# Calculus 1, Related Rates, Circle Problem?

Calculus 1. Related Rates. Chetna is racing around a circular track, starting at the easternmost edge and running counterclockwise at a constant pace of 40 seconds per lap. At the center of the track is a lamp, and 60 meters east of the lamp is a wall that runs north-south. When Chetna is one-twelfth of the way around the track, how fast is her shadow moving along the wall?

### 3 Answers

- 1 month ago
Full track is 2piRadius

2pi*60 = 377m

Circunference over 40 per lap is speed.

377/40 = 9.24 meters per second

- 1 month ago
She runs 2pi radians in 40 seconds

2pi/40 = pi/20

dT/dt = pi/20 radians/second

The wall is 60 meters away from the center of the circle and T = pi/6 (because 2pi/12 = pi/6)

If we say that the point on the wall closest to the lamp is O, then we can say that the shadow touches the wall at some point y, where y is y meters north or south of O.

So you have a right triangle. x = 60. y = ?. y/x = tan(T)

y = x * tan(T)

y = 60 * tan(T)

Derive with respect to time

dy/dt = 60 * sec(T)^2 * dT/dt

T = pi/6 , dT/dt = pi/20

dy/dt = 60 * sec(pi/6)^2 * (pi/20)

dy/dt = 3 * pi / cos(pi/6)^2

dy/dt = 3 * pi / (sqrt(3)/2)^2

dy/dt = 3 * pi / (3/4)

dy/dt = 3 * pi * 4 / 3

dy/dt = 4 * pi

The shadow is moving along the wall at 4pi meters per second at that moment.

- Jeff AaronLv 71 month ago
Let x = how many degrees Chetna has run around the track (2*pi radians per lap)

Let y = distance from the shadow to the point on the wall that's closest to the circletan(x) = y/60

y = 60*tan(x)

dy/dx = 60*sec^2(x)

dx/dt = 1/40 laps per second, that's 2*pi/40 = pi/20 radians per second

dy/dt = dy/dx * dx/dt

dy/dt = 60*sec^2(x) * (pi/20)

dy/dt = 3*pi*sec^2(x)

When Chetna is 1/12 of the way around the track, x = 2*pi/12 = pi/6 radians, so we have:

dy/dt = 3*pi*sec^2(pi/6)

dy/dt = 3*pi*(2/sqrt(3)) ^2

dy/dt = 3*pi*4/3

dy/dt = 4*pi

dy/dt =~ 12.566370614359172953850573533118 meters per second