Chapter 4: Applications of Derivatives

AP Calculus AB/BC · Updated February 2026 · 25 min read · AP Calculus Papers

Derivatives are not merely an abstract operation on functions. They are tools that let us solve concrete problems about rates of change, shape of curves, and optimal design. This chapter develops the major applications of derivatives that form the backbone of the AP Calculus AB and BC exams: related rates, linear approximation, curve analysis through the first and second derivative tests, optimization, and the Mean Value Theorem.

Table of Contents

  1. Related Rates
  2. Linear Approximation and Differentials
  3. Extreme Values and the First Derivative Test
  4. Concavity and the Second Derivative Test
  5. Optimization Problems
  6. Mean Value Theorem
  7. Practice Problems

Many real-world situations involve two or more quantities that change simultaneously. When these quantities are linked by an equation, we can use implicit differentiation with respect to time $t$ to find how fast one quantity changes given information about the other. These are called related rates problems.

Strategy for Related Rates Problems

  1. Draw a diagram and label all quantities that vary with time. Assign variables to each.
  2. Write an equation that relates the variables. Use geometry, trigonometry, or other known relationships.
  3. Differentiate both sides with respect to $t$ using implicit differentiation and the chain rule.
  4. Substitute all known values (including rates) at the instant in question.
  5. Solve for the unknown rate.

A common mistake is substituting specific numerical values before differentiating. Always differentiate first, then substitute.

Example 1: Expanding Circle

A stone is dropped into a still pond, producing a circular ripple whose radius increases at a constant rate of $3$ ft/s. How fast is the area of the circle increasing when the radius is $10$ ft?

Solution. Let $r$ be the radius and $A$ the area of the circle at time $t$. We are given $\frac{dr}{dt} = 3$ ft/s and want $\frac{dA}{dt}$ when $r = 10$ ft.

The relationship between area and radius is:

$$A = \pi r^2$$

Differentiate both sides with respect to $t$:

$$\frac{dA}{dt} = 2\pi r \, \frac{dr}{dt}$$

Substitute $r = 10$ and $\frac{dr}{dt} = 3$:

$$\frac{dA}{dt} = 2\pi(10)(3) = 60\pi \approx 188.5 \text{ ft}^2\text{/s}$$

Example 2: Sliding Ladder

A 13-foot ladder leans against a vertical wall. The base of the ladder slides away from the wall at $2$ ft/s. How fast is the top of the ladder sliding down the wall when the base is $5$ ft from the wall?

Solution. Let $x$ be the distance from the base of the ladder to the wall, and let $y$ be the height of the top of the ladder on the wall. The ladder length is constant at 13 ft, so by the Pythagorean theorem:

$$x^2 + y^2 = 169$$

Differentiate with respect to $t$:

$$2x\,\frac{dx}{dt} + 2y\,\frac{dy}{dt} = 0$$

When $x = 5$, we find $y = \sqrt{169 - 25} = \sqrt{144} = 12$. We are given $\frac{dx}{dt} = 2$ ft/s. Substitute:

$$2(5)(2) + 2(12)\frac{dy}{dt} = 0$$ $$20 + 24\frac{dy}{dt} = 0$$ $$\frac{dy}{dt} = -\frac{20}{24} = -\frac{5}{6} \text{ ft/s}$$

The negative sign confirms the top is sliding down at $\frac{5}{6}$ ft/s.

Example 3: Filling a Conical Tank

Water flows into an inverted conical tank at a rate of $8$ ft$^3$/min. The tank has a radius of $4$ ft at the top and a height of $10$ ft. How fast is the water level rising when the water is $5$ ft deep?

Solution. Let $r$ be the radius of the water surface and $h$ the depth of water at time $t$. We know $\frac{dV}{dt} = 8$ ft$^3$/min and want $\frac{dh}{dt}$ when $h = 5$ ft.

