calculus related rates problem

Question

Two people start from the same point. One walks east at 5 mi/h and the other walks northeast at 8 mi/h. How fast is the distance between the people changing after 15 minutes? (Round your answer to three decimal places.)

Answer 1

hi,

First make a coordinate plane then on x-axis choose a and north as b then connect these points

Now, you have a right triangle!

on x- axis speed is 5m/h and y-axis speed is 8m/h

you can write an equation such a^2 + b^2 = c^2

after 15 minutes =====> on x- axis you are on: 5m/h * h/60 * 15 = 1.25 BE CAREFUL a = 1.25

after 15 minutes ======> on y-axis you are on : 8m/h*h/60*15 = 2 BE CAREFUL b= 2

on the other hand base of Pythagorean a^2 + b^2 = c^2 ====> (1.25)^2 + (2)^2 = c^2 ====> c = 2.358

now, take a derivative of a^2 + b^2 = c^2 =====> 2aa' + 2bb' = 2cc'

now, you know a = 1.25 a' = 5m/h b= 2 b'= 8m/h and c =2.358 only you need to find c'

replace all information in equation 2aa' + 2bb' = 2cc' ====> 2(1.250)(5) + 2(2)(8) = 2 (2.358)c'====>

c' =9.436 m/h is speed between two people after 15 minutes.

I hope it is useful, let me know if you have any question.

Minoo

Answer 2

Let x = distance travelled by the eastbound walker in t hours

y = distance travelled by the northeastbound walker in t hours

z = distance between the walkers after t hours

When t = 15 min = 1/4 hr, x = (5 mi/hr)(1/4 hr) = 1.25 mi and y = (8 mi/hr)(1/4 hr) = 2 mi

By the Law of Cosines, z² = x² + y² - 2xycos45°

When x = 1.25 and y = 2, z² = (1.25)² + 2² - 2(1.25)(2)(√2/2) = 2.026966. So z = 1.4237.

2z(dz/dt) = 2x(dx/dt) + 2y(dy/dt) - √2[x(dy/dt) + y(dx/dt)]

2.8474(dz/dt) = 12.5 + 32 - √2[10 + 10}

dz/dt = 5.695 mi/hr