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

• Jun 12th 2009, 08:21 AM
mant1s
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?
• Jun 12th 2009, 08:43 AM
Rachel.F
• Jun 12th 2009, 08:45 AM
Chris L T521
Quote:

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 $\displaystyle y=4\tan x$. To find the slope for any x value, differentiate it. We now see that $\displaystyle y^{\prime}=4\sec^2x$. Since we're looking for the eqn of a tangent at $\displaystyle \left(\tfrac{\pi}{4},4\right)$, we want to find the slope of the function at $\displaystyle \frac{\pi}{4}$. So it follows that $\displaystyle y^{\prime}\!\left(\tfrac{\pi}{4}\right)=4\sec^2\le ft(\tfrac{\pi}{4}\right)=4\left(\sqrt{2}\right)^2= 4\cdot2=8$. This is your slope in your tangent line equation.

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

Does this make sense?
• Jun 12th 2009, 08:57 AM
mant1s
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