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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 t: __________. 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 your friend 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: ____________.

User Nael
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Answer:

Let x be the distance traveled by you to the north and y be the distance traveled by your friend to the west. Let d be the distance between you and your friend at time t. Then, we have:

d^2 = x^2 + y^2

Taking the derivative of both sides with respect to time, we get:

2d (dd/dt) = 2x(dx/dt) + 2y(dy/dt)

Simplifying this expression, we get:

dd/dt = (x(dx/dt) + y(dy/dt)) / d

Since you are driving north at a constant speed v and your friend is driving west at a constant speed w, we have:

dx/dt = v and dy/dt = -w

Substituting these values and simplifying, we get:

dd/dt = (v^2 + w^2)^0.5

Therefore, the speed of separation between you and your friend at time t is the square root of the sum of the squares of your speeds.

To find the second derivative of the distance between you and your friend, we take the derivative of dd/dt with respect to time:

d^2d/dt^2 = (v dv/dt + w dw/dt) / (v^2 + w^2)^0.5

Since v and w are constant, their derivatives with respect to time are zero, so the second derivative of the distance between you and your friend is zero.

Suppose you wait for an hour (i.e., t = 1 hour) before driving off to the north. Then, at time t > 1 hour, the distance between you and your friend is given by:

d = (vt)^2 + ((t-1)w)^2)^0.5

Taking the derivative of both sides with respect to time, we get:

dd/dt = 2v^2t / (2((vt)^2 + ((t-1)w)^2)^0.5)

Simplifying this expression, we get:

dd/dt = v / ((v^2 + ((t-1)w)^2)^0.5)

Therefore, the speed of separation between you and your friend at time t (greater than one hour) is given by v divided by the square root of the sum of the squares of your speed and your friend's speed.

The second derivative of the distance between you and your friend is given by:

d^2d/dt^2 = -vw(t-1) / ((v^2 + ((t-1)w)^2)^1.5)

User Eppsilon
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