Calculus I Related Rates Problem! (Differential Equations)

Question

A police car is located 70 feet to the side of a straight road.

A red car is driving along the road in the direction of the police car and is 110 feet up the road from the location of the police car. The police radar reads that the distance between the police car and the red car is decreasing at a rate of 70 feet per second. How fast is the red car actually traveling along the road?

The actual speed (along the road) of the red car is ____ feet per second

Answer 1

construct a right triangle, longer leg=y= 110 feet, smaller leg = x=70 feet

hypotenuse squared = y^2+ x^2 = 110^2+x^2= 110^2+70^2= h^2 = 12,100+4,900= 17,000

h= sqr17000

h= 10sqr170

h' = 70 ft/sec

x'=0

h^2 = y^2 + x^2

take the derivative with respect to time and solve for y' = speed of the car

hh'= yy' + xx', h'=70 ft/sec

10sqr170(70) = 110y' +70(0)

110y' = 700sqr170

11y' = 70sqr170

y' = (70sqr170)/11 = about 82.97 feet per second

Answer 2

Let s be the distance between the red car and the police car, which is the hypotenuse of a right triangle.

Let y be the distance along the road to the position directly across from the police car that is 70 ft off the road.

Given: ds/dt = 70 ft/s

s²=y²+70²

Taking the derivative of both sides of the equation with respect to time, t.

2s ds/dt = 2y dy/dt

dy/dt = s (ds/dt)/y

When y = 110, s= √(110²+70²) = 10√170

dy/dt = (10√170)(70)/110 ≈ 82.97 ft/sec