Yahoo Answers: Answers and Comments for Two cars start moving from the same point. One travels south at 64 mi/h and the other travels west at 48 mi/h.? [Mathematics]
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From LuckywithLeLe
enUS
Mon, 09 Nov 2009 22:38:34 +0000
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Yahoo Answers: Answers and Comments for Two cars start moving from the same point. One travels south at 64 mi/h and the other travels west at 48 mi/h.? [Mathematics]
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From ladaghini: Let x be the distance the car heading west tra...
https://answers.yahoo.com/question/index?qid=20091109223834AAfM8RD
https://answers.yahoo.com/question/index?qid=20091109223834AAfM8RD
Tue, 10 Nov 2009 07:26:03 +0000
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