# Simple differentiation.

1. Nov 7, 2009

### Sirsh

The tangent to the curve y=ax^3 at the point (-1,b) is perpendicular to the line y = 2x+3. Find the values of a and b.

Could someone show me how to answer that?

2. Nov 7, 2009

### HallsofIvy

Staff Emeritus
First, if (-1, b) is on the curve $y= ax^3$, what is b?

The two lines given by y= mx+ u and y= nx+ v are perependicular if and only if mn= -1. With y= 2x+ 3, m= 2 so what is n?

The slope, n, of a tangent line is equal to the derivative of the function at that point. What is the derivative there. What is the derivative of $y= ax^3$ at x= -1?

3. Nov 7, 2009

### Sirsh

y = 2x+3
y' = 2
m = 2
so, a perpendicular gradient to 2 is -1/2.
y = ax^3
y' = 3ax^2
-1/2 = 3ax^2
-1/2 = 3a(-1)^2
a = -1/6

y = -1/6(-1)^3
b = 1/6.

4. Nov 7, 2009

### HallsofIvy

Staff Emeritus
Exactly right! Congratulations.