The volume of a cone is $V = \frac{1}{3}\pi r^2 h$. By similar triangles with the tank dimensions, $\frac{r}{h} = \frac{4}{10} = \frac{2}{5}$, so $r = \frac{2h}{5}$. Substituting to eliminate $r$:

$$V = \frac{1}{3}\pi \left(\frac{2h}{5}\right)^2 h = \frac{1}{3}\pi \cdot \frac{4h^2}{25} \cdot h = \frac{4\pi h^3}{75}$$

Differentiate with respect to $t$:

$$\frac{dV}{dt} = \frac{4\pi}{75} \cdot 3h^2 \, \frac{dh}{dt} = \frac{4\pi h^2}{25}\,\frac{dh}{dt}$$

Substitute $\frac{dV}{dt} = 8$ and $h = 5$:

$$8 = \frac{4\pi(25)}{25}\,\frac{dh}{dt} = 4\pi\,\frac{dh}{dt}$$ $$\frac{dh}{dt} = \frac{8}{4\pi} = \frac{2}{\pi} \approx 0.637 \text{ ft/min}$$

4.2 Linear Approximation and Differentials

The tangent line to a curve at a point is the best linear approximation to the function near that point. This idea underpins one of the most practical uses of derivatives: estimating function values without a calculator.

Linear Approximation (Tangent Line Approximation)

If $f$ is differentiable at $x = a$, then for values of $x$ near $a$:

$$f(a + \Delta x) \approx f(a) + f'(a)\,\Delta x$$

Equivalently, the linearization of $f$ at $a$ is $L(x) = f(a) + f'(a)(x - a)$.

The quality of this approximation depends on how close $x$ is to $a$ and on the curvature of $f$. Functions with large second derivatives (high curvature) produce less accurate linear approximations for the same $\Delta x$.

Differentials

If $y = f(x)$ is differentiable, we define the differential $dy$ as:

$$dy = f'(x)\,dx$$

Here $dx$ is an independent variable (representing a small change in $x$), and $dy$ approximates the actual change $\Delta y = f(x + dx) - f(x)$. The differential $dy$ is the change along the tangent line, while $\Delta y$ is the change along the curve itself.

Example 4: Approximating a Square Root

Use linear approximation to estimate $\sqrt{50}$.

Solution. Let $f(x) = \sqrt{x}$, with $a = 49$ (the nearest perfect square). Then $f(a) = 7$ and $f'(x) = \frac{1}{2\sqrt{x}}$, so $f'(49) = \frac{1}{14}$.

With $\Delta x = 50 - 49 = 1$:

$$\sqrt{50} \approx f(49) + f'(49)(1) = 7 + \frac{1}{14} = 7.0714\ldots$$

The actual value is $\sqrt{50} = 7.0711\ldots$, so the approximation is accurate to three decimal places.

Example 5: Using Differentials to Estimate Error

The radius of a sphere is measured as $6$ cm with a possible error of $\pm 0.02$ cm. Estimate the maximum error in the calculated volume.

Solution. The volume is $V = \frac{4}{3}\pi r^3$. Differentiating:

$$dV = 4\pi r^2 \, dr$$

With $r = 6$ and $dr = \pm 0.02$:

$$dV = 4\pi(36)(0.02) = 2.88\pi \approx 9.05 \text{ cm}^3$$

The actual volume is $V = \frac{4}{3}\pi(216) = 288\pi \approx 904.8$ cm$^3$. The relative error is:

$$\frac{dV}{V} = \frac{2.88\pi}{288\pi} = 0.01 = 1\%$$

Notice this is exactly $3 \cdot \frac{dr}{r} = 3 \cdot \frac{0.02}{6} = 1\%$. In general, for $V = \frac{4}{3}\pi r^3$, the relative error in volume is three times the relative error in radius.

4.3 Extreme Values and the First Derivative Test

Absolute vs. Local Extrema

A function $f$ has an absolute maximum at $c$ if $f(c) \geq f(x)$ for all $x$ in the domain. It has a local maximum at $c$ if $f(c) \geq f(x)$ for all $x$ near $c$. The same definitions apply to minima with the inequality reversed. Absolute extrema are sometimes called global extrema.

