To determine
To find: The quantities in the given problem.
Answer
The quantities in the given problem is dxdt=5 ft/s.
Explanation
Given:
Given: a man 6 ft tall walks away from street light mounted on a 15-ft-tall pole at a rate of 5 ft/s. If we let t be time (in s) and x be the distance from the pole to the man ( in ft ), then we are given that dxdt=5 ft/s.
To determine
To find: The unknown.
Explanation
Formula used:
Properties of similar triangles
Unknown: the rate at which the tip of his shadow is moving when he is 40 ft from the pole. If we let y be the distance from the man to the tip of his shadow (in ft. ), then we want to find ddt(x+y) when x=40 ft .
To determine
To draw: The picture of the position of the man and his shadow from the pole.
Explanation
Draw the picture as,
From Figure 1, it is observed that position of the man and his shadow from the pole.
Notation:
- x - Distance to the man from the pole.
- y - Distance to the man from the tip of his shadow.
- (x+y) - Distance to the tip of his shadow from pole.
To determine
To write: The equation that relates the quantities.
Answer
The Equation is y =23x.
Explanation
Equation which relates the quantities: by using the properties of similar triangles in the triangle △CAB and △CDE
x+yy=15615y =6x+6y9y =6xy =23x
To determine
To find: How fast is the tip of his shadow moving when he is 40 ft from the pole.
Answer
The required rate at which the tip of his shadow moves is ddt(x+y)=253 ft/s.
Explanation
The tip of his shadow moving at a rate of
ddt(x+y)=ddt(x+23x) [Q y=23x]=ddt(53x)=53⋅dxdt
Substitutes 5 for dxdt in ddt(x+y)
ddt(x+y)=53(5)=253 ft/s
Therefore, the required rate at which the tip of his shadow moves is ddt(x+y)=253 ft/s.