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Geometry for Calculus

Calculus word problems start with a geometric equation.

Practice 1: Implementing Mathematical Processes

Part 1 · Hook

Why this matters

A ladder slides down a wall; water pours into a cone; a shadow stretches as someone walks from a lamppost. Unit 4 asks how fast those quantities change, and Unit 8 builds solids from slices. Before any calculus happens, each problem needs one geometric equation linking its variables. This page collects the formulas and the two big tools: the Pythagorean theorem and similar triangles.

Part 2 · Prerequisite check

Quick check before you start

1. Solve $\frac{4}{s} = \frac{12}{8 + s}$ for $s$ .

$4$
$\frac{8}{3}$
$2$
$8$

Show the answer

Cross-multiply: $4(8 + s) = 12s$ , so $32 = 8s$ and $s = 4$ .

Correct: $4$ :
$\frac{8}{3}$ :
$2$ :
$8$ :

2. In a right triangle with angle $\theta$ , $\tan\theta$ is

opposite over adjacent
adjacent over opposite
opposite over hypotenuse
adjacent over hypotenuse

Show the answer

$\tan\theta$ $= \frac{\sin\theta}{\cos\theta}$ $= \frac{\text{opp}}{\text{adj}}$ .

Correct: opposite over adjacent:
adjacent over opposite:
opposite over hypotenuse:
adjacent over hypotenuse:

Part 3 · See it

See it first

Left: a 10-foot ladder leans against a wall with its foot 6 feet out and its top 8 feet up, a right triangle with x squared + y squared = 10 squared. Right: a cone-shaped tank, point down, 10 feet tall with top radius 4 feet, holding water of depth h and surface radius r; similar triangles give r over h = 4 over 10. — Two equations that come up again and again: a ladder gives $x^2 + y^2 = 10^2$ ; a cone filling with water gives similar triangles, $\frac{r}{h} = \frac{4}{10}$ . LevlPrep original diagram.

Part 4 · Step by step

How it works, step by step

A right angle in the picture (wall and ground, two roads crossing)the Pythagorean theorem links the sides: $x^2 + y^2 = z^2$
Two triangles with the same angles (a shadow, water in a cone)similar triangles: corresponding sides are in the same ratio, such as $\frac{r}{h} = \frac{4}{10}$
Use the ratio to replace one variable by anothera formula like $V = \frac{1}{3}\pi r^2 h$ becomes a function of one variable, $V = \frac{4\pi}{75}h^3$

Part 5 · Key ideas

Key ideas

Pythagorean theorem: in a right triangle with legs $a$ , $b$ and hypotenuse $c$ , $a^2 + b^2 = c^2$ . The hypotenuse is opposite the right angle.
Similar triangles have equal angles, so their sides are in proportion: $\frac{a}{b} = \frac{A}{B}$ . Look for a small triangle inside a big one.
Know the area and volume formulas: circle $\pi r^2$ , semicircle $\frac{1}{2}\pi r^2$ , trapezoid $\frac{b_1 + b_2}{2}h$ , cylinder $\pi r^2 h$ , cone $\frac{1}{3}\pi r^2 h$ , sphere $\frac{4}{3}\pi r^3$ .
Draw the picture, label what changes with letters and what is fixed with numbers, then write one equation.

Part 6 · Misconception

A common mistake

The wrong idea: In the cone problem, the radius of the water is the tank's radius, 4, so $V = \frac{1}{3}\pi(4)^2 h$ .

What actually happens: The water's radius changes with its depth. Similar triangles give $r = \frac{4}{10}h$ ; substitute before doing anything else: $V = \frac{1}{3}\pi\left(\frac{2}{5}h\right)^2 h$ .

Part 7 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

1. A 13-foot ladder leans against a vertical wall. Its foot is 5 feet from the wall. How high up the wall does the top reach?

No calculator. Type the exact value (a fraction, π, e or a root is fine), then its unit.

Show the answer

The ladder, wall and ground form a right triangle with hypotenuse 13: $5^2 + y^2 = 13^2$ , so $y^2 = 144$ and $y = 12$ feet.

Answer: 12 feet

2. A conical tank, point down, is 9 feet deep with a top radius of 6 feet. When the water is $h$ feet deep, its surface has radius $r$ . Which equation relates $r$ and $h$ ?

$r = \frac{2}{3}h$
$r = \frac{3}{2}h$
$r = h - 3$
$r^2 + h^2 = 36$

Show the answer

The water forms a smaller cone similar to the tank, so the ratio of radius to depth is the same: $\frac{r}{h} = \frac{6}{9} = \frac{2}{3}$ .

