Friday

October 24, 2014

October 24, 2014

Posted by **Bonaventutre** on Wednesday, March 27, 2013 at 1:19am.

- calculus -
**Reiny**, Wednesday, March 27, 2013 at 7:52amdy/dx = 2sec^2 (2x)

when x = π/8

f(π/8) = tan π/4 = 1

so our point of contact is (π/8, 1)

slope of tangent = 2 sec^2 (π/4)

= 2(√2/1)^2 = 4

equation of tangent:

y - 1 = 4(x - π/8)

y - 1 = 4x - π/2

y = 4x + 1-π/2

normal has slope -1/4

y = (-1/4)x + b

but the point (π/8,1) lies on it, so

1 = (-1/4)(π/8) + b

b = 1 + π/32

y = (-1/4)x + 1+π/32

**Answer this Question**

**Related Questions**

Math - Write an equation of the tangent function with period (3π)/8, phase ...

Calculus - the curve: (x)(y^2)-(x^3)(y)=6 (dy/dx)=(3(x^2)y-(y^2))/(2xy-(x^3)) a...

calculus - Consider the curve given by the equation y^3+3x^2y+13=0 a.find dy/dx...

Ap Calc AB - Find the equation of the line normal to the curve at (0,0). (Normal...

Calculus - If the equation of the tangent line to the curve y=9cosx at the point...

Calculus - Consider line segments which are tangent to a point on the right ...

12th grade calculus - find the lines that are (a) tangent and (b) normal to the ...

calculus - Find an equation of the tangent line to the curve at the given point...

calculus - Find an equation of the tangent line to the curve at the given point...

Calculus - Please help this is due tomorrow and I dont know how to Ive missed a ...