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?
- 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