radian and degree in derivative trig functions

[bambu]

Hi everyone,

I am in example 4 of Calculus I book.
I dont understand when they suppose θ in radian, why not in degree?
Normally, I will put 45° directly in the function of s'. It will be s'(45°)=-100 feet/degree, but it is a wrong answer.
Could you please show me my misunderstanding?

Thank you.

 

[topsquark]

When you have Trig function in Calculus you have to remember that the formulas for derivatives are derived by assuming that the angles are in radians.

So if [math]s(t) = 50 ~ cot( \theta )[/math], then [math]s'(t) = - 50 ~ csc^2 ( \theta )[/math], where [math]\theta[/math] is in radians.

If [math]\theta[/math] angle is in degrees we have to rewrite it in terms of radians. So [math]s(t) = 50 ~ cot( \theta ) = 50 ~ cot \left ( \dfrac{ \pi }{180} \theta (rad) \right )[/math], where [math]\dfrac{ \pi }{180} \theta [/math] is the angle measure in radians. Then [math]s'(t) = -50 \cdot \dfrac{ \pi }{180} csc^2 \left ( \dfrac{ \pi }{180} \theta (rad) \right )[/math], or putting the angle back into degrees: [math]s'(t) = -50 \cdot \dfrac{ \pi }{180} ~ csc^2( \theta )[/math].

I find it alot easier to just convert degrees right from the start.

-Dan

 

[bambu]

Thank you Dan for your answer.
I didnt know we should assume the angle in radian.
Also, I tried to find reason why is radian on internet, it is still difficult to understand.

 

[Dr.Peterson]

I didnt know we should assume the angle in radian.
Also, I tried to find reason why is radian on internet, it is still difficult to understand.

I assume you mean, why radians are used in calculus, rather than degrees?

Were you shown proofs for the derivatives of trig functions? They depend on the argument being in radians. For example, the proof of the derivative of the sine commonly depends on the fact that [MATH]\lim_{x\to 0}\frac{\sin(x)}{x} = 1[/MATH], which is not true in terms of degrees.

When you use a different angle measure, you have a different function; for example sin(30), when the angle is measured in radians, would not be 1/2 as it is when you use degrees. So you have to define the function in terms of a specific unit, or you haven't defined the function fully.

It turns out that the trig functions in terms of radians have much simpler properties in calculus, so we stick with those definitions.

 

[bambu]

Yes that is I mean.
I understood. Thank you Dr. Peterson!