Critical Points

A number $c$ in the domain of $f$ is a critical point if either $f'(c) = 0$ or $f'(c)$ does not exist. Critical points are the only candidates for local extrema (by Fermat's Theorem).

Extreme Value Theorem

If $f$ is continuous on a closed interval $[a, b]$, then $f$ attains both an absolute maximum and an absolute minimum on $[a, b]$.

Closed Interval Method: To find the absolute extrema of a continuous function on $[a, b]$:

  1. Find all critical points of $f$ in $(a, b)$.
  2. Evaluate $f$ at each critical point and at the endpoints $a$ and $b$.
  3. The largest value is the absolute maximum; the smallest is the absolute minimum.

First Derivative Test

Suppose $c$ is a critical point of a continuous function $f$.

Example 6: Closed Interval Method

Find the absolute maximum and minimum of $f(x) = x^3 - 3x^2 + 1$ on $[-1, 4]$.

Solution. First, find critical points: $f'(x) = 3x^2 - 6x = 3x(x - 2)$. Setting $f'(x) = 0$ gives $x = 0$ and $x = 2$, both in $(-1, 4)$.

Evaluate $f$ at critical points and endpoints:

The absolute maximum is $17$ at $x = 4$. The absolute minimum is $-3$, attained at both $x = -1$ and $x = 2$.

Example 7: First Derivative Test

Find and classify all local extrema of $g(x) = x^4 - 4x^3$.

Solution. Compute $g'(x) = 4x^3 - 12x^2 = 4x^2(x - 3)$. The critical points are $x = 0$ and $x = 3$.

Sign analysis of $g'$:

At $x = 0$, $g'$ does not change sign (negative on both sides), so there is no local extremum. At $x = 3$, $g'$ changes from negative to positive, so $g$ has a local minimum of $g(3) = 81 - 108 = -27$.

4.4 Concavity and the Second Derivative Test

While the first derivative tells us where a function is increasing or decreasing, the second derivative reveals how it increases or decreases -- that is, the shape of the curve.

Concavity

A function $f$ is concave up on an interval if $f''(x) > 0$ on that interval. Geometrically, the curve lies above its tangent lines, and the slope is increasing. A function is concave down on an interval if $f''(x) < 0$, meaning the curve lies below its tangent lines.

Inflection Points

A point where the concavity changes is called an inflection point. At an inflection point, $f''(x) = 0$ or $f''(x)$ is undefined. However, $f''(c) = 0$ alone does not guarantee an inflection point -- the concavity must actually change. For instance, $f(x) = x^4$ has $f''(0) = 0$ but no inflection point at $x = 0$.

Second Derivative Test

Suppose $f'(c) = 0$ and $f''$ is continuous near $c$.

Example 8: Concavity and Inflection Points

Find the intervals of concavity and all inflection points of $f(x) = x^3 - 6x^2 + 9x + 2$.

Solution. We need the second derivative:

$$f'(x) = 3x^2 - 12x + 9, \qquad f''(x) = 6x - 12 = 6(x - 2)$$

Setting $f''(x) = 0$ gives $x = 2$. Test intervals:

Since concavity changes at $x = 2$, there is an inflection point at $(2, f(2)) = (2, 4)$.

Example 9: Second Derivative Test

Find and classify the critical points of $h(x) = 3x^5 - 5x^3$ using the Second Derivative Test.

Solution. Compute the derivatives:

$$h'(x) = 15x^4 - 15x^2 = 15x^2(x^2 - 1) = 15x^2(x - 1)(x + 1)$$

Critical points: $x = -1, 0, 1$.

$$h''(x) = 60x^3 - 30x = 30x(2x^2 - 1)$$

Evaluate $h''$ at each critical point:

4.5 Optimization Problems

Optimization problems ask you to find the maximum or minimum value of a quantity subject to given constraints. These are among the most heavily tested topics on the AP exam, and they require combining algebraic modeling with derivative techniques.

