[SOLVED] Differentiation & Intergration Question

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April 28th 2008, 07:58 AM
looi76

[SOLVED] Differentiation & Intergration Question

Question:

http://img293.imageshack.us/img293/4...9w06qp1op7.jpg

The diagrams shows the curve $y = x(x-1)(x-2)$ , which crosses the $x-axis$ at the points $O(0,0)$ , $A(1,0)$ and $B(2,0)$ .

(i) The tangents to the curve at the points A and B meet at the point C. Find the coordinates of C.
(ii) Show by integration that the area of the shaded region $R_1$ is the same as the area of the shaded region $R_2$ .

Attempt:

I can't see the point $C$ on the graph!(Sadsmile)
April 28th 2008, 08:13 AM
Danshader

1 Attachment(s)

Point C is the intersection between the tangent of the curve at A and the tangent of the curve at B:
April 28th 2008, 08:32 AM
looi76

Quote:

Originally Posted by Danshader

Point C is the intersection between the tangent of the curve at A and the tangent of the curve at B:

Thanks Danshader! (Rofl)

Can you also explain to me how to get the tangent from the equarion of the curve? (Thinking)
April 28th 2008, 08:50 AM
Isomorphism

Quote:

Originally Posted by looi76

Question:

The diagrams shows the curve $y = x(x-1)(x-2)$ , which crosses the $x-axis$ at the points $O(0,0)$ , $A(1,0)$ and $B(2,0)$ .

(i) The tangents to the curve at the points A and B meet at the point C. Find the coordinates of C.

First of all get the equations of the tangents.
$y - y_{A}= \left(\frac{dy}{dx}\right)_{A}(x - x_{A})$
$y - y_{B}= \left(\frac{dy}{dx}\right)_{B}(x - x_{B})$
After this solve the above two equations simultaneously to get C. You do this because C lies on both of them.

Quote:

Originally Posted by looi76

(ii) Show by integration that the area of the shaded region $R_1$ is the same as the area of the shaded region $R_2$ .

Where are you stuck here? Try writing the integrals...

Good luck :)
April 28th 2008, 09:18 AM
Danshader

well first you need to find the gradient of the lines AC and BC, remember that differentiating the equation of the curve will give you the gradient at a particular point:

$ \frac{d}{dx}y = x(x-1)(x-2) $
$ \frac{d}{dx}y = x^3 - 3x^2 +2x $
$ \frac{dy}{dx} = 3x^2 - 6x +2 $

substituting the coordinates of x given in A(1,0) and B(2,0) you can get the gradient at A and B respectively. :D

then using the general line equation:

$y=mx+c $

obtain the line equation for AC and BC

then solving the simultaneous equation(equation of line AC and BC) you can get the coordinates of C :p
April 28th 2008, 10:36 AM
looi76

Quote:

Originally Posted by Isomorphism

First of all get the equations of the tangents.
$y - y_{A}= \left(\frac{dy}{dx}\right)_{A}(x - x_{A})$
$y - y_{B}= \left(\frac{dy}{dx}\right)_{B}(x - x_{B})$
After this solve the above two equations simultaneously to get C. You do this because C lies on both of them.

Where are you stuck here? Try writing the integrals...

Good luck :)

Quote:

Originally Posted by Danshader

well first you need to find the gradient of the lines AC and BC, remember that differentiating the equation of the curve will give you the gradient at a particular point:

$ \frac{d}{dx}y = x(x-1)(x-2) $
$ \frac{d}{dx}y = x^3 - 3x^2 +2x $
$ \frac{dy}{dx} = 3x^2 - 6x +2 $

substituting the coordinates of x given in A(1,0) and B(2,0) you can get the gradient at A and B respectively. :D

then using the general line equation:

$y=mx+c $

obtain the line equation for AC and BC

then solving the simultaneous equation(equation of line AC and BC) you can get the coordinates of C :p

Thanks Isomorphism & Danshader (Smile)

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

$\frac{dy}{dx} = 3x^{3-1} - 3 \times 2x^{2-1} + 2$

$\frac{dy}{dx} = 3x^2 - 6x + 2$

$\frac{dy}{dx}_A | x = 1$ ███ $\frac{dy}{dx}_A = 3(1)^2 - 6(1) + 2 = -1$

$\frac{dy}{dx}_B | x = 2$ ███ $\frac{dy}{dx}_B = 3(2)^2 - 6(2) + 2 = -4$

Equation of tangent to the curve at the point $A(1,0)$
$y - y_1 = m(x - x_1)$
$y - 0 = -1(x - 1)$
$y = -x + 1$

Equation of tangent to the curve at the point $B(2,0)$
$y - y_1 = m(x - x_1)$
$y - 0 = -4(x - 2)$
$y = -4x + 8$

$y + x = 1 \longrightarrow$ equation 1
$y + 4x = 8 \longrightarrow$ equation 2
$-3x = -7$
$x = \frac{7}{3}$

Substitute $x$ in equation $1$
$y + x = 1$
$y + \frac{7}{3} = 1$
$y = 1 - \frac{7}{3}$
$y = -\frac{4}{3}$

$C\left(\frac{7}{3},-\frac{4}{3}\right)$ (Wondering)

