# [SOLVED] tangent/normal lines

• Mar 15th 2009, 02:25 PM
>_<SHY_GUY>_<
[SOLVED] tangent/normal lines
1. fint the lines that are tangent and normal to the curve $\displaystyle y=x^3$ 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?

• Mar 15th 2009, 02:29 PM
Plato
Quote:

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

The tangent line has slope $\displaystyle y'(1)$; the normal line has slope $\displaystyle \frac{-1}{y'(1)}$.
• Mar 15th 2009, 02:29 PM
e^(i*pi)
Quote:

Originally Posted by >_<SHY_GUY>_<
1. fint the lines that are tangent and normal to the curve $\displaystyle y=x^3$ 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 $\displaystyle y=x^3$ and solve $\displaystyle f'(1)$

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

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

$\displaystyle y-y_1 = m(x-x_1)$ where $\displaystyle (x_1,y_1)$ is a coordinate. In your case this is (1,1)
• Mar 15th 2009, 02:41 PM
>_<SHY_GUY>_<
Quote:

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

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

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

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

ok, by finding the derivative, i got $\displaystyle 3x^2$
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!
• Mar 15th 2009, 02:44 PM
e^(i*pi)
Quote:

Originally Posted by >_<SHY_GUY>_<
ok, by finding the derivative, i got $\displaystyle 3x^2$
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