Answer:
[tex]d'=\sqrt{v^2+w^2}[/tex]
Explanation:
Rate of Change
When an object moves at constant speed v, the distance traveled at time t is
[tex]x=v.t[/tex]
We know at time t=0 two friends are at the intersection of two perpendicular roads. One of them goes north at speed v and the other goes west at constant speed w (assumed). Since both directions are perpendicular, the distances make a right triangle. The vertical distance is
[tex]y=v.t[/tex]
and the horizontal distance is
[tex]x=w.t[/tex]
The distance between both friends is computed as the hypotenuse of the triangle
[tex]d^2=x^2+y^2[/tex]
We need to find d', the rate of change of the distance between both friends.
Plugging in the above relations
[tex]d^2=(v.t)^2+(w.t)^2[/tex]
[tex]d^2=v^2.t^2+w^2.t^2=(v^2+w^2)t^2[/tex]
Solving for d
[tex]d=\sqrt{(v^2+w^2)t^2}[/tex]
[tex]d=\sqrt{(v^2+w^2)}.t[/tex]
Differentiating with respect to t
[tex]\boxed{d'=\sqrt{v^2+w^2}}[/tex]
