Find any missing instantaneous measurement
A rider on a Ferris wheel
A Ferris wheel has a radius of 10 meters, and its center is 12 meters above the ground. A rider sits on the rim.
The orange segment u runs sideways from the center to the rider, and the orange segment v runs up from the center to the rider.
Drag the rider along the rim. The lengths u and v change, but the faint radius is 10 meters long at every moment. It is the longest side of a right triangle with sides u and v, so \(u^2 + v^2 = 100\).
At \(u = 6\), the relation gives \(v^2 = 100 - 36 = 64\). The rider is above the center, so \(v = 8\), and the height above the ground is \(12 + 8 = 20\) meters.
One known value fixes the others
The relation \(u^2 + v^2 = R^2\) holds at every moment, where R is the radius.
Drag the rider. Each value of u fixes the value of v, because the rider stays on the rim.
One value of u fits two points on the rim, one above the center and one below it. The problem shows which one is meant.
Because the relation is true at the moment of the question, it gives the missing value without any rate.
Notation
Write the relation that holds at every moment. List the values that the problem gives at the instant.
\[\cy{u}^2 + \cy{v}^2 = 10^2\]Use a relation, not a rate, to find a length.
Substitute the known value into the relation. Then solve for the missing value.
\[\cy{v}^2 = 100 - 6^2 = 64\]Both 8 and \(-8\) square to 64. The rider is above the center, so take the positive root.
\[\cy{v} = 8\]The root \(-8\) gives the point below the center.
Find an angle from the same triangle. Let \(\theta\) be the angle between the radius and the faint vertical line. Its sine is the opposite side \(u\) over the radius, and its cosine is the adjacent side \(v\) over the radius.
\[\begin{gathered} \sin\theta = \frac{6}{10} = \frac{3}{5} \\ \cos\theta = \frac{8}{10} = \frac{4}{5} \end{gathered}\]
When it fails
The known value must come from the same moment as the question. If you use \(u = 3\) from an earlier moment, you find \(v^2 = 100 - 9 = 91\).
This is \(v\) at the earlier moment. The red radius ends where the rider was then, not where the rider is at the moment of the question.
Example 1: A pane of glass
The four corners of a rectangular pane touch a circular frame of radius 5 feet. At one moment the width \(w\) of the pane is 6 feet. Find the height \(h\) and the area of the pane at that moment.
The corners touch the circle, so the diagonal of the pane is a diameter, 10 feet long. For every shape of the pane, the diagonal is the longest side of a right triangle with sides \(w\) and \(h\).
\[\cy{w}^2 + \cy{h}^2 = 10^2\]Substitute \(w = 6\) into the relation.
\[\cy{h}^2 = 100 - 36 = 64\]A height is positive, so \(h = 8\).
\[\cy{h} = 8\]The area is the width times the height.
\[\cb{A} = 6 \cdot 8 = 48\]
Example 2: A conical tank
Water fills a tank shaped like an upside-down cone. The tank is 6 m tall, and its open top has radius 3 m. At one moment the water is 4 m deep. Find the radius and the area of the water surface.
A tempting wrong move: The surface of the water is a circle. Use the radius of the top of the tank, \(r = 3\).
\[A = \pi \cdot 3^2 = 9\pi\]The red circle has the radius of the top of the tank, so at the water line it reaches past the walls of the cone. The water is only 4 m deep, so its surface is a smaller circle.
The water forms a smaller cone with the same shape as the tank, so r divided by h is the same at every depth.
\[\frac{\cy{r}}{\cy{h}} = \frac{3}{6}\]This gives \(r = h/2\) at every moment. With \(h = 4\), the radius is 2 m.
\[\cy{r} = \frac{4}{2} = 2\]The surface is a circle of radius 2 m.
\[\cb{A} = \pi \cdot 2^2 = 4\pi\]
Practice
A kite flies 12 m above the hand of the person who holds the string. At one moment 20 m of string is out, and the string is straight. Find the horizontal distance \(x\) to the point below the kite. Then find the cosine of the angle between the string and the ground.
In the picture, the string is the longest side of a right triangle. The height and x are the other two sides.
Solution
The string, the height and x form a right triangle at every moment.
\[x^2 + 12^2 = s^2\]Substitute \(s = 20\). A distance is positive, so take the positive root.
\[\begin{gathered} x^2 = 400 - 144 = 256 \\ x = 16 \end{gathered}\]The cosine is the side next to the angle over the string.
\[\cos\theta = \frac{16}{20} = \frac{4}{5}\]
Questions
The figure shows a ramp that rises 8 meters over a horizontal distance of 15 meters.
How long is the ramp?
- 17 m
- 23 m
- \(\sqrt{161}\) m
Answer
17 m
What is the sine of the angle between the ramp and the ground?
- \(\dfrac{8}{17}\)
- \(\dfrac{15}{17}\)
- \(\dfrac{8}{15}\)
Answer
\(\dfrac{8}{17}\)
The diagonal of a square is 10 m at one moment. What is the area of the square at that moment?
- \(50\ \text{m}^2\)
- \(100\ \text{m}^2\)
- \(25\ \text{m}^2\)
- \(5\sqrt{2}\ \text{m}^2\)
Answer
\(50\ \text{m}^2\)
Summary
You find a missing length, angle or area from the relation that holds at every moment, using the value you know at that moment.
This works because the relation is true at the moment of the question, so one known value fixes the others.
You use it before you substitute into the rate equation, which needs every value at the same instant.
This is one of four skills in topic 4.5 of Which & Why: AP Calculus.