Geometry for Calculus — AP® Calculus AB Notes

Some of the most memorable problems in calculus are pictures: a ladder sliding down a wall, a cone-shaped tank filling with water, a person walking away from a streetlight while the shadow grows. The calculus in them is short. What takes the time, and where most mistakes happen, is the geometry: writing the one equation that connects the quantities in the picture. This page gathers the geometry you need and shows the two tools that do most of the work.

Formulas to know by heart

The exam gives no formula sheet, so these must be automatic:

Shape	Area or volume	Other
Rectangle, triangle	$lw$ , $\;\frac{1}{2}bh$
Circle	$\pi r^2$	circumference $2\pi r$
Semicircle	$\frac{1}{2}\pi r^2$	$r$ is half the diameter
Trapezoid	$\frac{b_1 + b_2}{2}h$	average of the parallel sides times the height
Equilateral triangle, side $s$	$\frac{\sqrt{3}}{4}s^2$	height $\frac{\sqrt{3}}{2}s$
Box	$lwh$
Cylinder	$\pi r^2 h$	curved surface $2\pi rh$
Cone	$\frac{1}{3}\pi r^2 h$	one third of the cylinder
Sphere	$\frac{4}{3}\pi r^3$	surface $4\pi r^2$

A quick check catches many slips: areas have squared units and volumes have cubed units. If your "volume" formula has $r^2$ and nothing else with a length, something is missing.

The Pythagorean theorem

Whenever a picture has a right angle, the Pythagorean theorem links the three sides: $a^2 + b^2 = c^2$ , where $c$ is the hypotenuse, opposite the right angle. Calculus problems use it as an equation between changing quantities.

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. — **Figure 1.** A right triangle (left) and similar triangles (right): the two equations behind most picture problems. LevlPrep original diagram.

Worked example 1. A 10-foot ladder leans against a wall. Let $x$ be the distance from the wall to its foot and $y$ the height of its top. Write an equation, and find $y$ when $x = 6$ .

The wall meets the ground at a right angle, and the ladder is the hypotenuse: $x^2 + y^2 = 100$ .

When $x = 6$ : $36 + y^2 = 100$ , so $y = 8$ feet.

The equation $x^2 + y^2 = 100$ holds at every moment as the ladder slides; that is what Unit 4 will use.

The same theorem gives the distance between two cars moving along perpendicular roads, the length of a kite string, and the height of an equilateral triangle.

Similar triangles

Two triangles are similar when they have the same three angles. Then one is a scaled copy of the other, and every pair of matching sides has the same ratio. Look for a small triangle sitting inside a big one, sharing an angle.

Worked example 2. A cone-shaped tank, point down, is 10 feet tall with a top radius of 4 feet. Write the volume of water as a function of its depth $h$ .

The water is a small cone inside the tank. Viewed from the side, the water's triangle and the tank's triangle are similar: $\frac{r}{h} = \frac{4}{10}$ , so $r = \frac{2}{5}h$ .

Substitute into the cone formula: $V$ $= \frac{1}{3}\pi r^2 h$ $= \frac{1}{3}\pi\left(\frac{2}{5}h\right)^2 h$ $= \frac{4\pi}{75}h^3$ .

Now $V$ depends on one variable only. Replacing $r$ with $\frac{2}{5}h$ before doing any calculus is the key step.

Worked example 3. A person 6 feet tall stands 10 feet from a lamppost 15 feet tall. Find the length $s$ of the shadow.

The light, the top of the person's head and the tip of the shadow line up. The big triangle has height 15 and base $10 + s$ ; the small triangle has height 6 and base $s$ .

$\frac{6}{s} = \frac{15}{10 + s}$ , so $6(10 + s) = 15s$ , so $60 = 9s$ , and $s = \frac{20}{3}$ feet.

The most common slip is using 10 as the big triangle's base instead of $10 + s$ .

Right-triangle trigonometry

When an angle is part of the question, use the right-triangle ratios: $\sin\theta$ $= \frac{\text{opposite}}{\text{hypotenuse}}$ , $\cos\theta$ $= \frac{\text{adjacent}}{\text{hypotenuse}}$ , $\tan\theta$ $= \frac{\text{opposite}}{\text{adjacent}}$ . A balloon rising straight up from a point 50 meters from an observer makes an angle of elevation $\theta$ with $\tan\theta = \frac{h}{50}$ .

A routine for picture problems

Draw the picture, even if one is given.
Label every quantity that changes with a letter and every fixed quantity with its number.
Find the right angle or the similar triangles, and write one equation.
If the formula has too many variables, use a ratio or the Pythagorean theorem to eliminate one.

Substitute a specific moment's numbers (like $x = 6$ ) only at the end. If you put them in too early, the equation stops describing the changing picture.