Derivatives of Trigonometric Functions

In summary, when 80 m of string has been let out, the angle between the string and the horizontal decreases by 0.02 rad/sec.
  • #1
LadiesMan
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0
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 derivitive the angle.


Thanks very much
 
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  • #2
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 [itex]sin(\theta)= y/r)[/itex] 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 [itex]tan(\theta)= x/40[/itex], where x is the horizontal distance from directly over head. Then [itex]sec^2(\theta)(d\theta/dt)= x'/40= 3/40[/itex]. Use the fact that, at the moment in question, r= 80 and y= 40 to determine both x and [itex]\theta[/itex] at that moment.
 
  • #3
LadiesMan said:
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...]
 
Last edited:
  • #4
thanks :) that helped!
 

What are the derivatives of sine and cosine?

The derivative of sine is cosine, and the derivative of cosine is negative sine.

How do you find the derivative of tangent?

To find the derivative of tangent, use the quotient rule and the fact that tangent is equal to sine over cosine. The derivative of tangent is equal to the derivative of sine multiplied by cosine, minus the derivative of cosine multiplied by sine, all divided by cosine squared.

What is the derivative of the inverse sine function?

The derivative of the inverse sine function, also known as arcsine, is equal to 1 over the square root of 1 minus x squared.

Can you explain the chain rule in relation to derivatives of trigonometric functions?

When using the chain rule to find the derivative of a composite function involving a trigonometric function, you must first take the derivative of the outer function, and then multiply it by the derivative of the inner function. For example, when finding the derivative of sine squared, you would first take the derivative of sine, which is cosine, and then multiply it by the derivative of the inner function, 2x, resulting in a final answer of 2xcosx.

How are derivatives of trigonometric functions used in real-life applications?

Derivatives of trigonometric functions are commonly used in physics and engineering to model and analyze periodic motion, such as the motion of a pendulum or a spring. They are also used in navigation and surveying to calculate distances and angles, and in signal processing to analyze and manipulate waveforms.

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