How do you find the equation of the tangent and normal line to the curve #y=1+x^(2/3)# at (0,1)?

1 Answer
Douglas K.
Apr 13, 2017

Any line through the point #(0,1)# will have the form:
#y = mx+1#
Substitute, the slope of the tangent line, #m_t = y'(0)# or the slope of the normal line, #m_n = -1/(y'(0))#

Explanation:

Usually we would use the form given in the above answer:

#y = mx+1#

Compute #y'(x)#:

#y'(x)= 2/3x^(-1/3)#

Evaluate at the x coordinate:

#m_t=y'(0) = 2/3(0)^(-1/3)#

But we have an exception to the procedure. Please observe that that the evaluation causes a division by 0, which implies that the tangent is the vertical passing the point #(0,1)#

Therefore, the tangent is the line, #x = 0#

The procedure for the normal line is similar.

#y = mx+1#

Compute #y'(x)#:

#y'(x)= 2/3x^(-1/3)#

Evaluate at the x coordinate:

#m_n=-1/(y'(0)) = -1/((2/3)0^(-1/3)) = 0#

Substitute 0 for the slope:

#y = 1#

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