hba
A particle moves along x-axis.It starts from rest at origin O and its acceleration after t sec is given by a=3-t . Calculate the displacment from O when it is at rest.
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hba
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@saifoo.khan
hba
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Can you do it ?
jennychan12
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find antiderivative twice for a = 3-t
saifoo.khan
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integrate.
a=3-t
v = 3t - t^2/2 + C
Now set it equal to zero to get time.
Then integrate again, find the disp.
hba
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Therefore, the answer is 18.
Something which appeared in the (NUST) test.
hba
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@saifoo.khan
saifoo.khan
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REALLY? O_O
hba
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Yeah
saifoo.khan
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An O'levels student who have Add math can solve this too! D: D: D:
hba
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Hmm,good :)
hba
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What about hyperbolic functions ?
You've done that ?
saifoo.khan
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give me the problem, then i can tlel.
hba
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\[\int\limits_{}^{}e^x[\sinh^{-1} x+\frac{ 1 }{ \sqrt{1+x^2} }]dx=\]
hba
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@saifoo.khan Like this.
saifoo.khan
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INtegration by parts?
saifoo.khan
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idk how to deal with sinh^-1
hba
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So you don't know hyperbolic functions ?
experimentX
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use e^x(f(x) + f'(x)) rule
differentiation of arcsinh(x) = 1/sqrt(...)
saifoo.khan
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I guess no, then. :S
hba
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hmm
No need to worry about that it's easy,you can do that later.
What about finding the area from a diag ?
saifoo.khan
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We have to learn that parallelogram thing. But i can find the area/vol of a curve and stuff
hba
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Lets see.
find the Area A+B
|dw:1356722864399:dw|
saifoo.khan
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NOOO! Give me more difficult one.
hba
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Ok.
hba
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find the area A
|dw:1356723204131:dw|
saifoo.khan
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|dw:1356723438919:dw|
saifoo.khan
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Let me grab a pen and paper.
hba
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Hmm :)
saifoo.khan
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Is calc allowed?
hba
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No not allowed+You changed the question lol :P
saifoo.khan
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Sinx graph starts from zero. lol.
hba
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Oh okay then :/
saifoo.khan
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Haha.
hba
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Now solve it .
saifoo.khan
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isk what's sin(pi/4). i can only solve with the help of calc.
hba
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180/4
saifoo.khan
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area under cos is 1/sqrt2
hba
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Sin45=0.707
saifoo.khan
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0.707-0.293 ?
hba
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@saifoo.khan
Solve it and let me know :)
saifoo.khan
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0.414
hba
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Don't do it in decimals.
saifoo.khan
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dang.
saifoo.khan
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1/sqrt2 - (-1+sqrt2)/sqrt2
saifoo.khan
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2
Making:
(2-sqrt2)/sqrt2
saifoo.khan
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2
Making:
\[\frac{2\sqrt2-2}{2}\]
saifoo.khan
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sqrt2-1 final answer?
hba
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\[\sqrt{2}-1 \ sq.unit\]
Yeah right :D
saifoo.khan
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Wheo! B)
hba
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One more.
Find the area enclosed by x=y^2 and y=x-2
saifoo.khan
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umm okay. solving.
hba
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k.
saifoo.khan
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1,4
saifoo.khan
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area under line = 2 squnits
saifoo.khan
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trying to figure out about x=y^2
saifoo.khan
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1/2 final answer?
hba
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Let me give you options lol
A)9/2 sq.units
B)9/4 //
C)9/7 //
D)7/9 //
E)4/9 //
saifoo.khan
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Can't fig out.
hba
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I also couldn't figure this out lol :P
hba
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@hartnn
Can you solve it ?
hba
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Find the area enclosed by x=y^2 and y=x-2
hartnn
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this is double integration problem.. right ?
hba
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i don't think so.
hartnn
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plot both curves..
saifoo.khan
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|dw:1356725537575:dw|
hba
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@hartnn
Yes ?
hartnn
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sorry, got disconnected.
hartnn
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|dw:1356727367179:dw|
\(\int \limits_{y_1}^{y_2}f_1(x)-f_2(x)=\int \limits_{y_1}^{y_2}(y+2)-y^2dy|dw:1356727499321:dw|\)
hartnn
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\(\int \limits_{y_1}^{y_2}f_1(x)-f_2(x)=\int \limits_{y_1}^{y_2}(y+2)-y^2dy\)
hartnn
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find y2 and solve the integral
hartnn
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sorry, y1 and y2 limits should be
|dw:1356727654740:dw|