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Mathematics: Post your doubts here!

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can u tell me the answer.? :p
im getting 42.67 for the first part. and 0.5 for the second :p
Accurate! very nice indeed. You got it right for the first part! but not for the second though, it's the same as first both parts have answer 42.67
the answer is same for both the parts since you are to find the area of the same region in two (almost) different ways. For the first part, i guess u both have already done it, but fr the second one, i guess u guys are getting confused by the 'horizontal components'. All u need to do is, rotate your x-y plane by 90 degrees, now your x-axis will be your y axis, and your y axis will be your x axis. I guess I made it more confusing. Just replace the y with x, and x with y. and u get the equation x = sq rt (16-y).......( as given in second part).
integrate this eq wrt y. u get (-2 (16-y)^2/3)/3. Apply the limits y = 16 (when x = 0) and y = 0 on this and u get 128/3 = 42.67
hope i made it clear enough.
 
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the answer is same for both the parts since you are to find the area of the same region in two (almost) different ways. For the first part, i guess u both have already done it, but fr the second one, i guess u guys are getting confused by the 'horizontal components'. All u need to do is, rotate your x-y plane by 90 degrees, now your x-axis will be your y axis, and your y axis will be your x axis. I guess I made it more confusing. Just replace the y with x, and x with y. and u get the equation x = sq rt (16-y).......( as given in second part).
integrate this eq wrt y. u get (-2 (16-y)^2/3)/3. Apply the limits y = 16 (when x = 0) and y = 0 on this and u get 128/3 = 42.67
hope i made it clear enough.
Oh damn. I did exact same thing. Miscalculation of the power. I must've subtracted 1 instead of adding it. *facepalm* Moreover I was too lazy to redo it. :p
Thank you though! :D
 
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Oh wait. That k is a constant neha. They have just written k instead of the answer. I got 2ln(2 plus e^3x) So k is basically 2. Dont get confused. If you want me to solve it then I will.... ?
yes could you solve it please? :)
 
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the answer is same for both the parts since you are to find the area of the same region in two (almost) different ways. For the first part, i guess u both have already done it, but fr the second one, i guess u guys are getting confused by the 'horizontal components'. All u need to do is, rotate your x-y plane by 90 degrees, now your x-axis will be your y axis, and your y axis will be your x axis. I guess I made it more confusing. Just replace the y with x, and x with y. and u get the equation x = sq rt (16-y).......( as given in second part).
integrate this eq wrt y. u get (-2 (16-y)^2/3)/3. Apply the limits y = 16 (when x = 0) and y = 0 on this and u get 128/3 = 42.67
hope i made it clear enough.
I'd understood that though thank you!
 
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yes could you solve it please? :)
Dy/dx = 6ye^3x/(2+e^3x)

Dy/y = 6e^3x/(2+e^3x)dx

ʃ dy/y = ʃ 6e^3x/(2+e^3x)dx

lny = 6 ʃ e^3x/(2+e^3x)dx

lny = 6 (1/3)ln(2+e^3x) + C

Now you have y = 36 and x = 0. Use them to find C.

ln36 = 2ln(2+e^3(0)) + C

ln36 = 2 ln(2 +1) + C

ln36 = 2ln3 + C

ln 36 = ln3^2 + C

ln 36 =ln9 + C

ln36 – ln9 = C

ln(36/9) = C

C = ln4

Re-write.

lny = 2ln(2 + e^3x) + ln4

lny = ln(2 + e^3x)^2 + ln4

lny = ln[4((2 + e^3x)^2]

ln cancels out.

y= [4((2 + e^3x)^2] Ans.
 
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Dy/dx = 6ye^3x/(2+e^3x)

Dy/y = 6e^3x/(2+e^3x)dx

ʃ dy/y = ʃ 6e^3x/(2+e^3x)dx

lny = 6 ʃ e^3x/(2+e^3x)dx

lny = 6 (1/3)ln(2+e^3x) + C

Now you have y = 36 and x = 0. Use them to find C.

ln36 = 2ln(2+e^3(0)) + C

ln36 = 2 ln(2 +1) + C

ln36 = 2ln3 + C

ln 36 = ln3^2 + C

ln 36 =ln9 + C

ln36 – ln9 = C

ln(36/9) = C

C = ln4

Re-write.

lny = 2ln(2 + e^3x) + ln4

lny = ln(2 + e^3x)^2 + ln4

lny = ln[4((2 + e^3x)^2]

ln cancels out.

y= [4((2 + e^3x)^2] Ans.
Thankyou soooo much! I was taking the 6y to the other side in the first step so it was becoming all complicated and i wasn't getting the answer :p
Thanks a lot! :D:)
 
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ii) Maximum point. Put dy/dx = 0

dy/dx ;

y = e^(2sinx) * cosx

u = e^(2sinx)

u’ = 2cosx*e^(2sinx)

v = cosx

v’ = -sinx

dy/dx = uv’ + vu’

0 = [e^(2sinx)* -sinx ] + [cosx*2cosx*e^(2sinx)]

0 = e^2sinx [-sinx + 2cos^2(x)]

0 = -sinx + 2cos^2(x)

Sinx = 2(1 – sin^2(x))

sinx= 2 – 2sin^2(x)

2sin^2(x) + sinx -2 = 0

Apply the quadratic formulae. You will get a negative answer. Ignore that. The second answer you will get will be 0.78

Lastly,

Sinx = 0.78

x=sin^-1(0.78)

x=0.8959 Radians ̴̴ 0.896 Ans.
 
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(x-2)^2 + 3
I am asked to sketch this graph. All that is ok. I found its minimum point and y-intercept and sketched. But what is confusing is that when I equated the equation(that is, y=0) , I got two answers for x.
But how is that possible when the minimum point of the graph is (2,3) and it doesn't touch the x-axis at all.
Help please?
 
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(x-2)^2 + 3
I am asked to sketch this graph. All that is ok. I found its minimum point and y-intercept and sketched. But what is confusing is that when I equated the equation(that is, y=0) , I got two answers for x.
But how is that possible when the minimum point of the graph is (2,3) and it doesn't touch the x-axis at all.
Help please?
Hello? When y=0 , (x-2)^2 = -3, and that means no real roots for that eq. 2 values for x must be in terms of i
 
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