Author Topic: Help on calculus Q! (Read 734 times) Tweet Share

samiira · « **on:** June 15, 2010, 04:56:21 pm »

open attachment..thanks

.. any help much appreciated

moekamo · « **Reply #1 on:** June 15, 2010, 05:37:24 pm »

a) $RHS = (36\pi)^{1/3} \cdot v^{2/3} = (36\pi)^{1/3} \cdot \left ( \frac{4}{3} \pi r^3 \right )^{2/3} = (36\pi)^{1/3} \cdot \left ( \frac{4\pi}{3}\right )^{2/3} \cdot r^2 = 36^{1/3} \cdot 16^{1/3} \div 9^{1/3} \cdot \pi r^2 = 4\pi r^2 = A = LHS, QED$

b) $A= (36\pi)^{1/3} \cdot v^{2/3} \implies \frac{dA}{dV} = (36\pi)^{1/3} \cdot \frac{2}{3} v^{-1/3}$

c) so $\frac{dV}{dt} = -50$ we need $\frac{dA}{dt}= \frac{dV}{dt} \frac{dA}{dV} = -50 \cdot (36\pi)^{1/3} \cdot \frac{2}{3} v^{-1/3}$

when r=10, $V=\frac{4}{3} \pi 10^3 = 4000\pi / 3$ , sub that in as volume in las line to get the rate, i dont know how many DP you need or if you need axact answer...

d) if SA = 0, then r=0 and so V=0 too, so we want time for either to get to 0

since we know $\frac{dV}{dt} = -50 \implies V=\int -50 dt$ or we can do $t=\int_{250}^0 -\frac{1}{50} dV = -\frac{1}{50}[0-250] = 5 secs$

i probably made mistakes so just point them out so i can fix them :S

samiira · « **Reply #2 on:** June 15, 2010, 10:58:14 pm »

Thanks Alot

Juddinator · « **Reply #3 on:** July 08, 2010, 04:51:27 pm »

If anyone could help me on some of the questions in the attachments that would be awesome, thanks. I get close but I don't find the correct answer....

superflya · « **Reply #4 on:** July 08, 2010, 05:22:33 pm »

1) integrate twice, should get D
2) graph ontop-graph bottom to find enclosed area so (9-x^2)-(x^2-9)=18-2x^2 so clearly B
3)sub in u=1-x , change terminals (NOTE the -ve), integrate then a bit of addition and subtraction should yield 4/3. A
4) same deal as previous question make the 1-x^2 substitution. B

EDIT: just incase u get stuck, a -ve outside the integral is counter-acted by swapping the terminals.

brightsky · « **Reply #5 on:** July 08, 2010, 05:29:19 pm »

1. $\int 2\cos(x) + 1 dx = 2\sin(x) + x + C$

$\int 2\sin(x) + x + C dx = -2\cos(x) + \frac{x^2}{2} + Cx + D$

Hence D is one possible answer.

2. $\int \frac{x}{\sqrt{1 - x^2}} dx$

Substitute $u = 1-x^2$ , then $du = -2x dx$

$\int \frac{x}{\sqrt{u}} \frac{du}{-2x} = -\frac{1}{2} \int \frac{1}{\sqrt{u}} du = -\frac{1}{2} \cdot 2\sqrt{u} + C = -\sqrt{u} + C = -\sqrt{1-x^2} + C$

So $\int^{\frac{\sqrt{3}}{2}}_{0} \frac{x}{\sqrt{1 - x^2}} dx = -\sqrt{1 - \frac{3}{4}} + \sqrt{1} = -\frac{1}{2} + 1 = \frac{1}{2}$

So B.

3. $\int \frac{1}{(1-x)^2} dx$

Let $u = 1 - x$ , then $du = - dx$

$- \int \frac{1}{u^2} du$

$= \frac{1}{u}$

$= \frac{1}{1 - x}$

Hence $\int^{\frac{1}{2}}_{-\frac{1}{2}} \frac{1}{(1-x)^2} dx = \frac{1}{1 - \frac{1}{2}} - \frac{1}{1 + \frac{1}{2}} = \frac{4}{3}$

So A.

4. $\int^{3}_{-3} 9 - x^2 dx - \int^3_{-3} x^2 - 9 dx = \int^3_{-3} 18 - 2x^2 dx$

So B.

EDIT: Beaten xD

Juddinator · « **Reply #6 on:** July 08, 2010, 05:45:10 pm »

I now see the errors I made in my own work... for the first question i made it cos(x+1) rather then cos (x)+1, I wish the textbook was more clearer on things like that. I also made wrong substitutions lol. Thanks guys

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Author Topic: Help on calculus Q! (Read 734 times) Tweet Share

samiira

Help on calculus Q!

moekamo

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samiira

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Juddinator

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superflya

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brightsky

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Juddinator

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