# Thread: Rate of change question

1. ## Rate of change question

Hi Guys,

Am currently stuck with this rate of change question.

question is as per attached file.

2. up for help guys !

Thanks !

3. Hello, liukawa!

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

$\displaystyle \text{(a) Express the coordinates of }Q\text{ and }R\text{ in terms of }p.$
Code:
          |
|       *
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*    |     o P
*   |   */
- - - * - o - - - - - - -
|  /Q
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o R
/|

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

4. Originally Posted by Soroban
Hello, liukawa!

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Hi Soroban !