# Thread: Derivatives of Inverse Trig. Functions

1. ## Derivatives of Inverse Trig. Functions

Hey, I'm currently in Calc 2 on a section dealing with the Derivatives/Integrals of Inverse Trig. Functions. I ran across this problem, and it's just kind of completely out of the blue, and I'm not 100% sure what to do with it, but I think I have an idea.

q) Where should the point $\displaystyle P$ be chosen on the line segment $\displaystyle AB$ so as to maximize the angle $\displaystyle \theta$

My questions are: Could I assume that the triangles are right angles? I'm not sure I can. And, can anyone give me any hints on how to work the problem? I'd just like some guidance, don't want the answer

Thanks!

2. Originally Posted by Coco87
Hey, I'm currently in Calc 2 on a section dealing with the Derivatives/Integrals of Inverse Trig. Functions. I ran across this problem, and it's just kind of completely out of the blue, and I'm not 100% sure what to do with it, but I think I have an idea.

q) Where should the point $\displaystyle P$ be chosen on the line segment $\displaystyle AB$ so as to maximize the angle $\displaystyle \theta$

My questions are: Could I assume that the triangles are right angles? I'm not sure I can. Mr F says: Yes you can.

And, can anyone give me any hints on how to work the problem? I'd just like some guidance, don't want the answer

Thanks!
It is sufficient to maximise $\displaystyle \tan \theta$ (why?).

$\displaystyle \theta = \pi - (\alpha + \beta) \Rightarrow \tan \theta = - \tan (\alpha + \beta)$ (keep reading and it will be clear where the angles $\displaystyle \alpha$ and $\displaystyle \beta$ are).

Let $\displaystyle BP = x \Rightarrow PA = 3 - x$.

Then $\displaystyle \tan \alpha = \frac{2}{x}$ and $\displaystyle \tan \beta = \frac{5}{3-x}$.

3. Well, I did set it up the way you have, but I couldn't figure out the $\displaystyle \tan\theta=-\tan(\alpha+\beta)$ part, so I just set:

$\displaystyle \tan\alpha = \frac{2}{x} \Rightarrow \alpha = \arctan\frac{2}{x}$

$\displaystyle \tan\beta = \frac{5}{3-x} \Rightarrow \beta = \arctan\frac{5}{3-x}$

$\displaystyle \theta = \pi-\arctan\frac{2}{x}-\arctan\frac{5}{3-x}$

then:

$\displaystyle \frac{d\theta}{dx}=\frac{2}{x^2+4}-\frac{5}{x^2-6x+34}$

solve $\displaystyle \frac{d\theta}{dx}$ for $\displaystyle 0$

$\displaystyle x=-2+2\sqrt{5}$

$\displaystyle |AP|=3-x\Rightarrow3-x=-2+2\sqrt{5}$

$\displaystyle |AP|=x\Rightarrow x=5-2\sqrt{5}$

So $\displaystyle \theta$ is maxed when P is $\displaystyle \approx$0.5279 from $\displaystyle A$

That's the answer in the back of the book

Thanks for the help!

4. Sorry, I also forgot to ask, why would you use tangent? I know it looks to be more convenient, but just curious if that is really why.

5. Originally Posted by Coco87
Sorry, I also forgot to ask, why would you use tangent? I know it looks to be more convenient, but just curious if that is really why.
Since the angle you want lies between 0 and pi/2 degrees and tan is strictly increasing over that interval, the value of x that maximises $\displaystyle \theta$ will also maximise $\displaystyle \tan \theta$.

And well done getting the right answer.

6. Originally Posted by mr fantastic
Since the angle you want lies between 0 and pi/2 degrees and tan is strictly increasing over that interval, the value of x that maximises $\displaystyle \theta$ will also maximise $\displaystyle \tan \theta$.

And well done getting the right answer.
Ah, I see. Thanks again!