Wemazi

Topic 4.4

Write the relationship between them

Builds on identifying the changing quantities. 7 minutes.

A ladder against a wall

A 5-meter ladder leans against a wall. The foot is x meters from the wall, and the top is y meters above the floor.

Drag the foot away from the wall, or move the slider. Both x and y change, but the ladder, the wall and the floor always form a right triangle.

The readout shows that \(x^2 + y^2\) is 25 at every position. This equation links the two changing quantities, and it uses the fixed length 5 as a number.

Because the equation holds at every time, you can differentiate it later. An equation for one position only cannot be differentiated this way.

Geometry gives the equation

The shape of the scene decides how the changing quantities depend on each other.

Press each button. A right triangle gives the Pythagorean equation, and similar triangles give equal ratios of sides. An angle gives a ratio of sides such as \(\tan\theta\).

Choose the rule that matches the picture, and write it with the changing quantities and the fixed lengths.

Notation

  1. Find the geometry in the picture. It can be a right triangle, similar triangles, an angle, or a formula for an area or a volume. Here the ladder, the wall and the floor form the shaded right triangle.

  2. Write the equation with the changing quantities and the fixed lengths. The square of the long side equals the sum of the squares of the other two sides.

    \[x^2 + y^2 = \cq{5^2}\]

  3. Check that the equation holds at every time. Do not put in the values of one moment.

  4. If a variable has no given rate and no wanted rate, remove it. Use the geometry to write it with another variable.

When it fails

An equation that uses the values of one moment is true at that moment only. At \(x = 3\) the top is at \(y = 4\), so \(9 + y^2 = 25\) is true there.

The button moves the foot to \(x = 4\), where the top is at \(y = 3\). The red equation still gives \(y = 4\), the red dot, because it treats x as 3. Write the relationship with the letter \(x\).

Example 1: A shadow

A lamp is on a pole that is 5 meters tall. A person who is 1.8 meters tall stands \(x\) meters from the pole. The person’s shadow is \(s\) meters long. Write an equation that links \(s\) and \(x\).

  1. The ray from the lamp passes the top of the person’s head and meets the ground at the tip of the shadow. This makes two right triangles that share the angle at the tip.

  2. The two triangles are similar, so the ratio of height to base is the same in both.

    \[\frac{5}{x + s} = \frac{1.8}{s}\]

  3. Multiply both sides by \(s(x + s)\). Then collect the terms in \(s\).

    \[\begin{aligned} 5s &= 1.8(x + s) \\ 3.2s &= 1.8x \end{aligned}\]

  4. Divide by 3.2. This relationship holds at every distance \(x\).

    \[s = \frac{9}{16}x\]

Example 2: A boat on a rope

A rope runs from the top of a dock post, 6 meters above the water, to a boat. The boat is \(x\) meters from the base of the post, and the rope is \(s\) meters long. Write an equation that links \(s\) and \(x\).

  1. The post, the distance x and the rope form a right triangle. The rope is the long side.

  2. A tempting wrong move: The rope goes from the post to the boat, so add the two lengths.

    \[\cr{s = x + 6}\]

    The red dashed path down the post and along the water has length \(x + 6\), but the rope goes straight across. When \(x\) is 8, the rope is 10 meters long and \(x + 6\) is 14.

  3. The square of the long side equals the sum of the squares of the other two sides.

    \[s^2 = x^2 + 6^2\]

Practice

A camera is 4 kilometers from a launch pad and points at a rocket. The rocket is \(h\) kilometers above the ground, and the camera’s angle above the horizontal is \(\theta\). Write an equation that links \(h\) and \(\theta\).

In the right triangle in the picture, one side is the fixed 4 kilometers. Choose the ratio of sides that uses the angle.

Solution

  1. Relative to \(\theta\), the side \(h\) is opposite and the fixed side 4 is adjacent. The tangent uses these two sides.

    \[\tan\theta = \frac{h}{4}\]

  2. Multiply both sides by 4 to write \(h\) by itself.

    \[h = 4\tan\theta\]

Questions

A balloon rises straight up from the ground. A range finder on the ground is 200 meters from the point below the balloon. The balloon is \(y\) meters high, the line of sight to it is \(s\) meters long, and its angle above the horizontal is \(\theta\).

  1. Which equation links \(y\) and \(\theta\)?

    • \(\tan\theta = \dfrac{y}{200}\)
    • \(\sin\theta = \dfrac{y}{200}\)
    • \(y = 200 + \theta\)
    • \(y^2 + \theta^2 = 200^2\)
    Answer

    \(\tan\theta = \dfrac{y}{200}\)

  2. Which equation links \(s\) and \(y\)?

    • \(s^2 = y^2 + 200^2\)
    • \(s = y + 200\)
    • \(s^2 = y^2 - 200^2\)
    • \(y^2 = s^2 + 200^2\)
    Answer

    \(s^2 = y^2 + 200^2\)

  3. A box has a square base with side \(x\) and a height \(h = 2x\). Which equation gives the volume \(V\) with \(x\) alone?

    • \(V = 2x^3\)
    • \(V = x^3\)
    • \(V = 2x^2\)
    • \(V = 4x^3\)
    Answer

    \(V = 2x^3\)

Summary

You write one equation from the geometry of the picture, with the changing quantities as letters and the fixed lengths as numbers.

The equation must hold at every time, because the next step differentiates it with respect to t.

A right triangle, similar triangles, an angle or a volume formula gives the equation, and you remove any variable that you do not need.

This is one of four skills in topic 4.4 of Which & Why: AP Calculus.