# Thread: tangent to the curve with a given slope help

1. ## tangent to the curve with a given slope help

The question is:

Find x, such that the slope of the tangent to the curve $y=tanx$ is equal to 4. (over $[0,2\pi)$)

I've got:

$\frac{d}{dx}tanx=sec^2x$

I think I should set that equal to 4, but I'm not sure on how I'm supposed solve it. Any help would be appreciated.

2. ## Re: tangent to the curve with a given slope help

I presume you know that $sec(x)= \frac{1}{cos(x)}$. So that if $sec^2(x)= 4$, $sec(x)= \frac{1}{cos(x)}= \pm 2$. That is, $cos(x)= \pm\frac{1}{2}$. Can you solve that?

3. ## Re: tangent to the curve with a given slope help

Originally Posted by amthomasjr
The question is:

Find x, such that the slope of the tangent to the curve $y=tanx$ is equal to 4. (over $[0,2\pi)$)

I've got:

$\frac{d}{dx}tanx=sec^2x$

I think I should set that equal to 4, but I'm not sure on how I'm supposed solve it. Any help would be appreciated.
Good so far. So you know that $sec^2x = 4$ . Can you solve that for x?
(Hint: sec(x) = 1/cos(x).)

-Dan

4. ## Re: tangent to the curve with a given slope help

$x=\arccos(\frac{1}{2})$ correct?

Then I just find the other values that fit on the interval right?

5. ## Re: tangent to the curve with a given slope help

I wouldn't bother with using inverse trig functions:

Use:

$\cos\left(\frac{\pi}{3} \right)=\frac{1}{2}$

$\cos\left(\frac{2\pi}{3} \right)=-\frac{1}{2}$

Now use the identity $\cos(2\pi-x)=\cos(x)$ to find the other 2 roots, i.e., the quadrant III and IV roots.