How do you find the equation of the tangent and normal line to the curve y=1/x^3y=1x3 at (2,1/8)(2,18)?

1 Answer
Douglas K.
Apr 28, 2017

The slope of the tangent line is the first derivative evaluated at x = 2.
The slope of the normal line will be perpendicular to the tangent.
Use the point-slope form to find the equation.

Explanation:

To find the slope of the tangent line, m_tmt, we compute dy/dxdydx and then evaluate it at x = 2x=2

dy/dx = -3/x^4dydx=−3x4

m_t= -3/2^4= -3/16mt=−324=−316

The slope of the normal line, m_nmn, can be found using the following equation:

m_n= -1/m_tmn=−1mt

m_n= -1/(-3/16)mn=−1−316

m_n = 16/3mn=163

Using the point-slope form of the equation of a line, we find that the equation of any line passing through the point (2,1/8)(2,18) is:

y = m(x-2)+1/8" [1]"y=m(x−2)+18 [1]

Substitute the value of m_tmt into equation [1], to obtain the equation of the tangent line:

y = -3/16(x-2)+1/8y=−316(x−2)+18

Substitute the value of m_nmn into equation [1], to obtain the equation of the normal line:

y = 16/3(x-2)+1/8y=163(x−2)+18

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