Calculus Practice

Applications of Derivatives Practice

Practice critical numbers, increasing and decreasing intervals, extrema, concavity, optimization, motion, related rates, and linear approximation.

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Try each problem first, then click Show solution to check your work and see the complete step-by-step solution.

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Applications Essentials

First Derivative

\[ f'(x)>0 \Rightarrow \text{increasing} \] \[ f'(x)<0 \Rightarrow \text{decreasing} \]

Use \(f'\) to study direction and local extrema.

Second Derivative

\[ f''(x)>0 \Rightarrow \text{concave up} \] \[ f''(x)<0 \Rightarrow \text{concave down} \]

Use \(f''\) to study concavity and classify some critical points.

Motion

\[ v(t)=s'(t) \] \[ a(t)=s''(t) \]

Velocity is the derivative of position; acceleration is the derivative of velocity.

Download Practice PDF Review Applications

Level 1

Critical Numbers and Function Behavior

Find critical numbers and use the sign of the derivative to understand how a function behaves.

Problem 1

Critical Numbers

Find the critical numbers of \[ f(x)=x^3-3x^2. \]

Step-by-Step Solution

Differentiate:

\[ f'(x) = 3x^2-6x \] \[ = 3x(x-2). \]

Set the derivative equal to zero.

\[ 3x(x-2)=0. \] \[ x=0 \qquad \text{or} \qquad x=2. \]
Answer: \[ \boxed{x=0,\;2} \]

Problem 2

Increasing / Decreasing

Suppose \[ f'(x) = 3x(x-2). \] On which intervals is \(f\) increasing?

\((-\infty,0)\) and \((2,\infty)\)
\((0,2)\)
\((-\infty,2)\)

Step-by-Step Solution

The critical numbers are \(0\) and \(2\), so test the intervals

\[ (-\infty,0), \qquad (0,2), \qquad (2,\infty). \]

The sign pattern of \(f'(x)\) is

\[ + \qquad - \qquad + \]

A function is increasing where \(f'(x)>0\).

Answer: \[ \boxed{ (-\infty,0) \cup (2,\infty) } \]

Problem 3

First Derivative Test

At a critical number \(x=c\), suppose \(f'(x)\) changes from positive to negative. What occurs at \(x=c\)?

Local minimum
Local maximum
Inflection point

Step-by-Step Solution

Before \(c\), \(f'(x)>0\), so the function is increasing.

After \(c\), \(f'(x)<0\), so the function is decreasing.

\[ + \longrightarrow - \]
Answer: \(x=c\) is a local maximum.

Level 2

Extrema, Concavity, and Curve Analysis

Use first and second derivatives to classify extrema and analyze the shape of a graph.

Problem 4

Absolute Extrema

Find the absolute maximum and absolute minimum of \[ f(x)=x^3-3x^2 \] on \[ [-1,3]. \]

Step-by-Step Solution

The critical numbers are

\[ x=0,\;2. \]

Evaluate the function at the critical numbers and both endpoints.

\[ f(-1)=-4 \] \[ f(0)=0 \] \[ f(2)=-4 \] \[ f(3)=0. \]

The largest value is \(0\) and the smallest value is \(-4\).

Answer: Absolute maximum: \(0\) at \(x=0,3\). Absolute minimum: \(-4\) at \(x=-1,2\).

Problem 5

Concavity

Determine where \[ f(x)=x^3-3x^2 \] is concave up and concave down.

Step-by-Step Solution

Differentiate twice.

\[ f'(x) = 3x^2-6x \] \[ f''(x) = 6x-6. \]

Set the second derivative equal to zero.

\[ 6x-6=0 \] \[ x=1. \]

For \(x<1\), \(f''(x)<0\).

For \(x>1\), \(f''(x)>0\).

Answer: Concave down on \((-\infty,1)\); concave up on \((1,\infty)\).

Problem 6

Inflection Point

Find the inflection point of \[ f(x)=x^3-3x^2. \]

Step-by-Step Solution

From the previous problem,

\[ f''(x)=6x-6. \]

The possible inflection point occurs at

\[ x=1. \]

The second derivative changes from negative to positive there, so the concavity changes.

Find the corresponding \(y\)-value:

\[ f(1) = 1-3 = -2. \]
Answer: \[ \boxed{(1,-2)} \]

Problem 7

Second Derivative Test

Suppose \[ f'(4)=0 \] and \[ f''(4)=6. \] What can you conclude?

Local maximum at \(x=4\)
Local minimum at \(x=4\)
No conclusion is possible

Step-by-Step Solution

Since

\[ f'(4)=0, \]

\(x=4\) is a critical number.

Also,

\[ f''(4)=6>0. \]

A positive second derivative means the graph is concave up.

Answer: \(f\) has a local minimum at \(x=4\).

Level 3

Optimization

Build an objective function, reduce it to one variable, and use derivatives to find an optimal value.

Problem 8

Optimization Setup

A rectangle has perimeter \(40\) units. If one side is \(x\), write the area \(A\) as a function of \(x\).

Step-by-Step Solution

Let the other side be \(y\).

\[ 2x+2y=40. \]

Solve for \(y\):

\[ y=20-x. \]

Area is

\[ A=xy. \]

Substitute the expression for \(y\):

\[ A(x) = x(20-x) \] \[ = 20x-x^2. \]
Answer: \[ \boxed{ A(x)=20x-x^2 } \]

Problem 9

Maximum Area

Using \[ A(x)=20x-x^2, \] find the dimensions of the rectangle with maximum area.