Strategy for Optimization

  1. Understand the problem. Read carefully. Identify what quantity is to be maximized or minimized (the objective function).
  2. Draw a diagram if applicable. Label all relevant quantities.
  3. Assign variables and write the objective function in terms of those variables.
  4. Identify the constraint equation. Use it to express the objective function in terms of a single variable.
  5. Determine the domain of the objective function (what values of the variable are physically meaningful).
  6. Find the critical points by setting the derivative equal to zero. Test endpoints if the domain is a closed interval.
  7. Verify that your answer is indeed a maximum or minimum (using First or Second Derivative Test, or by checking endpoint values).

Optimization: maximize the area of a rectangle with fixed perimeter P. Drag x to find the optimal width.

Example 10: Maximize Enclosed Area

A farmer has $600$ meters of fencing and wants to enclose a rectangular field that borders a straight river. No fencing is needed along the river. What dimensions maximize the enclosed area?

Solution. Let $x$ be the length of the side perpendicular to the river and $y$ the length of the side parallel to the river. The fencing constraint is:

$$2x + y = 600 \quad \Longrightarrow \quad y = 600 - 2x$$

The area to maximize is:

$$A(x) = x \cdot y = x(600 - 2x) = 600x - 2x^2$$

The domain is $0 \leq x \leq 300$ (since $y \geq 0$). Differentiate:

$$A'(x) = 600 - 4x$$

Setting $A'(x) = 0$: $x = 150$. Since $A''(x) = -4 < 0$, this is a maximum. Then $y = 600 - 300 = 300$.

The maximum area is $A(150) = 150 \times 300 = 45{,}000$ m$^2$, with the field being $150$ m by $300$ m.

Example 11: Minimize Cost

An open-top rectangular box with a square base must have a volume of $32{,}000$ cm$^3$. The material for the base costs $\$10$ per cm$^2$ and the material for the sides costs $\$5$ per cm$^2$. Find the dimensions that minimize the total cost.

Solution. Let the base have side length $x$ and the height be $h$. The volume constraint is:

$$x^2 h = 32{,}000 \quad \Longrightarrow \quad h = \frac{32{,}000}{x^2}$$

The cost function is (base area times $\$10$) + (four sides times $\$5$):

$$C(x) = 10x^2 + 5 \cdot 4xh = 10x^2 + 20x \cdot \frac{32{,}000}{x^2} = 10x^2 + \frac{640{,}000}{x}$$

The domain is $x > 0$. Differentiate:

$$C'(x) = 20x - \frac{640{,}000}{x^2}$$

Set $C'(x) = 0$:

$$20x = \frac{640{,}000}{x^2} \quad \Longrightarrow \quad 20x^3 = 640{,}000 \quad \Longrightarrow \quad x^3 = 32{,}000 \quad \Longrightarrow \quad x = \sqrt[3]{32{,}000}$$

Since $32{,}000 = 32 \times 1000$, we get $x = \sqrt[3]{32} \times 10 = 2\sqrt[3]{4} \times 10 \approx 31.75$ cm.

Verify: $C''(x) = 20 + \frac{1{,}280{,}000}{x^3} > 0$ for all $x > 0$, confirming a minimum. The height is $h = \frac{32{,}000}{x^2} \approx 31.75$ cm. The box is (approximately) a cube with side $\approx 31.75$ cm.

Example 12: AP-Style Optimization

A particle moves along the $x$-axis so that its velocity at time $t \geq 0$ is given by $v(t) = t^2 - 6t + 8$. For what value of $t$ does the particle achieve its minimum velocity, and what is the particle's position at that time if $x(0) = 3$?

