You say goodbye to your friend at the intersection of two perpendicular roads. A

Question

You say goodbye to your friend at the intersection of two perpendicular roads. At time t=0 you drive off North at a (constant) speed v and your friend drives West at a (constant) speed w. You badly want to know: how fast is the distance between you and your friend increasing at time t ? Enter here the derivative of the distance from your friend with respect to : [ ? ] Being scientifically minded you ask yourself how does the speed of separation change with time. In other words, what is the second derivative of the distance between you and your friend? [ ? ] Suppose that after your friend takes off (at time t=0 ) you linger for an hour to contemplate the spot on which he or she was standing. After that hour you drive off too (to the North). How fast is the distance between you and your friend increasing at time t (greater than one hour)? [ ? ] Again, you ask what is the second derivative of your separation: [ ? ]

Explanation / Answer

1) The distance at time 't' is (v^2 + w^2)t
So, CHANGE IN DISTANCE = (v2 + w2)

the second derivative of the distance between you and your friend is ZERO

2) x = sqrt((vt)^2 + (wt + w)^2)
x = sqrt(v^2t^2 + w^2t^2 + 2w^2t + w^2)
x = sqrt((v^2 + w^2)t^2 + 2w^2t + w^2)
dx/dt = (1/2)((v^2 + w^2)t^2 + 2w^2t + w^2)(-1/2)(2(v^2 + w^2)t + 2w^2)
CHANGE IN DISTANCE = [(v2 + w2)t + w2] / [(v2 + w2)t2 + 2w2t + w2]

Now, again differentiate to get SPEED OF CHANGE IN DISTACE...

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