# Connected Rates Of Change (Calculus) - Need an Answer.

• Sep 4th 2012, 08:36 AM
zikcau25
Approximate Changes (Calculus) - Need right Answer
I am unable to get the right answer, closest answer : -0.01(r-1)
Trying again and again it keeps raising a lot more of other questions, that is simply boring for me.
Would be more than grateful if someone could kindly show me a detailed steps to the solution please.

(JF Talbert Textbook, Exercise 16D, Q.22)

For the function A=1/(r-1)^2 , a small change in r when r=2(r-1)^2 produces a 2% reduction in the value of A.
Find the approximate change in r.
• Sep 4th 2012, 10:41 AM
skeeter
Re: Connected Rates Of Change (Calculus) - Need an Answer.
$\displaystyle \frac{dA}{dr} = -\frac{2}{(r-1)^3}$

$\displaystyle -\frac{dA \cdot (r-1)^3}{2} = dr$

when $\displaystyle r = 2(r-1)^2$ , $\displaystyle r = \frac{1}{2}$ or $\displaystyle r = 2$

note: any info given on the range of values for r ?

for $\displaystyle r = \frac{1}{2}$ ...

$\displaystyle -\frac{(-0.02) \cdot \left(\frac{1}{2}-1 \right)^3}{2} = dr$

$\displaystyle -\frac{1}{800} = dr$ ... a decrease in r at 1/8 of a percent

for $\displaystyle r = 2$ ...

$\displaystyle -\frac{(-0.02) \cdot \left(2-1 \right)^3}{2} = dr$

$\displaystyle 0.01 = dr$

there is your 1% increase in r
• Sep 4th 2012, 01:24 PM
zikcau25
Re: Connected Rates Of Change (Calculus) - Need an Answer.
Thank you so much for that help.

you put an end to my day of agony and open a day of paradise :)
• Sep 8th 2012, 08:39 AM
zikcau25
Re: Connected Rates Of Change (Calculus) - Need an Answer.
I have reviewed the problem, it comprises of 3 stages:

I. Solving for $\displaystyle r$ from $\displaystyle r=2(r-1)^2}$, gives$\displaystyle \hspace{2mm}r=2\hspace{2mm}or\hspace{2mm} r=\frac{1}{2}$

II. Solving for $\displaystyle \delta A$ from $\displaystyle \frac{\delta A}{A}=-2\%$, gives:
$\displaystyle \frac{\delta A}{A}=-2\%$

$\displaystyle \frac{\delta A}{A}=\frac{-2}{100}$

$\displaystyle \delta A=-0.02A$

III. And finally, Solving for $\displaystyle \delta r}$ from $\displaystyle \delta A\approx (\frac{\mathrm{d}A }{\mathrm{d} r})_{r=2(r-1)^2}\times \delta r$, that is a)$\displaystyle \delta r\approx \frac{\delta A}{(\frac{\mathrm{d}A }{\mathrm{d} r})_{r=2}}$ or b)$\displaystyle \delta r\approx \frac{\delta A}{(\frac{\mathrm{d}A }{\mathrm{d} r})_{r=\frac{1}{2}}}$

Having given,

$\displaystyle A=(r-1)^{-2}$

$\displaystyle \frac{\mathrm{d} A}{\mathrm{d} r}=-2(r-1)^{-3}$

Therefore,
a) When $\displaystyle r=2, A=(2-1)^{-2}=1$

$\displaystyle \delta r\approx \frac{\delta A}{(\frac{\mathrm{d}A }{\mathrm{d} r})_{r=2}}$

$\displaystyle \delta r\approx \frac{-0.02(1)}{-2(2-1)^{-3}}$

$\displaystyle r\approx +0.01$ (increasing) (Happy)

b) when $\displaystyle r=\frac{1}{2}, A=(\frac{1}{2}-1)^{-2}=4$

$\displaystyle \delta r\approx \frac{\delta A}{(\frac{\mathrm{d}A }{\mathrm{d} r})_{r=\frac{1}{2}}}$

$\displaystyle \delta r\approx \frac{-0.02(4)}{-2(\frac{1}{2}-1)^{-3}}$

$\displaystyle r\approx -0.005$ (decreasing) (Wondering)

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Sir, hope you take into consideration stage II. that lead to a different solution. :)