Solution. The velocity is $v(t) = t^2 - 6t + 8$. To minimize velocity, find where $v'(t) = 0$:

$$v'(t) = 2t - 6 = 0 \quad \Longrightarrow \quad t = 3$$

Since $v''(t) = 2 > 0$, this is a minimum. The minimum velocity is $v(3) = 9 - 18 + 8 = -1$ (the particle is moving to the left at speed 1).

For the position, integrate the velocity:

$$x(t) = \int_0^t v(s)\,ds + x(0) = \int_0^t (s^2 - 6s + 8)\,ds + 3$$ $$x(t) = \left[\frac{s^3}{3} - 3s^2 + 8s\right]_0^t + 3 = \frac{t^3}{3} - 3t^2 + 8t + 3$$

At $t = 3$:

$$x(3) = \frac{27}{3} - 27 + 24 + 3 = 9 - 27 + 24 + 3 = 9$$

4.6 Mean Value Theorem

Mean Value Theorem (MVT)

If $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$, then there exists at least one number $c$ in $(a, b)$ such that:

$$f'(c) = \frac{f(b) - f(a)}{b - a}$$

Geometrically, this says there is a point where the tangent line is parallel to the secant line connecting $(a, f(a))$ and $(b, f(b))$. In other words, the instantaneous rate of change equals the average rate of change at some point in the interval.

The Mean Value Theorem is a cornerstone of calculus theory. It justifies why a function with a positive derivative on an interval must be increasing: if $f'(x) > 0$ for all $x$ in $(a, b)$, then for any two points $x_1 < x_2$ in $[a, b]$, the MVT gives $f(x_2) - f(x_1) = f'(c)(x_2 - x_1) > 0$.

Rolle's Theorem (Special Case)

If $f$ is continuous on $[a, b]$, differentiable on $(a, b)$, and $f(a) = f(b)$, then there exists at least one $c$ in $(a, b)$ with $f'(c) = 0$.

Rolle's Theorem is the MVT with a horizontal secant line (slope $= 0$). It guarantees a horizontal tangent between any two points at the same height.

Example 13: Applying the Mean Value Theorem

Let $f(x) = x^3 - x$ on the interval $[0, 2]$. Verify that the hypotheses of the MVT are satisfied and find all values of $c$ guaranteed by the theorem.

Solution. The function $f(x) = x^3 - x$ is a polynomial, so it is continuous on $[0, 2]$ and differentiable on $(0, 2)$. The hypotheses are satisfied.

Compute the average rate of change:

$$\frac{f(2) - f(0)}{2 - 0} = \frac{(8 - 2) - (0)}{2} = \frac{6}{2} = 3$$

We need $f'(c) = 3$. Since $f'(x) = 3x^2 - 1$:

$$3c^2 - 1 = 3 \quad \Longrightarrow \quad 3c^2 = 4 \quad \Longrightarrow \quad c^2 = \frac{4}{3} \quad \Longrightarrow \quad c = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3} \approx 1.155$$

We discard the negative root since $c$ must lie in $(0, 2)$. The value $c = \frac{2\sqrt{3}}{3}$ is the point where the tangent line is parallel to the secant line.

Mean Value Theorem: the tangent line at x = c is parallel to the secant line through the endpoints. Drag c to find where they match.

4.7 Practice Problems

Test your understanding with these problems modeled on AP Calculus exam questions. Work through each one before checking the solution.

Problem 1 (Related Rates)

A spherical balloon is being inflated so that its volume increases at a constant rate of $100\pi$ cm$^3$/s. How fast is the radius increasing when the radius is $5$ cm?

Show Solution

$V = \frac{4}{3}\pi r^3$, so $\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$. With $\frac{dV}{dt} = 100\pi$ and $r = 5$:

$$100\pi = 4\pi(25)\frac{dr}{dt} = 100\pi\,\frac{dr}{dt}$$ $$\frac{dr}{dt} = 1 \text{ cm/s}$$

Problem 2 (Related Rates)

Two cars start from the same intersection. Car A travels north at $60$ mph and Car B travels east at $80$ mph. How fast is the distance between them increasing after $2$ hours?