Correct: $r = \frac{2}{3}h$ : Similar triangles: $\frac{r}{h} = \frac{6}{9}$ .
$r = \frac{3}{2}h$ : That is depth over radius; the ratio is $\frac{r}{h} = \frac{6}{9}$ .
$r = h - 3$ : Similar shapes scale by a factor; they do not differ by a constant.
$r^2 + h^2 = 36$ : $r$ and $h$ are legs of the side-view triangle, not a radius and a slant length of 6.

3. For the tank above ( $r = \frac{2}{3}h$ ), write the volume of water $V$ as a function of the depth $h$ alone. Use $V = \frac{1}{3}\pi r^2 h$ .

Type an expression for $V =$ .

Show the answer

Substitute $r = \frac{2}{3}h$ : $V$ $= \frac{1}{3}\pi\left(\frac{2}{3}h\right)^2 h$ $= \frac{1}{3}\pi \cdot \frac{4}{9}h^2 \cdot h$ $= \frac{4\pi h^3}{27}$ .

Answer: $V =$ $\frac{4\pi h^3}{27}$

4. A person 6 feet tall stands 10 feet from a lamppost 15 feet tall. How long is the person's shadow on the ground?

No calculator. Type the exact value (a fraction, π, e or a root is fine), then its unit.

Show the answer

Two similar right triangles share the angle at the shadow's tip: $\frac{6}{s} = \frac{15}{10 + s}$ . Cross-multiply: $60 + 6s = 15s$ , so $9s = 60$ and $s = \frac{20}{3} \approx 6.67$ feet.

Answer: $\frac{20}{3}$ ≈ 6.667 feet

5. Write the area $A$ of an equilateral triangle in terms of its side length $s$ .

Type an expression for $A =$ .

Show the answer

The height splits the base in half: $h^2 + \left(\frac{s}{2}\right)^2 = s^2$ , so $h = \frac{\sqrt{3}}{2}s$ . Then $A$ $= \frac{1}{2} \cdot s \cdot \frac{\sqrt{3}}{2}s$ $= \frac{\sqrt{3}}{4}s^2$ .

Answer: $A =$ $\frac{\sqrt{3}}{4}s^2$

6. A box with a square base of side $x$ and height $y$ has volume 32 cubic inches. Which expresses $y$ in terms of $x$ ?

$y = \frac{32}{x^2}$
$y = \frac{32}{x}$
$y = 32 - x^2$
$y = \sqrt{32} - x$

Show the answer

Volume $= x \cdot x \cdot y = x^2 y = 32$ , so $y = \frac{32}{x^2}$ .

Correct: $y = \frac{32}{x^2}$ : $x^2 y = 32$ .
$y = \frac{32}{x}$ : The base has area $x^2$ , not $x$ .
$y = 32 - x^2$ : Volume multiplies; it does not subtract.
$y = \sqrt{32} - x$ : $x^2 y = 32$ does not give a square root.

7. A balloon rises straight up from a point 50 meters from an observer on level ground. When the balloon is $h$ meters high, the angle of elevation from the observer is $\theta$ . Which equation is true?

$\tan\theta = \frac{h}{50}$
$\sin\theta = \frac{h}{50}$
$\tan\theta = \frac{50}{h}$
$\cos\theta = \frac{h}{50}$

Show the answer

In the right triangle, $h$ is opposite $\theta$ and 50 is adjacent: $\tan\theta$ $= \frac{\text{opposite}}{\text{adjacent}}$ $= \frac{h}{50}$ .

Correct: $\tan\theta = \frac{h}{50}$ : Opposite over adjacent.
$\sin\theta = \frac{h}{50}$ : Sine uses the hypotenuse, $\sqrt{h^2 + 2500}$ , not 50.
$\tan\theta = \frac{50}{h}$ : That is adjacent over opposite, which is $\cot\theta$ .
$\cos\theta = \frac{h}{50}$ : Cosine is adjacent over hypotenuse.

Part 8 · Summary

Summary

Calculus word problems start with a geometric equation. Right angles call for the Pythagorean theorem; nested triangles call for similar triangles and a ratio; shapes call for their area or volume formulas. Label changing quantities with letters, fixed ones with numbers, and use a ratio to reduce a formula to one variable.

Part 9 · Up next

What comes next

Part 10 · Connections

Connections

Next: Introduction to Related Rates (4.4)