Author Topic: Unit 3 differentiation help plz (Read 656 times) Tweet Share

Maz · « **on:** December 04, 2015, 07:01:58 pm »

hey, can someone please help me?

if y=3x^2,
use differentiation to determine the approximate percentage change in x when t increases by 5%...
The answer in approximately 10% if that helps
thanks i advance

wyzard · « **Reply #1 on:** December 04, 2015, 08:34:00 pm »

This is a small change approximation problem, given a function $y=f(x)$ , the small change in $y$ can be approximated by:

$\Delta y\quad \approx \quad \frac { dy }{ dx } \Delta x$

Hence for this problem, since $\frac { dy }{ dx } =6x$ , the approximate change in $y$ given any change in $x$ is given to be:

$\Delta y\quad \approx \quad 6x\Delta x$

Now here's the trick, divide both sides by $y$ to obtain:
$\frac { \Delta y }{ y } \quad \approx \quad \frac { 6x\Delta x }{ y }$

The quantity $\frac { \Delta y }{ y }$ is the percentage change in $y$ , now for the left hand side we note that since $y=3{ x }^{ 2 }$ , the left hand side is reduced to where the $x$ cancels out:

$\frac { \Delta y }{ y } \quad \approx \quad 2\frac { \Delta x }{ x }$

The quantity $\frac { \Delta x }{ x }$ is the percentage change of $x$ . From here we can see that the percentage change in $y$ is twice the percentage change in $x$ , so if $x$ increases by 5%, $y$ will approximately increase by 10%.

Hope this helps

Maz · « **Reply #2 on:** December 05, 2015, 02:54:01 pm »

Firstly thanks...just a couple of questions...
why do you divide it by y...like is there a reason or is it just something you do?
and why do u only sub in y=3x^2 on the left hand side, because isn't there that rule that if you do something on
the left hand side u do it to the right hand side?
Thankyou

wyzard · « **Reply #3 on:** December 05, 2015, 06:16:13 pm »

The reason I divide by y on both sides is to obtain the term $\frac { \Delta y }{ y }$ on the right, which is the percentage change of y (or to be more precise fractional change, as the percentage change is just multiplying it by a 100). It just so happens that by doing so on the right side the $\frac { \Delta x }{ x }$ is also obtained, hence allowing us to relate the approximate percentage change in x and y

Maz · « **Reply #4 on:** December 05, 2015, 06:29:01 pm »

cool, Thankyou

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Author Topic: Unit 3 differentiation help plz (Read 656 times) Tweet Share

Maz

Unit 3 differentiation help plz

wyzard

Re: Unit 3 differentiation help plz

Maz

Re: Unit 3 differentiation help plz

wyzard

Re: Unit 3 differentiation help plz

Maz

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