# [SOLVED] Differentiation &amp; Intergration Question

• Apr 28th 2008, 08:58 AM
looi76
[SOLVED] Differentiation &amp; 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)
• Apr 28th 2008, 09:13 AM
Point C is the intersection between the tangent of the curve at A and the tangent of the curve at B:
• Apr 28th 2008, 09:32 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:

Can you also explain to me how to get the tangent from the equarion of the curve? (Thinking)
• Apr 28th 2008, 09:50 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 :)
• Apr 28th 2008, 10:18 AM
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
• Apr 28th 2008, 11:36 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?
• Apr 28th 2008, 11:49 AM
well you got the idea there but you made a little mistake in calculations :D

$\frac{dy}{dx}_B = 3(2)^2 - 6(2) + 2 = 2$
• Apr 28th 2008, 12:10 PM
looi76
Quote:

Originally Posted by looi76
Thanks Isomorphism &amp; 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?
• Apr 28th 2008, 12:14 PM
$y + x = 1 \longrightarrow$ 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.
• Apr 28th 2008, 12:24 PM
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?
• Apr 28th 2008, 12:28 PM
• Apr 28th 2008, 12:53 PM
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 $\int_0^{1}x(x-1)(x-2)dx=\bigg|\int_1^{2}x(x-1)(x-2)dx\bigg|$