Do I Need to Convert Theta to Radians for Taking the Derivative?

[jonesyxd]

Hi all,

The following is part of a part, of an assignment for my MATLAB course for uni.

For most of the ODE solvers in MATLAB you also have to be able to solve the question numerically - and this is the component of the question I'm doing here..


The equilibrium equation of an arch type structure is the function below..

P = 4 x K x L[cos(alpha -theta ) - cos(alpha)] x tan(alpha -theta );
GIVEN L=1m, alpha=30degrees, K=100kN/m

Substituting in unknowns we get..

P = 400[cos(30 - theta) - cos(30)] x tan(30 - theta)

Now I have a function of P(force) and theta(angle in degs), for the question though I have to solve for the roots using 'Newtons Method'(need the derivative of the function) and this is where my questions comes in..

To find the derivative of this function with respect to theta, do I first need to convert theta to radians? ie, alpha = Pi / 6, Theta = Pi / 180(theta) ?? ..should I be taking the derivative of the above function? or the one below

P = 400[cos(Pi/6 - Pi/180(theta)) - cos(Pi/6)] x tan(Pi/6 - Pi/180(theta))

Thanks for reading!

 

[szynkasz]

Formally you should convert it to radians.

 

[jonesyxd]

What do you mean 'formally'?

If they are not in radians when I get to the chain rule, this will happen

derivative of cos(30 - theta) = -sin(30 - theta) * (-1) = sin(30 - theta)

but if I convert to rads..

derivative of cos(∏/6 - ∏/180(theta)) = -sin(∏/6 - ∏/180(theta)) * (-∏/180) = ∏/180 sin(∏/6 - ∏/180(theta))

Or am I way off here?

 

[szynkasz]

I mean that arguments of trigonometric function are in radians. But the only difference is that derivative is mulitplied by ∏/180.

 

[HallsofIvy]

What do you mean 'formally'?

If they are not in radians when I get to the chain rule, this will happen

derivative of cos(30 - theta) = -sin(30 - theta) * (-1) = sin(30 - theta)

but if I convert to rads..

derivative of cos(∏/6 - ∏/180(theta)) = -sin(∏/6 - ∏/180(theta)) * (-∏/180) = ∏/180 sin(∏/6 - ∏/180(theta))

Or am I way off here?

Way off! If you look at the derivation of the derivatives of the trig functions in any Calculus book, you will use the fact that $\lim_{x\to 0} sin(x)/x= 1$. That is only true if x is measured in radians.

To differentiate sin(x) where x is in degrees, you would have to write $sin(x)= sin(180u/\pi)$ and use the chain rule: $(sin(x))'= (180/\pi)cos(180u/\pi)= (180/\pi)cos(x)$.

 

[jonesyxd]

To differentiate sin(x) where x is in degrees, you would have to write $sin(x)= sin(180u/\pi)$ and use the chain rule: $(sin(x))'= (180/\pi)cos(180u/\pi)= (180/\pi)cos(x)$.

I really don't consider myself good at calculus at all, but isn't that what i did?


to find derivative of cos(30 - x) when x and '30' are both in degrees..

= cos(∏/6 - ∏/180(x)) now to find derivative..

=-sin(∏/6 - ∏/180(x)) * (-∏/180)

=∏/180*sin(∏/6 - ∏/180(x))

?

 

[HallsofIvy]

That may be what you intended but it was not what you said. You said:

What do you mean 'formally'?

If they are not in radians when I get to the chain rule, this will happen

derivative of cos(30 - theta) = -sin(30 - theta) * (-1) = sin(30 - theta)

and this is completely wrong.

but if I convert to rads..

derivative of cos(∏/6 - ∏/180(theta)) = -sin(∏/6 - ∏/180(theta)) * (-∏/180) = ∏/180 sin(∏/6 - ∏/180(theta))

Or am I way off here?

You appeared to be thinking that sine and cosine, in different units, were different functions and so would have different derivatives. That is not at all true.

Indeed, in many Pre-Calculus and Calculus texts, the sine and cosine are defined so that the variables are measured around the circumference of a unit circle rather than as angles. The advantage of that is that the variables do NOT have any units at all, they are just numbers- like all other functions. But engineers, who design calculators still think in terms of angle so we have to talk about "radians" which, on a unit circle, are the same as the distance on the circumference.