How do you build a function for this problem?

Question

Two cars are approaching an intersection. One is 1 mile south of the intersection and is moving at a constant speed of 40 mph. At the same time, the other car is 2 miles east of the intersection and is moving at a constant speed of 10 mph.

Express the distance (D) between the cars as a function of time (t).

For what value of T is D smaller?

Answer 1

What you have is a right triangle that starts with sides of 2 and 1

 

The intial distance between the cars is SQRT(2² + 1²) = SQRT(5)

 

But the cars are moving so the sides of our right triangle are changing

 

One side equals (2 - 10t)  where t is in hours

 

the other side equals (1 - 40t)  t in hours

 

the distance D = SQRT((2-10t)² + (1-40t)²)

 

expand the terms and simplfiy

 

D = SQRT( 4 - 40t + 100t² + 1 - 80t + 1600t²)

 

D = SQRT(1700t² - 120t +5)  This is your distance function

 

now we want to find the point where the distance is at a minimum

 

we can use derivatives and set the deravitive to 0 to find the minimum

 

D = (1700t² - 120t + 5)^1/2

 

The derivative of this is

 

D' = (3400t - 120)/2(1700t² - 120t + 5)^1/2

 

This will be zero when th numberator is zero

 

So we set 3400t - 120 = 0

 

the minimum will be at t =120/3400  or t = .035294

 

you can try values of t larger and smaller than this number to see that in fact the cars are closest at that time.

 

Hope this helps