Show Solution

Let $a$ be Car A's distance north, $b$ be Car B's distance east, and $d$ the distance between them. Then $d^2 = a^2 + b^2$.

Differentiating: $2d\frac{dd}{dt} = 2a\frac{da}{dt} + 2b\frac{db}{dt}$.

At $t = 2$: $a = 120$, $b = 160$, $d = \sqrt{120^2 + 160^2} = \sqrt{40000} = 200$.

$$200\frac{dd}{dt} = 120(60) + 160(80) = 7200 + 12800 = 20000$$ $$\frac{dd}{dt} = 100 \text{ mph}$$

Problem 3 (Linear Approximation)

Use linear approximation to estimate $(1.02)^5$.

Show Solution

Let $f(x) = x^5$ with $a = 1$ and $\Delta x = 0.02$. Then $f(1) = 1$ and $f'(x) = 5x^4$, so $f'(1) = 5$.

$$(1.02)^5 \approx 1 + 5(0.02) = 1.10$$

The actual value is $(1.02)^5 = 1.10408...$, so the approximation is accurate to two decimal places.

Problem 4 (Extreme Values)

Find the absolute maximum and minimum of $f(x) = 2\sin x + \cos 2x$ on $[0, 2\pi]$.

Show Solution

$f'(x) = 2\cos x - 2\sin 2x = 2\cos x - 4\sin x \cos x = 2\cos x(1 - 2\sin x)$.

$f'(x) = 0$ when $\cos x = 0$ (i.e., $x = \frac{\pi}{2}, \frac{3\pi}{2}$) or $\sin x = \frac{1}{2}$ (i.e., $x = \frac{\pi}{6}, \frac{5\pi}{6}$).

Evaluate $f$ at critical points and endpoints:

  • $f(0) = 0 + 1 = 1$
  • $f(\pi/6) = 2(\frac{1}{2}) + \cos(\pi/3) = 1 + \frac{1}{2} = \frac{3}{2}$
  • $f(\pi/2) = 2 + \cos\pi = 2 - 1 = 1$
  • $f(5\pi/6) = 2(\frac{1}{2}) + \cos(5\pi/3) = 1 + \frac{1}{2} = \frac{3}{2}$
  • $f(3\pi/2) = -2 + \cos 3\pi = -2 - 1 = -3$
  • $f(2\pi) = 0 + 1 = 1$

Absolute maximum: $\frac{3}{2}$ at $x = \frac{\pi}{6}$ and $x = \frac{5\pi}{6}$. Absolute minimum: $-3$ at $x = \frac{3\pi}{2}$.

Problem 5 (First Derivative Test)

Find all local extrema of $f(x) = xe^{-x}$.

Show Solution

$f'(x) = e^{-x} + x(-e^{-x}) = e^{-x}(1 - x)$.

$f'(x) = 0$ when $x = 1$ (since $e^{-x} > 0$ always).

For $x < 1$: $f'(x) > 0$ (increasing). For $x > 1$: $f'(x) < 0$ (decreasing).

By the First Derivative Test, $f$ has a local maximum at $x = 1$ with $f(1) = e^{-1} = \frac{1}{e}$.

Problem 6 (Second Derivative Test)

Find and classify the critical points of $f(x) = x^4 - 8x^2 + 3$.

Show Solution

$f'(x) = 4x^3 - 16x = 4x(x^2 - 4) = 4x(x-2)(x+2)$. Critical points: $x = -2, 0, 2$.

$f''(x) = 12x^2 - 16$.

  • $f''(-2) = 48 - 16 = 32 > 0$: local minimum, $f(-2) = 16 - 32 + 3 = -13$
  • $f''(0) = -16 < 0$: local maximum, $f(0) = 3$
  • $f''(2) = 48 - 16 = 32 > 0$: local minimum, $f(2) = 16 - 32 + 3 = -13$

Problem 7 (Concavity)

Find the intervals of concavity and inflection points of $g(x) = x^4 - 6x^2 + 4$.

