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# Two cars start moving from the same point. One travels south at 64 mi/h and the other travels west at 48 mi/h.?

Two cars start moving from the same point. One travels south at 64 mi/h and the other travels west at 48 mi/h. At what rate is the distance between the cars increasing four hours later?

### 1 Answer

- ladaghiniLv 51 decade agoFavorite Answer
Let x be the distance the car heading west travels in a given time t.

Let y be the distance the car heading east travels in that given time t.

let z be the hypotenuse of the triangle.

Then, by the Pythagorean theorem,

z^2 = x^2 + y^2

Differentiate implicitly, with respect to t (time):

2z dz/dt = 2x dx/dt + 2y dy/dt

For , refer to dz/dt as z', dx/dt as x' and dy/dt as y'.

This gives:

zz' = xx' + yy'

The question asks the rate at which z is changing, which is z', so solve for z':

z' = (xx' + yy')/z

Now x at a certain time is simply the rate times time

x = x' t

and

y = y' t

So that gives:

z' = [ t(x')(x') + t(y')(y') ] / z

z' = t[(x')^2 +(y')^2] / z

z = sqrt(x^2 + y^2), by Pythagoras again

= sqrt[ (x' t)^2 + (y' t)^2 ]

= sqrt( t^2 [ (x')^2 + (y')^2 ] )

= t * sqrt( (x')^2 + (y')^2 )

So

z' = t[(x')^2 +(y')^2] / z

= [(x')^2 +(y')^2] / sqrt( (x')^2 + (y')^2 )

Since time t cancels out, it shows that regardless of how much time has passed, the rate at which z changes will remain the same.

Now it might be apparent that z' is of the form: u/sqrt(u), where

u = (x')^2 + (y')^2.

u/ sqrt(u) = sqrt(u)

so

z' = sqrt( (x')^2 + (y')^2 )

Moving on, now plug in the values of x' and y':

z' = sqrt( (48)^2 + (64)^2 )

z' = sqrt( (16*3)^2 + (16*4)^2 )

z' = sqrt( 16^2 [ 3^2 + 4^2 ] )

z' = 16 sqrt (25) = 16 * 5 = 80

So the rate of change of the distance between the two cars is 80 mph

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