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

Question

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

levurdondishav4 · Accepted Answer

Step 1    Step 2  At the moment, dxdt=8fts and we want to find dθdt. We can find the current angle first  θ=arcsin⁡100200=π6  Step 3  Get an equation to relate the x and θ. Differentiate with respect to time t  x100=cot⁡θ  001dxdt=−csc2θdθdt  dθdt=0.01dxdt−csc2θ  =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 (≈1.15∘s)

Mary Nicholson · Accepted Answer

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  Step 2 Let the angle between the string and the horizontal be θ Then cos⁡tjη=xs Differentiate to get dθdt×(−sin⁡θ)=sdxdt−xdsdts2 dθdt×−hs=8s−8x2ss2 [Use eqn(1)] dθdt×−hs=8s2−8x2s3 Note that s2−x2=h2 dθdt×−hs=8h2s3 dθdt−8(100)2002=−800200×1200 dθdt=−8(100)2002=−150

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

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