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

Question

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

Ezekiel Alexander · Accepted Answer

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.

To find the slope of the tangent line, #m_t#, we compute #dy/dx# and then evaluate it at #x = 2# #dy/dx = -3/x^4# #m_t= -3/2^4= -3/16# The slope of the normal line, #m_n#, can be found using the following equation: #m_n= -1/m_t# #m_n= -1/(-3/16)# #m_n = 16/3# 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)# is: #y = m(x-2)+1/8" [1]"# Substitute the value of #m_t# into equation [1], to obtain the equation of the tangent line: #y = -3/16(x-2)+1/8# Substitute the value of #m_n# into equation [1], to obtain the equation of the normal line: #y = 16/3(x-2)+1/8#

Ezekiel Alexander · Answer

undefined To find the equation of the tangent line to the curve $ y = \frac{1}{x^3} $ at the point $ (2, \frac{1}{8}) $, first, find the derivative of the function $ y = \frac{1}{x^3} $ using the power rule. Then evaluate the derivative at $ x = 2 $ to find the slope of the tangent line. Next, use the point-slope form to write the equation of the tangent line. Similarly, to find the equation of the normal line, find the negative reciprocal of the slope of the tangent line and use it to write the equation of the normal line using the point-slope form.

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