Author Topic: Related Rates - Need Help (Read 1375 times) Tweet Share

ElephantStew · « **on:** June 24, 2008, 09:22:39 pm »

Hey ppl, im having trouble with a question...

When the depth of liquid in a container is xcm the volume is $x(x^2+36)$ . Liquid is added to the container at a rate of 3cm^3/s. Find the rate of change of the depth of liquid at the instant when x= 11

my answer was 1/121, but the answers in the book got 1/133.

Mao · « **Reply #1 on:** June 24, 2008, 09:32:56 pm »

$\frac{dx}{dt}=\frac{dV}{dt}\cdot \frac{dx}{dV}$

$\frac{dV}{dx}=3x^2+36$

$\implies \frac{dx}{dV}=\frac{1}{3x^2+36}$

$\implies \frac{dx}{dt}=\frac{1}{x^2+12}$ , x=11

$\implies \frac{dx}{dt}=\frac{1}{121+12}=\frac{1}{133}$

bucket · « **Reply #2 on:** June 24, 2008, 09:34:05 pm »

We're looking for $\dfrac{dx}{dt}$ when $x=11$

$\dfrac{dV}{dt}= \dfrac{dx}{dt} \cdot \dfrac{dV}{dx}$

We know that:
$\dfrac{dV}{dt}=3cm^3/s$

and that:
$V=x(x^2+36)$

deriving that gives us:
$\dfrac{dV}{dx}= 3x^2+36$ When x=11

$\dfrac{dV}{dx}= 399$

Now subbing it back into that equation at the top.

$3 = 399 \cdot \dfrac{dx}{dt}$

$\dfrac{3}{399} = \dfrac{dx}{dt}$

$\dfrac{1}{133} = \dfrac{dx}{dt}$

bucket · « **Reply #3 on:** June 24, 2008, 09:34:34 pm »

Damn you mao =[. This is why I never post answers to questions! someone always beats me!

Flaming_Arrow · « **Reply #4 on:** June 24, 2008, 09:35:36 pm »

Quote from: bucket on June 24, 2008, 09:34:34 pm

Damn you mao =[. This is why I never post answers to questions! someone always beats me!

GGed

ElephantStew · « **Reply #5 on:** June 24, 2008, 09:38:23 pm »

oh gay.. i diff'd x(x^2+36) wrong...sigh at such simple mistakes
cheers

Glockmeister · « **Reply #6 on:** June 24, 2008, 10:13:18 pm »

Quote from: bucket on June 24, 2008, 09:34:34 pm

Damn you mao =[. This is why I never post answers to questions! someone always beats me!

Still its pretty good, I never got related rates when I did methods.

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Author Topic: Related Rates - Need Help (Read 1375 times) Tweet Share

ElephantStew

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