calculus

Calculate the equation of the tangent and normal lines for the curve f(x) = tan 2x, at the point where the x-coordinate is: x = π/8.

asked by Bonaventutre
  1. dy/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

    posted by Reiny

Respond to this Question

First Name

Your Answer

Similar Questions

  1. Calculus

    For the question "Determine the equation of the tangent to the curve y = xtanx at the point with x-coordinate π." how is the answer -πx + y + π2 = 0?
  2. Calculus

    The line that is normal to the curve x^2=2xy-3y^2=0 at(1,1) intersects the curve at what other point? Please help. Thanks in advance. We have x2=2xy - 3y2 = 0 Are there supposed to be 2 equal signs in this expression or is it x2 +
  3. Math

    Write an equation of the tangent function with period (3π)/8, phase shift -π/5, and vertical shift -2. a. y=tan((8x)/3 - (8π)/15) -2 b. y=tan((8x)/3 + (8π)/15) -2 c. y=tan((16x)/3 +(3π)/80) -2 d.
  4. Calculus

    Lines tangent to the curve y-1=2x^2+3x pass through the point A(2,-1). Find the x-coordinate(s) of the point(s) of tangency. The answer is 2+2^1/2, 2-2^1/2, I just don't know how to get there. Please help! Thank you!
  5. Calculus

    the curve: (x)(y^2)-(x^3)(y)=6 (dy/dx)=(3(x^2)y-(y^2))/(2xy-(x^3)) a) find all points on the curve whose x-coordinate is 1 and write an equation for the tangent line of each of these points b)find the x-coordinate of each point on
  6. calculus

    Consider the curve given by the equation y^3+3x^2y+13=0 a.find dy/dx b. Write an equation for the line tangent to the curve at the point (2,-1) c. Find the minimum y-coordinate of any point on the curve. the work for these would
  7. Ap Calc AB

    Find the equation of the line normal to the curve at (0,0). (Normal lines are perpendicular to the tangent lines) 2y + sinx = xcoxy I found the derivative to be (cosy - cosx)/ (2+ xsiny) then the slope tangent is 0...then the
  8. Calculus

    a)The curve with equation: 2y^3 + y^2 - y^5 = x^4 - 2x^3 + x^2 has been linked to a bouncing wagon. Use a computer algebra system to graph this curve and discover why. b)At how many points does this curve have horizontal tangent
  9. Calculus

    If the equation of the tangent line to the curve y=9cosx at the point on the curve with x-coordinate 3pi/4 is written in the form y=mx+b then m=? and b=?
  10. 12th grade calculus

    find the lines that are (a) tangent and (b) normal to the curve at the given point x^2 + xy - y^2 = 1

More Similar Questions