Show Solution

$g'(x) = 4x^3 - 12x$, $g''(x) = 12x^2 - 12 = 12(x^2 - 1) = 12(x-1)(x+1)$.

$g''(x) = 0$ at $x = \pm 1$.

  • $x < -1$: $g''(-2) = 12(3) = 36 > 0$ (concave up)
  • $-1 < x < 1$: $g''(0) = -12 < 0$ (concave down)
  • $x > 1$: $g''(2) = 36 > 0$ (concave up)

Inflection points at $(-1, g(-1)) = (-1, -1)$ and $(1, g(1)) = (1, -1)$.

Problem 8 (Optimization)

A rectangular poster must contain $150$ in$^2$ of printed area. The margins at the top and bottom are $3$ in each, and the side margins are $2$ in each. Find the dimensions of the poster that minimize the total area of the poster.

Show Solution

Let $x$ and $y$ be the width and height of the printed area. Then $xy = 150$, so $y = \frac{150}{x}$.

The total poster width is $x + 4$ and total height is $y + 6$. The total area is:

$$A(x) = (x + 4)(y + 6) = (x + 4)\left(\frac{150}{x} + 6\right) = 150 + 6x + \frac{600}{x} + 24$$ $$A(x) = 174 + 6x + \frac{600}{x}$$

$A'(x) = 6 - \frac{600}{x^2} = 0 \Rightarrow x^2 = 100 \Rightarrow x = 10$.

$A''(x) = \frac{1200}{x^3} > 0$, confirming a minimum. Then $y = 15$.

Poster dimensions: $10 + 4 = 14$ in wide and $15 + 6 = 21$ in tall.

Problem 9 (Mean Value Theorem)

A car travels along a straight highway. At time $t = 0$ the odometer reads $0$ miles and at $t = 2$ hours it reads $140$ miles. Prove that at some instant the car's speed was exactly $70$ mph.

Show Solution

Let $s(t)$ be the position function. Assuming $s$ is continuous on $[0, 2]$ and differentiable on $(0, 2)$ (reasonable for a car), the Mean Value Theorem guarantees a $c$ in $(0, 2)$ with:

$$s'(c) = \frac{s(2) - s(0)}{2 - 0} = \frac{140 - 0}{2} = 70 \text{ mph}$$

Therefore, the car's instantaneous speed was exactly $70$ mph at time $t = c$.

Problem 10 (Mixed: Optimization + Related Rates)

A trough is $8$ ft long with a cross section in the shape of an isosceles trapezoid that is $2$ ft wide at the bottom, $4$ ft wide at the top, and $2$ ft tall. Water is poured in at a rate of $3$ ft$^3$/min. How fast is the water level rising when the water is $1$ ft deep?

Show Solution

At water depth $h$, the water surface width by similar triangles is $w = 2 + 2 \cdot \frac{h}{2} = 2 + h$ (since the trapezoid widens by $1$ ft on each side over $2$ ft of height).

The cross-sectional area of water is a trapezoid with bases $2$ (bottom) and $2 + h$ (water surface) and height $h$:

$$A_{\text{cross}} = \frac{1}{2}(2 + 2 + h)h = \frac{1}{2}(4 + h)h$$

The volume is $V = 8 \cdot A_{\text{cross}} = 4h(4 + h) = 16h + 4h^2$.

Differentiate: $\frac{dV}{dt} = (16 + 8h)\frac{dh}{dt}$.

At $h = 1$ with $\frac{dV}{dt} = 3$:

$$3 = (16 + 8)\frac{dh}{dt} = 24\frac{dh}{dt}$$ $$\frac{dh}{dt} = \frac{3}{24} = \frac{1}{8} = 0.125 \text{ ft/min}$$
← Prev: Ch 3 — Differentiation Rules Chapter 4 of 10 Next: Ch 5 — Integrals →