Calc homewrok...please help!?

Question

2. 2 cars start at the same place and at the same time. One car travels west at a constant velocity of 50 mi/ hr and the 2nd car goes S at 60/ mph. How fast is distance between the 2 changing 1/2 hr later?

Answer 1

Because the angle between their direction of travel is 90 degrees, you can create an expression for the distance between them using the Pythagorean Theorem, a^2 + b^2 = c^2.

a = 50 , b = 60, t is the variable, c is the distance between them.

(50t)^2 + (60t)^2 = c^2
Solve for c.
[(50t)^2 + (60t)^2]^(1/2) = c

Simplify:

(6100t^2)^(1/2) = 78.1t = c; with t expressed in hours, and c expressed in miles.
dc/dt = 78.1 is the rate of change in the distance.

In other words the distance is growing at the rate of 78.1t
(You can see that the shape of the triangle formed by the two cars paths and the line (distance) connecting them never chages. The distance is growing at a constant rate, and that rate is 78.1. It doesn't matter whether it is one minute, 30 minutes, or two days later... the rate of change is the same.

Hence, at .5 hours, the distance c is 39.05 miles, and the rate of change is:

dc/dt = 78.1 miles per hour

Answer 2

Given
dw/dt = 50 mph
ds/dt = 60 mph

Let
D = distance between the two cars
t = time in hours

Find dD/dt.

We have:
w = 50t
s = 60t

D² = w² + s² = (50t)² + (60t)² = 2500t² + 3600t² = 6100t²
D = √(6100t²) = 10t√61
dD/dt = 10√61 mph ≈ 78.102497 mph

This rate of change is constant for all t.

Answer 3

If x(t) is the position of car 1 and y(t) is the position of car 2, then the distance between them is

s(t) = √[x(t)^2 + y(t)^2]

x(t) = v1*t and y(t) = v2*t; putting those into the above,

s(t) = √[v1^2*t^2 + v2^2*t*2] = t*√[v1^2 + v2^2].

The rate of change of distance is d s(t) /dt = √[v1^2 + v2^2] and is constant with time.