By evaluating the derivate of the distance in t = 10s, we will see that at t = 10s the distance is changing at a rate of 464.2 ft/s
We can think of the situation as in a right triangle, one cathetus is equal to 2000ft, and the other cathetus increases at a rate of 500ft/sec.
The hypotenuse would be the distance between the radar and the rocket.
Then the distance function is:
[tex]D(t) = \sqrt{(2000ft)^2 + (500ft/s*t)^2}[/tex]
Now we need to get the rate of change, this is given by:
[tex]\frac{dD(t)}{dt} = (1/2)*\frac{2*(500ft)^2*t}{\sqrt{(2000ft)^2 + (500ft/s*t)^2} }[/tex]
Now we evaluate this in t = 10s to get:
[tex]\frac{dD(10s)}{dt} = (1/2)*\frac{2*(500ft/s)^2*10s}{\sqrt{(2000ft)^2 + (500ft/s*10s)^2} } = 464.2 ft/s[/tex]
If you want to learn more, you can read:
https://brainly.com/question/18904995
