.

A simple calculus homework problem is finding the min and max of a polynomial. For example, the local max of the cubic function f(x) = x^3 - 3x^2 - 3x + 5 (graphed below) is at x = 1 - sqrt(2) and the local min is at x = 1 + sqrt(2). These values are found by solving the equation f'(x) = 3x^3 - 6x - 3 = 0 and testing if the critical points are minimum or maximum values. If you look at the function on a finite interval such as [-5, 5], the global min and global max are at x = -5 and x = 5 respectively.

Hard derivative problems like those found on calculus exams involve more complicated functions, or require you to deduce the function yourself, as in calculus word problems. Here are several examples of differential calculus test questions you may encounter.

Easy calculus problem: find the local min and max of f(x) = x^3 - 3x^2 - 3x + 5
Easy calculus problem: find the local min and max of f(x) = x^3 - 3x^2 - 3x + 5

Example 1: Calculus in Geometry

A right triangle has a perimeter of 8. What is the maximum area of the triangle?

Solution: This problem may seem at first like a three-variable problem since a triangle has three sides, but we can easily simplify it to a two-variable problem and even further to a single-variable calculus problem. If we call the leg lengths x and y, and the hypotenuse z, then the hypotenuse must be z = 8 - x - y. That eliminates z. And using the Pythagorean Theorem, we can also eliminate y:

x^2 + y^2 = (8 - x - y)^2
0 = 64 - 16x - 16y + 2xy
y = (64 - 16x)/(16 - 2x)

The two legs of the triangle are x and (64 - 16x)/(16 - 2x). The area of a right triangle is one half of the product of the two perpendicular legs, thus we have the area function

A(x) = (32x - 8x^2)/(16 - 2x)

The derivative of this function is

A'(x) = (4x^2 - 64x + 128)/(8 - x)^2

Setting the derivative equal to 0 and solving for x gives us

0 = (4x^2 - 64x + 128)/(8 - x)^2
0 = 4x^2 - 64x + 128
0 = x^2 - 16x + 32
x = 8 + 4*sqrt(2)
x = 8 - 4*sqrt(2)

The first value of x is greater than the perimeter of the triangle, which is physically impossible in the context of this problem. Therefore, the only critical value we need to check is x = 8 - 4*sqrt(2). The way we check if x = 8 - 4*sqrt(2) is a the max is by plugging it into the second derivative of A(x) and looking at the sign. If it is negative, it is a max.

A''(x) = 256/(x - 8)^3

A''(8 - 4*sqrt(2)) = 256/(-4*sqrt(2))^3 = a negative number

Therefore, the maximum area of the triangle is obtained when

x = 8 - 4*sqrt(2)
y = 8 - 4*sqrt(2)
z = 8*sqrt(2) - 8

This is an isosceles right triangle with an area of 48 - 32*sqrt(2).

Example 2: Which Path to Take

Martina wants to visit her friend who lives on the other side of a snow-covered field that is 920 feet wide, and across frozen river that is 770 feet long, as shown in the diagram below.

Martina can walk at a pace of 85 feet per minute through the field, but when she reaches the river she can ice skate at a speed of 450 feet per minute along the frozen surface. What path should she take to minimize the time it takes to get to her friend's house?

Solution: Let x be the length along the river that Martina ice skates. Following the diagram below, the length she walks across the field is found by applying the Pythagorean Theorem. That length L is a function of x,

L(x) = sqrt[ 920^2 + (770-x)^2].

The total time it takes to reach her friend's house is the time it takes to cross the field along some diagonal path plus the time it takes to skate along the frozen river vertically according to the diagram. Since time = distance ÷ rate, the time it takes to cross the field is L(x)/85, and the time it takes to skate along the river is x/450. Adding these two expressions gives the total time as a function of x,

T(x) = L(x)/85 + x/450
= sqrt[ 920^2 + (770-x)^2]/85 + x/450

The derivative of T(x) is

T'(x) = (1/85)(x-770)(920^2 + (770-x)^2)^(1/2) + 1/450

Setting the derivative equal to zero and solving for x gives you

0 = (1/85)(x-770)(920^2 + (770-x)^2)^(-1/2) + 1/450

(-1/450)sqrt[920^2 + (770-x)^2] = (1/85)(x-770)

(1/450^2)(920^2 + (770-x)^2) = (1/85^2)(x-770)^2

(920/450)^2 = [1/85^2 - 1/450^2](x-770)^2

244609600/7811 = (x-770)^2

x = 770 ± sqrt(244609600/7811)

x ≈ 593 and 947

Since x cannot be greater than 770, the length of the river between Martina and her friend's house, the only solution is x = 593 feet. Plugging this into the equation for T(x) gives a total time of about 12.3 minutes.

