# Rate of change question

• December 13th 2010, 04:14 PM
liukawa
Rate of change question
Hi Guys,

Am currently stuck with this rate of change question.

question is as per attached file.

• December 13th 2010, 08:57 PM
liukawa
up for help guys !

Thanks !
• December 13th 2010, 09:24 PM
Soroban
Hello, liukawa!

Quote:

$\text{In the diagram, }P(p,\,p^2)\text{ is a variable point on the curve }y\,=\,x^2$
$\text{where }p > 0.\:\text{ The tangent at }P\text{ intersects the }x\text{-axis and }y\text{-axis}$
$\text{at }Q\text{ and }R\text{, respectively.}$

$\text{(a) Express the coordinates of }Q\text{ and }R\text{ in terms of }p.$
Code:

          |           |      *           |        /           |        /           |      */           |      /     *    |    o P       *  |  */     - - - * - o - - - - - - -           |  /Q           | /           |/           o R         /|

The slope of the tangent is: . $y' \,=\,2x$

At $\,P$, the slope is: . $m \,=\,2p$

The equation of the tangent is:
. . $y = p^2 \:=\:2p(x-p) \quad\Rightarrow\quad y \:=\:2px - p^2$

When $y = 0,\;2px-p^2 \:=\:0 \quad\Rightarrow\quad x \,=\,\frac{1}{2}p$ .[1]

. . The coordinates of $\,Q$ are: . $\left(\frac{1}{2}p,\:0\right)$

When $x = 0,\;y \:=\:2p(0)-p^2 \quad\Rightarrow\quad y \,=\,-p^2$ .[2]

. . The coordinates of $\,R$ are: . $(0,\:-p^2)$

Quote:

$\text{(b) When }p = 2\text{, the rate of increase of }p\text{ is 3 units/sec.}$

$\text{Find the corresponding rate of change of:}$

. . $\text{(i) the }x\text{-coordinate of }Q.$

We are told: when $p = 2,\;\dfrac{dp}{dt} \,=\,3$

From [1]: . $x \:=\:\frac{1}{2}p$

Differentiate with respect to time: . $\dfrac{dx}{dt} \:=\:\dfrac{1}{2}\!\cdot\!\dfrac{dp}{dt}$

. . Therefore: . $\dfrac{dx}{dt} \:=\:\frac{1}{2}(3) \:=\:\frac{3}{2}\text{ units/sec}$

Quote:

$\text{(ii) the area of triangle }OQR.$

The base of the triangle is $\,OQ\!:\;x \,=\,\frac{1}{2}p$

The height of the triangle is $|OR|\!:\;|y| \,=\,p^2$

The area is: . $A \;=\;\frac{1}{2}\left(\frac{1}{2}p\right)\left(p^2 \right) \;=\;\frac{1}{4}p^3$

Differentiate with respect to time: . $\dfrac{dA}{dt} \;=\;\dfrac{3}{4}p^2\left(\dfrac{dp}{dt}\right)$

Therefore: . $\dfrac{dA}{dt} \;=\;\dfrac{3}{4}(2^2)(3) \;=\;9\text{ units}^2\!\text{/sec}$

• December 13th 2010, 09:36 PM
liukawa
Quote:

Originally Posted by Soroban
Hello, liukawa!

The slope of the tangent is: . $y' \,=\,2x$

At $\,P$, the slope is: . $m \,=\,2p$

The equation of the tangent is:
. . $y = p^2 \:=\:2p(x-p) \quad\Rightarrow\quad y \:=\:2px - p^2$

When $y = 0,\;2px-p^2 \:=\:0 \quad\Rightarrow\quad x \,=\,\frac{1}{2}p$ .[1]

. . The coordinates of $\,Q$ are: . $\left(\frac{1}{2}p,\:0\right)$

When $x = 0,\;y \:=\:2p(0)-p^2 \quad\Rightarrow\quad y \,=\,-p^2$ .[2]

. . The coordinates of $\,R$ are: . $(0,\:-p^2)$

We are told: when $p = 2,\;\dfrac{dp}{dt} \,=\,3$

From [1]: . $x \:=\:\frac{1}{2}p$

Differentiate with respect to time: . $\dfrac{dx}{dt} \:=\:\dfrac{1}{2}\!\cdot\!\dfrac{dp}{dt}$

. . Therefore: . $\dfrac{dx}{dt} \:=\:\frac{1}{2}(3) \:=\:\frac{3}{2}\text{ units/sec}$

The base of the triangle is $\,OQ\!:\;x \,=\,\frac{1}{2}p$

The height of the triangle is $|OR|\!:\;|y| \,=\,p^2$

The area is: . $A \;=\;\frac{1}{2}\left(\frac{1}{2}p\right)\left(p^2 \right) \;=\;\frac{1}{4}p^3$

Differentiate with respect to time: . $\dfrac{dA}{dt} \;=\;\dfrac{3}{4}p^2\left(\dfrac{dp}{dt}\right)$

Therefore: . $\dfrac{dA}{dt} \;=\;\dfrac{3}{4}(2^2)(3) \;=\;9\text{ units}^2\!\text{/sec}$

Hi Soroban !