A kite 100 ft above the ground moves horizontally at a speed of 8 \frac{ft}{s

widdonod1t

widdonod1t

Answered question

2021-12-09

A kite 100 ft above the ground moves horizontally at a speed of 8fts. When 200 feet of string have been released, how quickly does the angle between the string and the horizontal start to decrease?

Answer & Explanation

levurdondishav4

levurdondishav4

Beginner2021-12-10Added 38 answers

Step 1

Step 2
At the moment, dxdt=8fts and we want to find dθdt. We can find the current angle first
θ=arcsin100200=π6
Step 3
Get an equation to relate the x and θ. Differentiate with respect to time t
x100=cotθ
001dxdt=csc2θdθdt
dθdt=0.01dxdtcsc2θ
=0.01(8)csc2π6 plug in known values
=0.08(2)2
=0.02rads
The negative means it is decreasing at 0.02 radians per second (1.15s)
Mary Nicholson

Mary Nicholson

Beginner2021-12-11Added 38 answers

Step 1
The angle between the horizontal and the kite is θ
Let Height of kit be h
Horizontal distance of kite from where the string is held be x
Then the length of the string be s and will be given by the pythagoras teorem as follows:
s2=h2+x2
Differentiate throughout to get
2sdsdt=2hdhdt+2xdxdt
2sdsdt=2×(0)+2x×8[h is constant and horizontal speed is 8fts]
1) sdsdt=8x
image
Step 2
Let the angle between the string and the horizontal be θ
Then costjη=xs
Differentiate to get
dθdt×(sinθ)=sdxdtxdsdts2
dθdt×hs=8s8x2ss2 [Use eqn(1)]
dθdt×hs=8s28x2s3
Note that s2x2=h2
dθdt×hs=8h2s3
dθdt8(100)2002=800200×1200
dθdt=8(100)2002=150

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