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avatar+956 

Okay, so what I did was first plugged in π/3 into y=x-tan x and got: (π/3, -0.68) 

Then I found the derivative of y and got y' = 1-sec^2x 

I plugged in π/3 in y' and got -3 

 

So using my slope -3 and my point (π/3, -0.68)  I did the following: 

-0.68 = -3(π/3)+b 

b = 2.5 

So my equation is: y = -3x+2.5 

But as you can see it doesn't go through π/3 

Am I doing something wrong? 

 

Thanks for all your help!!!

 Apr 16, 2018
 #1
avatar+101761 
+2

y  = x - tan x

y' = 1 - sec^2x

At  pi/3   the slope is  1 - sec^2 (pi/3)   =  1  - (2)^2  =   1 - 4  = -3

 

And...at pi/3, y  =   pi/3  - tan (pi/3)    =  pi/3  - √3

 

So....the equation  of the tangent line at    (pi/3, pi/3 - √3 )  is

 

y  = -3 (x - pi/3) + pi/3 - √3      simplify

 

y = -3x + pi + pi/3  - √3

 

y = -3x + (4/3)pi - √3

 

Here's the graph, Julius :   https://www.desmos.com/calculator/aheivaxqtb

 

 

cool cool cool

 Apr 16, 2018
 #2
avatar+18433 
+1

Looks like you got it.....it is tangent at pi/3 , -0.68

 Apr 16, 2018

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