# Math Help - Find the equation of the tangent line to the curve.

1. ## Find the equation of the tangent line to the curve.

Hi guys,

Could someone take a look at my work.. im lost.

Problem:
Code:
Find the equation of the tangent line to the curve
y= 4 tan x  at the point ( pi/4 , 4).

The equation of this tangent line can be written in the form y = mx+b
my work..

Code:
using y-y1 = m(x-x1)

y - 4tan(x) = 4tan(x- (pi/4))
y - 4tan(x) = 4tan(x) - (pi tan (x))
y = 8tan(x) - pi tan(x)

m = 8tan(x)
b = pi tan (x)
Where did I go wrong? How do I get the equation from the point and the tangent line?

3. Originally Posted by mant1s
Hi guys,

Could someone take a look at my work.. im lost.

Problem:
Code:
Find the equation of the tangent line to the curve
y= 4 tan x  at the point ( pi/4 , 4).

The equation of this tangent line can be written in the form y = mx+b
my work..

Code:
using y-y1 = m(x-x1)

y - 4tan(x) = 4tan(x- (pi/4))
y - 4tan(x) = 4tan(x) - (pi tan (x))
y = 8tan(x) - pi tan(x)

m = 8tan(x)
b = pi tan (x)
Where did I go wrong? How do I get the equation from the point and the tangent line?
The function is $y=4\tan x$. To find the slope for any x value, differentiate it. We now see that $y^{\prime}=4\sec^2x$. Since we're looking for the eqn of a tangent at $\left(\tfrac{\pi}{4},4\right)$, we want to find the slope of the function at $\frac{\pi}{4}$. So it follows that $y^{\prime}\!\left(\tfrac{\pi}{4}\right)=4\sec^2\le ft(\tfrac{\pi}{4}\right)=4\left(\sqrt{2}\right)^2= 4\cdot2=8$. This is your slope in your tangent line equation.

So using the point-slope equation, we have $y-4=8\left(x-\tfrac{\pi}{4}\right)\implies y=8x-2\pi+4$

Does this make sense?

4. Rachel and Chris,

You guys rock! It makes sense to me now. I wasn't differentiating and it was screwing me up royally. Thanks for your time!

-M