Example 3: Optimizing a Class of Functions

Consider the class of functions fn(x) = (x^n)*2^(-x), where n is positive. Question 1: For an arbitrary value of n, what is the maximum value of the function fn(x)? Question 2: For which value of n does fn(x) have the smallest maximum?

Solution: This is a strange and tricky problem about minimizing a maximum. The first step in answering the first question is to take the derivative of fn(x), which is

fn'(x) = [nx^(n-1) - log(2)x^n]*2^(-x)

where log(2) is the natural logarithm of 2. Remember that when taking the derivative of this function, x is the variable and n is a parameter to be treated as a constant. Setting fn'(x) equal to 0 and solving for x in terms of n gives us

0 = [nx^(n-1) - log(2)x^n]*2^(-x)
0 = nx^(n-1) - log(2)x^n
0 = n - log(2)x
x = n/log(2)

Plugging this value into the function fn(x) and applying rules of logarithms gives us

fn(n/log(2)) = [n/log(2)]^n * 2^(-n/log(2))
= {n / [log(2)*2^(1/log(2))] }^n
= [n / (e*log(2))]^n

This is the global maximum of fn(x) for a given value of n, which can be verified by taking the second derivative, plugging in x = n/log(2), and checking that the result is negative. This answers Question 1. For the second question, we have to find the value of n that minimizes the function

k(n) = [n / (e*log(2))]^n

To find the derivative of k(n), we follow the method of finding derivatives of functions of the form f(x)^g(x). Taking the natural log of both sides gives us

log(k(n)) = n * log[n / (e*log(2))]
= n*log(n) - n*log(e) - n*log(log(2))
= n*log(n) - n*[1 + log(log(2))]

Taking the derivative of both sides gives us

k'(n)/k(n) = log(n) - log(log(2))

k'(n) = [log(n) - log(log(2))] * [n / (e*log(2))]^n

Setting the derivative of k(n) equal to 0 and solving for n gives us

0 = [log(n) - log(log(2))] * [n / (e*log(2))]^n

0 = log(n) - log(log(2))]

n = log(2)

Is this value a max or a min of the function k(n)? Looking at the graph of k(n) = [n / (e*log(2))]^n, it has a global min at n = log(2). You can also apply the second derivative test. So this answers Question 2.

In summary, among the class of functions fn(x), the function with n = log(2) has the smallest global maximum, and that maximum is

k(log(2)) = (1/e)^log(2) = 1/2

Example 4: A Piecewise Function

Let the piecewise function h(x) be defined as

h(x) = {
2x + 5, for x < 6
ax^3 + bx^2 + cx + d, for 6 ≤ x ≤ 20
x + 50, for 20 < x

What must be the values of a, b, c, and d so that h(x) is continuous and has a continuous first derivative?

Solution: For h(x) to be continuous, we have to satisfy the conditions

2x + 5 = ax^3 + bx^2 + cx + d, evaluated at x = 6


x + 50 = ax^3 + bx^2 + cx + d, evaluated at x = 20

For h'(x) to be continuous, we have to satisfy the conditions

d/dx [2x + 5] = d/dx [ax^3 + bx^2 + cx + d], evaluated at x = 6


d/dx [x + 50] = d/dx [ax^3 + bx^2 + cx + d], evaluated at x = 20

Taking derivatives and making the required substitutions gives us a system of four linear equations in the four unknowns a, b, c, and d.

17 = 216a + 36b + 6c + d
2 = 108a + 12b +c
70 = 8000a + 400b + 20c + d
1 = 1200a + 40b + c

This can be solved with elimination, substitution, or matrices. To save space, we just present the final answer of the system:

a = -8/343
b = 1199/1372
c = -2047/343
d = 9050/343

Below is a graph of the solution. The two linear functions are shown in red and green with the cubic connector function in between shown in blue. Notice that the slopes match at the connection points x = 6 and x = 20.

To connect two curves (including lines) with a gap between them, the minimum degree of the connector polynomial must be 3 in order to ensure that the resulting piecewise function is continuous with a continuous derivative.


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