Derivatives of Trigonometric Functions

[LadiesMan]

1.A kite 40 m above the ground moves horizontally at the rate of 3 m/s. At what rate is the angle between the string and the horizontal decreasing when 80 m of string has been let out. Answer is 0.02 m/s



2. What I did was:

-Drew a triangle as prescribed above
-I found the unknown length of the ground at that instaneous time.
-I found the derivative of the vertical height.
-Then I took derivative of sin theta which gave:
cos theta (d theta/t) = ry'-yr'(becomes 0 since the string is 80 m all the time) all divided by r^2
Then I divide by cos theta to get the derivative the angle.

Thanks very much

 

[HallsofIvy]

I'm not sure why you have included y and y'. You are told "A kite 40 m above the ground moves horizontally" so y= 40 and y'= 0. (You didn't actually say that y represented the height above the ground (which you should have) but I assume that from the fact that you say $sin(\theta)= y/r)$ and I assume (because, again, you didnt say that) that r is the hypotenuse, the length of the kites string. Since you are told "moves horizontally at the rate of 3 m/s", it might be better to use $tan(\theta)= x/40$, where x is the horizontal distance from directly over head. Then $sec^2(\theta)(d\theta/dt)= x'/40= 3/40$. Use the fact that, at the moment in question, r= 80 and y= 40 to determine both x and $\theta$ at that moment.

 

[dynamicsolo]

At what rate is the angle between the string and the horizontal decreasing when 80 m of string has been let out. Answer is 0.02 m/s

Incidentally, if the question is asking about the rate at which the angle is changing, the answer would have to be in radians/sec (or perhaps degrees/sec), rather than a linear velocity such as meters/sec.

[EDIT: Just finished working this out. The given answer is rounded-off to one significant figure, but would give 0.02 rad/sec. The "mathematically exact" answer is 3/160 rad/sec. Now you get to show why...]

 

[LadiesMan]

thanks :) that helped!