# [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 $\displaystyle y = x(x-1)(x-2)$, which crosses the $\displaystyle x-axis$ at the points $\displaystyle O(0,0)$, $\displaystyle A(1,0)$ and $\displaystyle B(2,0)$.

(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 $\displaystyle R_1$ is the same as the area of the shaded region $\displaystyle R_2$.

Attempt:

I can't see the point $\displaystyle C$ 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 $\displaystyle y = x(x-1)(x-2)$, which crosses the $\displaystyle x-axis$ at the points $\displaystyle O(0,0)$, $\displaystyle A(1,0)$ and $\displaystyle B(2,0)$.

(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.
$\displaystyle y - y_{A}= \left(\frac{dy}{dx}\right)_{A}(x - x_{A})$
$\displaystyle y - y_{B}= \left(\frac{dy}{dx}\right)_{B}(x - x_{B})$
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 $\displaystyle R_1$ is the same as the area of the shaded region $\displaystyle R_2$.

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:

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

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:

$\displaystyle y=mx+c$

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.
$\displaystyle y - y_{A}= \left(\frac{dy}{dx}\right)_{A}(x - x_{A})$
$\displaystyle y - y_{B}= \left(\frac{dy}{dx}\right)_{B}(x - x_{B})$
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:

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

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:

$\displaystyle y=mx+c$

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

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

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

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

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

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

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

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

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

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

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

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

Originally Posted by looi76

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

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

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

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

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

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

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

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

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

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

Is it correct?

Quote:

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

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

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

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

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

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

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

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

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

Is it correct?
• Apr 28th 2008, 11:14 AM
$\displaystyle y + x = 1 \longrightarrow$ equation 1
$\displaystyle y + 2x = -4 \longrightarrow$ equation 2
$\displaystyle 2x = -4$ <======== where did this came from??
$\displaystyle x = \frac{-4}{2}$
$\displaystyle x = -2$

this part is wrong.
• Apr 28th 2008, 11:24 AM
looi76
Quote:

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

this part is wrong.

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

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

$\displaystyle C(-\frac{5}{3},-\frac{2}{3})$(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 $\displaystyle y = x(x-1)(x-2)$, which crosses the $\displaystyle x-axis$ at the points $\displaystyle O(0,0)$, $\displaystyle A(1,0)$ and $\displaystyle B(2,0)$.

(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 $\displaystyle R_1$ is the same as the area of the shaded region $\displaystyle R_2$.

Attempt:

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

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