# What am I doing wrong with Related Rates?

Ok, so the first thing I did was assess the x-values at 30°, 60°, & 80° through $y tan(\theta°) = x$.

$@30°, x = \frac{\sqrt3}{120}$

$@60°, x = \frac{\sqrt{3}}{40}$

$@80°, x = 40 cot(\frac{\pi}{18})$

Then I converted 30 revolutions/min into radians/second.

$\frac{dx}{dt} = \frac{30rev.}{min} = \frac{\pi}{second}$

Then find the values of $\frac{dy}{dt}$.

$x^2 + y^2 = hypotenuse$

$2x(\frac{dx}{dt}) + 2y(\frac{dy}{dt}) = 0$

$x(\pi) + 40(\frac{dy}{dt}) = 0$

$@30°, \frac{dy}{dt} = \frac{-\pi}{1600\sqrt3}$

$@60°, \frac{dy}{dt} = \frac{-\pi\sqrt{3}}{1600}$

$@80°, \frac{dy}{dt} = -\pi cot(\frac{\pi}{18})$

Now I can use the related rates equation:

$tan(\theta) = \frac{x}{y}$

$\frac{d\theta}{dt} sec^2(\theta) = \frac{(\frac{dx}{dt})(y) - (x)(\frac{dy}{dt})}{y^2}$

Is this right so far?

• Use the relation $x = 40\tan \theta$ to relate $x$ and $\theta$ (the hypotenuse is immaterial in this problem), then differentiate both sides w.r.t. $t$. – Jon Dec 9 '14 at 6:13
• Huh, I'd never thought of it like that way. So $\frac{dx}{dt} = 40 sec^2(\theta)\frac{d\theta}{dt}$. And plugging-in values would get me $\pi = 40 sec^2(30°/60°/80°)\frac{d\theta}{dt}$. Are the answers: .0589ft/sec,.0196, & .002368? – Sentient Dec 9 '14 at 6:38
• Carefull... Revolutions per minute count the number of times around the circle per minute, which is the change in $\theta$ with respect to time, and the distance the light moves is measured on the $x$-axis. Your equation is correct, but you're assigning values to the wrong rate. – Jon Dec 9 '14 at 6:55

Your final equation is correct, but you would be plugging incorrect values of $\frac{dx}{dt}$ and $\frac{dy}{dt}$ into it, so it will not give you the correct answer until you fix those earlier errors.

The first error (credit for this one really should go to Jon, who pointed it out in a comment) is that you set $\frac{dx}{dt}$ equal to the rate of rotation. The rate of rotation is $\frac{d\theta}{dt}$, not $\frac{dx}{dt}$.

In fact, $\frac{dx}{dt}$ is the answer to the question, that is, it is the last thing you need to calculate. It's good to remember that.

The second error is not noticing that by definition, since $y = 40$ (a constant), $\frac{dy}{dt} = 0.$ Keeping this in mind, you can easily see that something went wrong in your calculations involving the hypotenuse, since they gave non-zero values of $\frac{dy}{dt}.$

The third error is setting the rate of change of the hypotenuse to zero. (I presume this is why you wrote $2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0.$) The hypotenuse changes quite a bit, in fact.

If you take your final equation and plug in the correct value of $\frac{dy}{dt}$ (that is, set it to zero) and perform some obvious simplifications (including setting $y$ to $40$), you will find that it is just the same thing you would get by differentiating both sides of $40 \tan \theta = \frac{dx}{dt}$.

What a quaint problem, by the way. It's been several decades since I've seen a patrol car with a revolving light.