# find the slope of the curve:

• Apr 20th 2009, 03:23 PM
craigm
find the slope of the curve:
hi i have no idea where to start on this.

find the slope on the line tangent to the curve
y = (x^3 -1)(x^2+3x+2) at the point (1,0)
• Apr 20th 2009, 05:42 PM
pickslides
Quote:

Originally Posted by craigm
hi i have no idea where to start on this.

find the slope on the line tangent to the curve
y = (x^3 -1)(x^2+3x+2) at the point (1,0)

$\displaystyle y= (x^3-1)(x^2+3x+2)$

Should be expanded to give

$\displaystyle y= x^5+3x^4+2x^3-x^2-3x-2$

Using differentiation to help find the gradient (or tangent) with respect to x

$\displaystyle \frac{dy}{dx} = x^5+3x^4+2x^3-x^2-3x-2$

Now substituting x=1 into the derivative will give you the gradient of the tangent.

$\displaystyle (1)^5+3(1)^4+2(1)^3-(1)^2-3(1)-2 = 0$

As the result is zero, this implies a stationary point.
• Apr 20th 2009, 05:50 PM
TheEmptySet
Quote:

Originally Posted by craigm
hi i have no idea where to start on this.

find the slope on the line tangent to the curve
y = (x^3 -1)(x^2+3x+2) at the point (1,0)

first we need to take the derivative.(product rule)

$\displaystyle y'=(3x^2)(x^2+3x+2)+(x^3-1)(2x+3)$

Now if we evaluate at x=1 we get

$\displaystyle y'(1)=(3)(1+3+2)+(1-1)(2+3)=18$
• Apr 20th 2009, 05:59 PM
pickslides
Quote:

Originally Posted by pickslides

$\displaystyle y= (x^3-1)(x^2+3x+2)$

Should be expanded to give

$\displaystyle y= x^5+3x^4+2x^3-x^2-3x-2$

Using differentiation to help find the gradient (or tangent) with respect to x

$\displaystyle \frac{dy}{dx} = x^5+3x^4+2x^3-x^2-3x-2$

Now substituting x=1 into the derivative will give you the gradient of the tangent.

$\displaystyle (1)^5+3(1)^4+2(1)^3-(1)^2-3(1)-2 = 0$

As the result is zero, this implies a stationary point.

Sorry craigm, I did not actually find the derivative to your function

$\displaystyle y= (x^3-1)(x^2+3x+2)$

Should be expanded to give

$\displaystyle y= x^5+3x^4+2x^3-x^2-3x-2$

Using differentiation to help find the gradient (or tangent) with respect to x

$\displaystyle \frac{dy}{dx} = 5x^4+12x^3+36x^2-2x-3$

now substitute x = 1 into this equation! This should yield 18.

My method is a little more primative to that supplied by the previous poster (thanks for the correction!) but my method is great if you have not been introduced to the product rule