Related Rates - Sector/Cosine

1. Jan 3, 2008

rocomath

A runner sprints around a circular track of radis 100m at a constant speed of 7m/s. The runner's friend is standing at a distance 200m from the center of the track. How fast is the distance between the friends changing when the distance between them is 200m.

Equations: Area of a Triangle, Law of Cosines, and Sector of a Circle.

$$A = \frac {1}{2}bh \ (\sin\theta = hr) \rightarrow A = \frac {1}{2}br\sin\theta$$

$$A = \frac {1}{2}r^{2}\theta \ (s = r\theta) \rightarrow A = \frac {s^2}{2\theta}$$

Setting Area of the Triangle and Sector of the Circle equal to each other ...

$$\frac {1}{2}br\sin\theta = \frac {s^2}{2\theta} \rightarrow 200\sin\theta = s$$

Using Law of Cosines to relate l with theta ...

$$l^2 = b^2 + r^2 - 2br\cos\theta \rightarrow l^2 = 50,000 - 40,000\cos\theta$$

$$\theta = \cos^{ - 1}\left(\frac {b^2 + r^2 - l^2}{2br}\right)$$

Taking the derivative ...

$$\frac {d}{dt}(200\sin\theta) = \frac {d}{dt}(s) \rightarrow \frac {d\theta}{dt} = \frac {1}{200\cos\theta}\frac {ds}{dt}$$

$$\frac {d}{dt}(l^2) = \frac {d}{dt}(50,000 - 40,000\cos\theta) \rightarrow \frac {dl}{dt} = 100\sin\theta\frac {d\theta}{dt}$$

Substituting ...

$$\frac {dl}{dt} = \frac {1}{2}\tan\theta\frac {ds}{dt} = \frac {1}{2}\tan\left[\cos^{ - 1}\left(\frac {b^2 + r^2 - l^2}{2br}\right)\right]\frac {ds}{dt} \approx 13.56m/s$$

Last edited: Jan 3, 2008
2. Jan 3, 2008

dynamicsolo

Why not just start from the Law of Cosines?
I'll give you a tip: implicit differentiation is the heart of related rates problems. If you differentiate your result here with respect to time, you have

$$\frac {d}{dt}(l^2) = \frac {d}{dt}(50,000 - 40,000\cos\theta) \rightarrow 2l \frac {dl}{dt} = 40,000\sin\theta\frac {d\theta}{dt}$$ .

They were nice to you in this problem and told you l = 200 m right off, giving you

$$\frac {dl}{dt} = 100\sin\theta\frac {d\theta}{dt}$$ ,

as you found. The Law of Cosines gives us

$$l^2 = 40,000 = 50,000 - 40,000\cos\theta$$,

so $$\cos\theta = \frac{1}{4}$$ , making $$\sin\theta$$ equal to [sqrt(15)]/4 .

The arclength of the runner's path is increasing at 7 m/sec, so the angular rate is

$$\frac {d\theta}{dt} = \frac{7}{100} rad/sec$$.

So at last we have

$$\frac {dl}{dt} = 100 (srqt 15)(1/4)(7/100) = 6.78 m/sec$$.

I'd avoid dealing with the angle explicitly, since the derivatives get messy and leave one more open to error (especially on a timed exam).

I think I see where the extra factor of 2 got into your analysis. I don't think the areas argument is actually helpful in this problem; you will need the runner's angular speed around the circular path, but that comes directly from the definition of angular velocity. The trigonometric equation for the sides of the triangle is sufficient to work from

Last edited: Jan 3, 2008
3. Jan 3, 2008

rocomath

Ah, you are a miracle worker :-]

Question: Did you find the angular rate through the question? When it states that "the runner sprints around a circular track of radius 100m at a constant speed of 7m/s" ... so 7/100?

nvm, you used D=rtheta

Last edited: Jan 3, 2008
4. Jan 4, 2008

HallsofIvy

Staff Emeritus
Not quite. The 7 m is measured around the circumference of a circle while the 100 m is a radius. You have to compare circumferences, not radii. Radian measure can be defined by "$\theta$= "arc length" / circumference of circle" so a distance of 7 meters around a circle of radius 100m, diameter 200 m and so circumference $2000\pi$. The radian measure of the angle is $7/(200\pi)$.