1. ## [SOLVED] tangent/normal lines

1. fint the lines that are tangent and normal to the curve $y=x^3$ at (1,1)

im guessing that i have to use the definition of a derivative, and substitute f(x) and x in the equation with one, and solve it?

2. Originally Posted by >_<SHY_GUY>_<
1. fint the lines that are tangent and normal to the curve $y=x^3$ at (1,1)?
The tangent line has slope $y'(1)$; the normal line has slope $\frac{-1}{y'(1)}$.

3. Originally Posted by >_<SHY_GUY>_<
1. fint the lines that are tangent and normal to the curve $y=x^3$ at (1,1)

im guessing that i have to use the definition of a derivative, and substitute f(x) and x in the equation with one, and solve it?

Correct, take the first derivative of $y=x^3$ and solve $f'(1)$

The normal is always perpendicular to the tangent and so the product of their gradients must be -1: $m_1m_2 = -1$ where $m_1 = f'(1)$ = the gradient at that point and $m_2 = \frac{-1}{m_1}$

In both cases use the equation of a straight line formula:

$y-y_1 = m(x-x_1)$ where $(x_1,y_1)$ is a coordinate. In your case this is (1,1)

4. Originally Posted by e^(i*pi)
Correct, take the first derivative of $y=x^3$ and solve $f'(1)$

The normal is always perpendicular to the tangent and so the product of their gradients must be -1: $m_1m_2 = -1$ where $m_1 = f'(1)$ = the gradient at that point and $m_2 = \frac{-1}{m_1}$

In both cases use the equation of a straight line formula:

$y-y_1 = m(x-x_1)$ where $(x_1,y_1)$ is a coordinate. In your case this is (1,1)
ok, by finding the derivative, i got $3x^2$
and substituting 1 gives me 3.

i know that the slope of the tangent line is 3 and the normal is -1/3.
so i use y=mx+b to find the formula correct?

Thanks!

5. Originally Posted by >_<SHY_GUY>_<
ok, by finding the derivative, i got $3x^2$
and substituting 1 gives me 3.

i know that the slope of the tangent line is 3 and the normal is -1/3.
so i use y=mx+b to find the formula correct?

Thanks!
Correct. You can use either y = mx+b or the formula I used above, both work