Calculus ( Find the equation of a tangent line help) ?

Find the equation of the tangent line to the graph of f(x) = csc 2x at the point where

x =π/8

. Exact trigonometric values must be used.

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  • 1 month ago
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    f(x) = csc(2x) ← this is a function, i.e. a curve

    f(x) = 1/sin(2x) → when: x = π/8

    f(π/8) = 1/sin(π/4)

    f(π/8) = 1/[(√2)/2] = 2/(√2) = (2 * √2)/(√2 * √2) = √2 → the curve passes throught A (π/8; √2)

    ...so the tangent line to the curve at x = π/8 too.

    The function f looks like (u/v), so the derivative looks like: [(u'.v) - (v'.u)]/v² → where:

    u = 1 → u' = 0

    v = sin(2x) → v' = 2.cos(2x)

    f'(x) = [(u'.v) - (v'.u)]/v² → where: u' = 0

    f'(x) = - (v'.u)/v² → where: u = 1

    f'(x) = - v'/v²

    f'(x) = - 2.cos(2x)/sin²(2x) ← this is the derivative

    …but the derivative is too the slope of the tangent line to the curve at x

    f'(π/8) = - 2.cos(π/4)/sin²(π/4)

    f'(π/8) = - 2.[(√2)/2]/[(√2)/2]²

    f'(π/8) = - 2/[(√2)/2]

    f'(π/8) = - 4/√2

    f'(π/8) = - (4 * √2)/(√2 * √2)

    f'(π/8) = - (4 * √2)/2

    f'(π/8) = - 2√2 ← this is the slope of the tangent line to the curve at x = π/8

    The typical equation of a line is: y = mx + y₀ → where m: slope and where y₀: y-intercept

    The slope of the tangent line is (- 2√2), so the equation of this tangent line becomes: y = - (2√2).x + y₀

    The tangent line passes through A, so the coordinates of this point must verify the equation of the tangent line.

    y = - (2√2).x + y₀

    y₀ = y + (2√2).x → you substitute x and y by the coordinates of the point A (π/8; √2)

    y₀ = √2 + [(2√2) * (π/8)]

    y₀ = √2 + [(π√2)/4]

    → The equation of the tangent line at x = π/8 is: y = - (2√2).x + √2 + [(π√2)/4)

  • 1 month ago

    Given curve :

    f(x) = csc 2x

    f'(x) = -  2 (csc 2x * cot 2x )

    Hence the tangent line has slope (m)  =  - 2 (csc 2x * cot 2x )

    =>  m  =  - 2 [ csc π/4 * cot π/4 ]  =  - 2 [ √2 * 1 ]  =  - 2√2 = - 2.8284 

    Hence the equation is :

    y = (- 2√2) x + b ............................... (1)

    when x = π/8 ,   y  =  √2

    From (1)    ---

    √2  =   (- 2√2) *  π/8  +  b

    =>  b  =  √2  +  (2√2) * π/8  =  1.41421 +  1.110720  =  2.52493

    Equation of the tangent :

    y  =   - 2.8284 x + 2.52493  ......................  Answer

  • 1 month ago

    f(x) = csc(2x) = 1 / sin(2x)

    f'(x) = -2cos(2x) / (sin(2x))^2 

    f'(π/8) = -2cos(π/4) / (sin(π/4))^2 = -√2 / (1/2) = -2√2

    f(π/8) = √2

    In point-slope form, the line's equation is:

    y - √2 = -2√2(x - π/8)

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