How do you find the equation for the lines that are tangent and normal to the curve $y = 1 + cos x$ $y=1+cosx$ at the point $(\frac{π}{2}, 1)$ $(pi/2, 1)$ ?

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How do you find the equation for the lines that are tangent and normal to the curve $y = 1 + cos x$ $y=1+cosx$ at the point $(\frac{π}{2}, 1)$ $(pi/2, 1)$ ?

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Answer 1 · 2017-09-01T20:05:48

Start with the derivative

Explanation:

For $y = 1 + cos x$ $y = 1 + cosx$ ,
$\frac{d y}{d x} = - sin x$ $dy/dx = -sinx$ .
At $\frac{π}{2}$ $pi/2$ , we have
$\frac{d y}{d x} = - sin (\frac{π}{2}) = - 1$ $dy/dx = -sin(pi/2) = -1$ .
This is the slope of the tangent line.

Use the point-slope form of a line to find its equation.
$y - y_{1} = m (x - x_{1})$ $y - y_1 = m(x - x_1)$

In this case
$y - 1 = - 1 (x - \frac{π}{2})$ $y - 1 = -1(x - pi/2)$
Solve for y.
$y = - x + \frac{π}{2} + 1$ $y = -x + pi/2 + 1$

The normal is perpendicular to the tangent.
So, if $m$ $m$ is the slope of the tangent line, then $- \frac{1}{m}$ $-1/m$ is the slope of the normal.

In this case, the normal slope is $1$ $1$ .

Use the same point, and the point-slope form of a line, to obtain the equation of the normal line.

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How do you find the equation for the lines that are tangent and normal to the curve $y = 1 + cos x$ $y=1+cosx$ at the point $(\frac{π}{2}, 1)$ $(pi/2, 1)$ ?

1 Answer

Explanation:

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How do you find the equation for the lines that are tangent and normal to the curve $y = 1 + cos x$ $y=1+cosx$ at the point $(\frac{π}{2}, 1)$ $(pi/2, 1)$ ?