# [SOLVED] Differentiation &amp; Intergration Question

• Apr 28th 2008, 07:58 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, 08:13 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, 08:32 AM
looi76
Quote:

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, 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 :)
• Apr 28th 2008, 09:18 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, 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:

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

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

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

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

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

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

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, 10:49 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, 11:10 AM
looi76
Quote:

Originally Posted by looi76

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

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

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

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

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

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:

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, 11:14 AM
$y + x = 1 \longrightarrow$ 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, 11:24 AM
looi76
Quote:

$y + x = 1 \longrightarrow$ 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, 11:28 AM
• Apr 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 $\int_0^{1}x(x-1)(x-2)dx=\bigg|\int_1^{2}x(x-1)(x-2)dx\bigg|$