Algebra.Com's Answer #814126 by ikleyn(52786)![\"\"](\"/images/stars/stars-13.gif\")  
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document.write( "A rocket is being launched vertically over a point 𝐴 on the ground with
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document.write( "a velocity of 550 𝑚𝑖Τℎ𝑟. Twenty five miles away from point 𝐴 on the
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document.write( "ground, there is a photographer video-taping the launch. At what rate
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document.write( "is the angle of elevation of the camera changing when the rocket
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document.write( "achieves an altitude of 25 miles?
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document.write( " The solution by Edwin is good conceptually, but his implementation has an error, which leads to wrong answer.\r
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document.write( " In this my post, I fix this error and bring a correct solution and answer to you.\r
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document.write( "h = the height of the rocket\r\n" );
document.write( "θ = the angle of elevation of the camera.\r\n" );
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document.write( "Take derivatives of both sides with respect to time t:\r\n" );
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document.write( " the upward speed of the rocket, which is given as 550 miles/hour\r\n" );
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document.write( "Divide both sides by 25\r\n" );
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document.write( "Change  to  \r\n" );
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document.write( "Multiply through by  :\r\n" );
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document.write( "Next we calculate the value of  when\r\n" );
document.write( "the rocket is 25 miles high.\r\n" );
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document.write( "The figure we are \"freezing\" the rocket at looks like this, so that's when\r\n" );
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document.write( "Since  , we substitute:\r\n" );
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document.write( " =  = 11 radians/hour\r\n" );
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document.write( "That's 11 radians per hour.\r\n" );
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document.write( "That's about 10.5 degrees per minute.\r\n" );
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document.write( "@ikleyn\r\n" );
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document.write( "Solved correctly.\r
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