A kite is stuck in a tree 40 ft above the ground I am reeling in the string at a rate of 5 ft/min

Question

a) how fast is my distance from the tree changing when i am 30 ft from the tree?

b) how fast is the angle of elevation changing at the same time?

Answer 1

A)If you have a distance of 30 feet away from tree and the height of tree, make a right triangle and set those as your x and y values. Or

40^2+30^2=c^2

Solve for C

C=50 feet

After that set up a differential equation where you use the Pythagorean theorem equation and implicitly differentiate the equation. You should get this.

a^2 + b^2 = c^2

2aa'+2bb' = 2cc'

The tree rate is 0 I gave that a, so 2aa' goes away completely. b' is what we are looking for, and c' is the rate at which you are reeling in the kite.

You should get this.

2(30)b'=2(50)(5)

Solve for b'

b'=25/3 ft/min

B) use Tanθ=opp/adj and differentiate. Solve for θ'

Tanθ=40/30, θ=53.1

Now take derivative

Tanθ=40/b

b=40/tanθ

b'=-40(tan^2θ/sec^2θ)*dθ/dt

Solve for dθ/dt