Curve

**gracey** · November 6th 2008, 08:06 AM

A curve y= (2x-3)³. Find the x coordinates of the two points on the curve which have gradient 6.
The region under the x axis and to the right of the y axis is bounded by the curve. What is the area of the region?
thanks

**Soroban** · November 6th 2008, 09:06 AM

Hello, gracey!

Given: . $y\:=\: (2x-3)^3$

(a) Find the x-coordinates of the two points on the curve which have gradient 6.

We have: . $y' \:=\:6(2x-3)^2 \:=\:6 \quad\Rightarrow\quad(2x-3)^2 \:=\:1 \quad\Rightarrow\quad 2x-3 \:=\:\pm1$

. . $\begin{array}{cccccc}2x-3 \:=\:1 & \Rightarrow & 2x\:=\:4 & \Rightarrow & \boxed{x \:=\:2} \\ \\[-3mm] 2x-3\:=\:\text{-}1 & \Rightarrow & 2x \:=\:2 & \Rightarrow & \boxed{x \:=\:1} \end{array}$

Code:

        |
        |                         *
        |
        |                        *
        |                       *
        |           1         *
    - - + - - - - - + - - o - - + - - -
        |:::::::::::::*   1½    2
        |:::::::::::*
        |::::::::::*
        |::::::::::
        |:::::::::*
        |::::::::

(b) The region under the x-axis and to the right of the y-axis
is bounded by the curve. .What is the area of the region?

$\text{Area} \;=\;\int^{\frac{3}{2}}_0 \text{-}\,(2x-3)^3\,dx \;=\;\text{-}\,\frac{1}{8}(2x-3)^4\,\bigg]^{\frac{3}{2}}_0 \quad\hdots\quad\text{etc.}$

**gracey** · November 6th 2008, 10:09 AM

thanks so much

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