Topic 4.5: Solving Related Rates Problems

Clocking the spot

Let’s go back to the harbor and clock that spot. From straight above, the lamp, the foot of the wall and the spot make a right triangle, with the lamp a hundred meters from the wall.

These flashes come at equal times, every one and a quarter seconds, and yet the marks keep spreading apart. The farther the spot gets from the foot, the more wall it covers between flashes. We’ve already solved three of these with the same five steps, so let’s name them as we go.

Step one is to draw it and name what changes. The wall is always a hundred meters away, while the angle theta and the spot’s distance from the foot, x, both change. Step two is an equation that’s true at every instant. The tangent of theta is the opposite side over the adjacent side, x over a hundred, so x equals a hundred tan theta.

Step three, differentiate both sides with respect to time. The derivative of tangent is secant squared, and the chain rule hands us d theta dt. That makes dx dt equal a hundred secant squared theta, times d theta dt.

Step four is the snapshot, and this one hides a trap. The lamp turns twice a minute, and the rule that tangent’s derivative is secant squared only works in radians. That’s twelve degrees per second, or two times two pi radians every sixty seconds, pi over fifteen radians per second. The formula needs the pi over fifteen, never the twelve, and that’s the point two oh nine you saw on the dial.

Right at the foot, theta is zero, and secant squared of zero is one. So the spot moves at a hundred times pi over fifteen, about twenty-one meters per second, as fast as that car from the beginning.

Now picture the spot at sixty degrees, much farther down the wall. How many times faster do you think it’s moving there than at the foot? Your turn to clock it. When the spot is a hundred meters from the foot, how fast is it moving, in meters per second, to three decimal places?

Step five is to say what it means, with the moment, the quantity and the units. When the spot is a hundred meters from the foot, its distance from the foot is increasing at about forty-one point nine meters per second.

Secant squared is one plus tan squared, and tan theta is x over a hundred. That lets us write the speed in terms of the distance, and the cold open’s numbers fall right out. At the foot it’s about twenty-one meters a second, a car. Three hundred ninety-two meters out, it’s the speed of sound. Near the end of the wall, it’s about seven point seven kilometers a second.

Far down the wall, the beam hits the stones at a steep slant, so each tiny turn of the lamp sweeps the spot across a lot more wall. Nothing on that wall is moving. The spot is just where the light lands, so nothing has to push it. That’s how it outruns a bullet.

Something very like this happened in the eclipse of August twenty-first, twenty seventeen. The Moon’s shadow hit the Oregon coast at about twenty-four hundred miles an hour, then slowed to about fourteen hundred fifty over Kentucky. On the coast it struck the curved Earth at a slant, a lot like our spot far down the wall, and over Kentucky it came down much more squarely.

Keep the boat in the light

Now it’s your turn to run the light. A patrol boat is about to race along the wall at a steady twenty meters per second, about seventy-two kilometers an hour. The keeper wants it in the beam the whole way.

Drag the ring around the lamp to keep the boat lit from one end of its run to the other. If it gets away from you, you can try again. This curve shows how fast the lamp turned during the run. So where does it have to turn fastest?

Drag the boat along its path and drop a few dots. Each dot shows how fast the lamp must turn right there. Then find the spot on the wall where it has to turn fastest. As the boat passes the foot, the lamp has to whip around, and out at the ends it hardly moves.

Let’s see why, with the same equation we used to clock the spot. This time we know the speed along the wall, twenty meters per second, and we want the turning rate, so we solve for d theta d t. Dividing, and remembering that one over secant squared is cosine squared, gives d theta dt equals zero point two times cosine squared theta.

It’s biggest when theta is zero, right at the foot, and only a quarter as big at sixty degrees, five hundredths of a radian per second. It’s the tap all over again. To keep the boat’s speed steady, the lamp’s speed has to keep changing. When the boat is a hundred meters past the foot, how fast does the lamp need to turn, in radians per second?

Lay the formula’s curve over the run and the dots, and the dots sit right on it. It gives the turning rate for perfect tracking, fastest at the foot and gentle at the ends, and any hand that keeps the boat lit has to follow it. Hand the lamp the formula and let it steer by itself. It turns fastest right at the foot and eases off at the ends, and the boat stays in the light the whole way.

That’s the promise from the beginning. From one number, how fast the lamp turns, you clocked the spot anywhere on the wall. Then you turned the same equation around and found exactly how the lamp has to turn to keep a racing boat in the light.

Tied together

Every problem in this lesson worked the same way. An equation that’s true at every instant tied two quantities together, and differentiating that equation with respect to time tied their rates together too. You draw and name, relate, differentiate with respect to time, plug in the snapshot, and say what it means. Save the snapshot for after the derivative, and keep your angles in radians.

On an exam, the context might be a tank, a camera or a particle on a curve. Name the changing quantities, find the equation that ties them together, and differentiate it to tie their rates. The next time a spotlight races along a wall, you’ll know why it speeds up, and with the lamp’s turning rate and its distance from the wall, you can work out exactly how fast the spot is going.