HELP Find The Equation Of The Tangent To The Curve At The Point Where t=π?

Question

This is how I started
The original equations are x = t + sin(t) and y = t - cos(t)
dx/dt = 1 + Cos(t) and dy/dt =1 + sin(t)
dy/dx = (1 + sin(t))/(1 + cos(t))
now because t=π I substitute that in
so dy/dx = (1 + sin(π))/(1 + cos(π)) and it gives me 1/0 or maths error on calculator
To get the x and y values I put t=π into the two original equations 
x=π + sin(π) so x = π and y=π - cos(π) so y = π + 1
however I cant figure out how to find the equation for the tangent when the gradient is 1/0, what do I do from here. 

Thanks In Advance

Geronimo · Accepted Answer

The slope (dy/dx) of the tangent line at { t  =  π } is infinity which means the
 tangent line at that time is a vertical line . . . .
 . . . as you have correctly determined:  dy/dx at { t  =  π } = 1 ⁄ 0 = ∞

 The x_coordinate at { t  =  π } is:  x = t + sin(t)  =  π + sin(π)  =  π

 The equation of the tangent line to the function y{x} at { t  =  π } is:  x = π ◀◀

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HELP Find The Equation Of The Tangent To The Curve At The Point Where t=π?

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