# Find the equation of the tangent line to the parametric curve defined by y = 2cos(t), y=2sin(t), at the point t= pi/4.Please show me the work so I can understand.

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### 1 Answer

You need to use the equation of the tangent line, such that:

`y - y(pi/4) = (dy)/(dx)|_(t = pi/4)(x - x(pi/4))`

You need to evaluate x and y at `t = pi/4` , such that:

`x(pi/4) = 2cos(pi/4) = 2*(sqrt2/2) => x(pi/4) = sqrt 2`

`y(pi/4) = 2sin(pi/4) = sqrt2 => y(pi/4) = sqrt2`

You need to evaluate `(dy)/(dx)` such that:

`(dy)/(dx) = ((dy)/(dt))/((dx)/(dt))`

`(dy)/(dx) = (2cos t)/(-2sin t) => (dy)/(dx) = -cot t`

You need to evaluate `(dy)/(dx)` at `t = pi/4` such that:

`(dy)/(dx)|_(t = pi/4) = -cot(pi/4) = -1`

You may write the equation of the tangent line at `t = pi/4` , such that:

`y - sqrt2 = (-1)(x - sqrt2) => y = -x + 2sqrt2`

**Hence, evaluating the equation of the tangent line to the parametric curve, at `t = pi/4` , yields `y = -x + 2sqrt2` .**