Find the equation of the tangent line to the curve.

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June 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?
June 12th 2009, 08:43 AM
Rachel.F

http://img38.imageshack.us/img38/2411/73889329.th.jpg

is it the answer?
June 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 $y=4\tan x$ . 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?
June 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