Step-by-Step Solution

Differentiate:

\[ A'(x) = 20-2x. \]

Set the derivative equal to zero:

\[ 20-2x=0. \] \[ x=10. \]

The other dimension is

\[ y = 20-x = 10. \]
Answer: \[ \boxed{ 10 \text{ units} \times 10 \text{ units} } \]

Problem 10

Optimization

A farmer has \(200\) meters of fencing to enclose a rectangular field against a straight river. No fence is needed along the river. Find the dimensions that maximize the area.

Step-by-Step Solution

Let \(x\) be each side perpendicular to the river and \(y\) the side parallel to the river.

\[ 2x+y=200. \]

So

\[ y=200-2x. \]

The area is

\[ A=xy. \] \[ A(x) = x(200-2x). \] \[ = 200x-2x^2. \]

Differentiate:

\[ A'(x) = 200-4x. \]

Set \(A'(x)=0\):

\[ 200-4x=0 \] \[ x=50. \]

Then

\[ y = 200-2(50) = 100. \]
Answer: \(50\) m by \(100\) m.

Level 4

Motion and Related Rates

Interpret derivatives as rates of change in physical situations.

Problem 11

Motion

An object's position is \[ s(t) = t^3-6t^2+9t. \] Find its velocity and acceleration.

Step-by-Step Solution

Velocity is the derivative of position.

\[ v(t) = s'(t) = 3t^2-12t+9. \]

Acceleration is the derivative of velocity.

\[ a(t) = v'(t) = 6t-12. \]
Answer: \(v(t)=3t^2-12t+9\), \(a(t)=6t-12\).

Problem 12

At Rest

For \[ s(t) = t^3-6t^2+9t, \] find the times when the object is at rest.

Step-by-Step Solution

The object is at rest when velocity equals zero.

\[ v(t) = 3t^2-12t+9. \]
\[ 3t^2-12t+9=0 \] \[ 3(t^2-4t+3)=0 \] \[ 3(t-1)(t-3)=0. \]
Answer: \[ \boxed{ t=1,\;3 } \]

Problem 13

Related Rates

The radius of a circle is increasing at \[ \frac{dr}{dt} = 3 \text{ cm/s}. \] How quickly is the area increasing when \(r=5\) cm?

Step-by-Step Solution

Start with the area formula.

\[ A=\pi r^2. \]

Differentiate with respect to time.

\[ \frac{dA}{dt} = 2\pi r \frac{dr}{dt}. \]

Substitute \(r=5\) and \(\frac{dr}{dt}=3\).

\[ \frac{dA}{dt} = 2\pi(5)(3). \] \[ = 30\pi. \]
Answer: \[ \boxed{ 30\pi \text{ cm}^2/\text{s} } \]

Problem 14

Related Rates

The side length \(s\) of a square is increasing at \[ \frac{ds}{dt} = 2 \text{ cm/s}. \] How quickly is the area increasing when \(s=6\) cm?

Step-by-Step Solution

Area of a square is

\[ A=s^2. \]

Differentiate with respect to time.

\[ \frac{dA}{dt} = 2s \frac{ds}{dt}. \]

Substitute \(s=6\) and \(\frac{ds}{dt}=2\).

\[ \frac{dA}{dt} = 2(6)(2) = 24. \]
Answer: \[ \boxed{ 24 \text{ cm}^2/\text{s} } \]

Level 5

Linear Approximation and Mixed Review

Finish by using tangent lines for approximation and combining several applications concepts.

Problem 15

Linear Approximation

Use linear approximation at \(x=4\) to estimate \[ \sqrt{4.1}. \]

Step-by-Step Solution

Let

\[ f(x)=\sqrt{x}. \]

Use \(a=4\):

\[ f(4)=2. \]

Differentiate:

\[ f'(x) = \frac{1}{2\sqrt{x}}. \] \[ f'(4) = \frac14. \]

Build the linearization:

\[ L(x) = f(4) + f'(4)(x-4) \] \[ L(x) = 2 + \frac14(x-4). \]

Evaluate at \(x=4.1\):

\[ L(4.1) = 2 + \frac14(0.1). \] \[ = 2.025. \]
Answer: \[ \boxed{ \sqrt{4.1} \approx 2.025 } \]

Problem 16

Mixed Analysis

Suppose a function satisfies \[ f'(x)<0 \] and \[ f''(x)>0 \] on an interval. Describe the behavior of \(f\).

Increasing and concave up
Decreasing and concave up
Decreasing and concave down

Step-by-Step Solution

Since

\[ f'(x)<0, \]

the function is decreasing.

Since

\[ f''(x)>0, \]

the graph is concave up.

Answer: The function is decreasing and concave up.

Before You Finish

Applications of Derivatives Checklist

  1. Find critical numbers from \(f'(x)=0\) and points where \(f'\) is undefined.
  2. Use the sign of \(f'\) to identify increasing and decreasing intervals.
  3. Use the First Derivative Test to classify local maxima and minima.
  4. Check critical numbers and endpoints when finding absolute extrema on a closed interval.
  5. Use \(f''\) to determine concavity.
  6. Verify a change in concavity before calling a point an inflection point.
  7. Use the Second Derivative Test when \(f'(c)=0\).
  8. Build optimization problems with one objective function in one variable.
  9. Remember \(v(t)=s'(t)\) and \(a(t)=s''(t)\).
  10. Set \(v(t)=0\) to find when an object is at rest.
  11. In related-rates problems, differentiate with respect to time before substituting the values for the instant in question.
  12. Use \(L(x)=f(a)+f'(a)(x-a)\) for linear approximation.
  13. Include appropriate units in applied problems.

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