Is it correct?
April 28th 2008, 10:49 AM
Danshader

well you got the idea there but you made a little mistake in calculations :D

$\frac{dy}{dx}_B = 3(2)^2 - 6(2) + 2 = 2$
April 28th 2008, 11:10 AM
looi76

Quote:

Originally Posted by looi76

Thanks Isomorphism & Danshader (Smile)

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

$\frac{dy}{dx} = 3x^{3-1} - 3 \times 2x^{2-1} + 2$

$\frac{dy}{dx} = 3x^2 - 6x + 2$

$\frac{dy}{dx}_A | x = 1$ ███ $\frac{dy}{dx}_A = 3(1)^2 - 6(1) + 2 = -1$

$\frac{dy}{dx}_B | x = 2$ ███ $\frac{dy}{dx}_B = 3(2)^2 - 6(2) + 2 = -4$

Equation of tangent to the curve at the point $A(1,0)$
$y - y_1 = m(x - x_1)$
$y - 0 = -1(x - 1)$
$y = -x + 1$

Equation of tangent to the curve at the point $B(2,0)$
$y - y_1 = m(x - x_1)$
$y - 0 = -4(x - 2)$
$y = -4x + 8$

$y + x = 1 \longrightarrow$ equation 1
$y + 4x = 8 \longrightarrow$ equation 2
$-3x = -7$
$x = \frac{7}{3}$

Substitute $x$ in equation $1$
$y + x = 1$
$y + \frac{7}{3} = 1$
$y = 1 - \frac{7}{3}$
$y = -\frac{4}{3}$

$C\left(\frac{7}{3},-\frac{4}{3}\right)$ (Wondering)

Is it correct?

Quote:

Originally Posted by Danshader

well you got the idea there but you made a little mistake in calculations :D

$\frac{dy}{dx}_B = 3(2)^2 - 6(2) + 2 = 2$

$\frac{dy}{dx}_A | x = 1$ ███ $\frac{dy}{dx}_A = 3(1)^2 - 6(1) + 2 = -1$

$\frac{dy}{dx}_B | x = 2$ ███ $\frac{dy}{dx}_B = 3(2)^2 - 6(2) + 2 = 2$

Equation of tangent to the curve at the point $A(1,0)$
$y - y_1 = m(x - x_1)$
$y - 0 = -1(x - 1)$
$y = -x + 1$

Equation of tangent to the curve at the point $B(2,0)$
$y - y_1 = m(x - x_1)$
$y - 0 = 2(x - 2)$
$y = 2x - 4$

$y + x = 1 \longrightarrow$ equation 1
$y + 2x = -4 \longrightarrow$ equation 2
$2x = -4$
$x = \frac{-4}{2}$
$x = -2$

Substitute $x$ in equation $1$
$y + x = 1$
$y + (-2) = 1$
$y = 1 + 2$
$y = 3$

$C(-2,3)$ (Wondering)

Is it correct?
April 28th 2008, 11:14 AM
Danshader

$y + x = 1 \longrightarrow$ equation 1
$y + 2x = -4 \longrightarrow$ equation 2
$2x = -4$ <======== where did this came from??
$x = \frac{-4}{2}$
$x = -2$

this part is wrong.
April 28th 2008, 11:24 AM
looi76

Quote:

Originally Posted by Danshader

$y + x = 1 \longrightarrow$ equation 1
$y + 2x = -4 \longrightarrow$ equation 2
$2x = -4$ <======== where did this came from??
$x = \frac{-4}{2}$
$x = -2$

this part is wrong.

$y + x = 1 \longrightarrow$ equation 1
$y - 2x = -4 \longrightarrow$ equation 2
$-3x = 5$
$x = \frac{5}{-3}$
$x = -\frac{5}{3}$

Substitute $x$ in equation $1$
$y + x = 1$
$y + -\frac{5}{3} = 1$
$y = 1 -\frac{5}{3}$
$y = -\frac{2}{3}$

$C(-\frac{5}{3},-\frac{2}{3})$ (Wondering)

Is it correct?
April 28th 2008, 11:28 AM
Danshader

lol you missed out on a small negative sign :D

http://www.mathhelpforum.com/math-he...9404909a-1.gif equation 1
http://www.mathhelpforum.com/math-he...e1048e1b-1.gif equation 2

http://www.mathhelpforum.com/math-he...3543ccb0-1.gif here

http://www.mathhelpforum.com/math-he...4e9cb047-1.gif
http://www.mathhelpforum.com/math-he...47a30f6e-1.gif

you should get 5/3 for x and -2/3 for y :p
April 28th 2008, 11:53 AM
Mathstud28

Quote:

Originally Posted by looi76

Question:

http://img293.imageshack.us/img293/4...9w06qp1op7.jpg

The diagrams shows the curve $y = x(x-1)(x-2)$ , which crosses the $x-axis$ at the points $O(0,0)$ , $A(1,0)$ and $B(2,0)$ .

(i) The tangents to the curve at the points A and B meet at the point C. Find the coordinates of C.
(ii) Show by integration that the area of the shaded region $R_1$ is the same as the area of the shaded region $R_2$ .

Attempt:

I can't see the point $C$ on the graph!(Sadsmile)

For part two show that $\int_0^{1}x(x-1)(x-2)dx=\bigg|\int_1^{2}x(x-1)(x-2)dx\bigg|$

[SOLVED] Differentiation &amp; Intergration Question

[SOLVED] Differentiation & Intergration Question