Question

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

Answer 1

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`