Calculus Question

Question

At noon, ship A is 10 miles due west of ship B. Ship A is sailing west at 19 mph and ship B is sailing north at 25 mph. How fast (in mph) is the distance between the ships changing at 4 PM?

Answer 1

The distance between the ships can be found using the Pythagorean Theorem. d_NORTH = 25t where "t" is the time in hours after noon.. d_WEST = 19t + 10.

s = √(d_NORTH² + d_WEST²)

s = √((25t)² + (19t + 10)²)

s = √(625t² + 361t² + 380t + 100)

s = √(986t² + 380t + 100)

s = (986t² + 380t + 100)^0.5

ds/dt = 0.5(986t² + 380t + 100)^-0.5•(1972t + 380)

When the time is 4 pm, plug in t = 4:

ds/dt = 0.5(986(4)² + 380(4) + 100)^-0.5•(1972(4) + 380) <-- Notice this requires the chain rule.

ds/dt = 0.5(17396)^-0.5•(8268)

ds/dt = 31.34 mph

Answer 2

d^2 = a^2 + b^2

2dd'= 2aa' + 2bb'

dd' = aa' + bb'

divide both sides by d

d'= (aa' +bb')/d where a=66, a'=19,b=100, b'=25, d =sqr(66^2+100^2)

d'= about 31.33 mph = rate of change of the distance between the 2 ships at 4 pm

construct a right triangle

use the Pythagorean theorem

take the derivative with respect to time

plug in values for a,a', b, b' and d, solve for d'

a=4(19)-10= 66

b=4(25)= 100

a'=19

b'=25

d = sqr(66^2+100^2)=about 119.82