Question

Related Rates question! please help!

Answer 1

\[x^2 + y^2 =r^2\]
let x = distance in x direction, y = distance in y direction
and r = true distance [hypotenuse formed]

take derivative of both sides [you'll use chain rule]:
\[\frac{ d }{ dt }(x^2 + y^2) = \frac{ d }{ dt }(r^2)\]\[2x \frac{ dx }{ dt } + 2y \frac{ dy }{ dt } = 2r \frac{ dr }{ dt }\]
plug in what you have. you're looking for dr/dt, and realize that dy/dt = 0 since the observer stays still and the biker is just going eastward:

Answer 2

2(30)(10) + 2(16)(0) = 2(r found using pythagorean)(dr/dt)

solve for dr/dt

Answer 3

answer is:

3: 8.82

Answer 4

@Euler271 that's what I got! thank you!