Math  /  Calculus

QuestionYou say goodbye to your friend at the intersection of two perpendicular roads. At time t=0t=0 you drive off North at a (constant) speed vv and your friend drives West at a (constant) speed ww. You badly want to know. how fast is the distance between you and your friend increasing at time tt ?
Enter here the derivative with respect to tt of the distance between you and your friend: \square Note: the next part will be MUCH easier if you simplify your answer to this part as much as possible.
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? \square Suppose that after your friend takes off (at time t=0t=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 tt (greater than one hour)? \square Again, you ask what is the second derivative of your separation: \square Notice how lingering can make things harder, mathematically speaking.

Studdy Solution
The rate of change of the distance between the friends is v2+w2\sqrt{v^2 + w^2}.
The second derivative is 00.
With the one-hour wait, the rate of change of distance is v2(t1)+w2tv2(t1)2+w2t2\frac{v^2(t-1) + w^2t}{\sqrt{v^2(t-1)^2 + w^2t^2}}, and the second derivative is v2w2(v2(t1)2+w2t2)3/2\frac{v^2w^2}{(v^2(t-1)^2 + w^2t^2)^{3/2}}.

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