anonymous
  • anonymous
The gradient at any point (x,y) on a curve is sqrt {1+2x}. The curve passes through the point (4,11). Find (i) the equation of the curve. (ii) the point at which the curve intersects the y-axis.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
Take the anti-derivative of the gradient to find out the equation of the curve. Once the equation has been figured out, put x as zero and get the answer to the second part.
campbell_st
  • campbell_st
(i) requires you to find the primitive or indefinite integral \[f'(x) = ( 1 + 2x)^{1/2}\] then \[F(x) = 1/3(1 + 2x)^{3/2} + c\] substitute x = 4 and F(x) = 11 to evaluate c this will also be the y-intercept
anonymous
  • anonymous
Won't it be times two, since it's (1 over 2/3)(1+2x)^3/2

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More answers

anonymous
  • anonymous
dY/dX=sqrt(1+2x)..integerate both sides Y=1/3(1+2X)^3/2+c.....put the points on this curve and find c constant.. (a)Y=1/3(1+2x)+c...put the value of c. (b) put X=0 in the curve equation and get that point...(0,y)..find Y .
anonymous
  • anonymous
How do you integrate?
anonymous
  • anonymous
@order the coefficient of X is 2 the divide by 2 here.
anonymous
  • anonymous
Why is the final answer not \[{(1+2x)^{3/2} \over {3\over2}}?\]
anonymous
  • anonymous
@order do you have any problem..
anonymous
  • anonymous
Yes... How do you integrate it? I did it like this: \[\int\limits 2 + \sqrt 2x^{1/2}\]\[y= x +{ \sqrt2x^{3/2} \over {3\over2}}\]\[y=x+ {2\sqrt2x^{3/2} \over 3}\]
anonymous
  • anonymous
\[\int\limits 1~~~ sorry\]
anonymous
  • anonymous
Would that be correct?
anonymous
  • anonymous
@Taufique
anonymous
  • anonymous
@FoolForMath
anonymous
  • anonymous
@kropot72
anonymous
  • anonymous
here think that you have integrate it with respect to X .not 2X .if you integrate it with respect to 2X then you must divide by 2 here.
anonymous
  • anonymous
I expanded the bracket though.. Would that be wrong?
anonymous
  • anonymous
no, it is not wrong.There is many way to integrate any function with respect to variable
anonymous
  • anonymous
where is your confusion??
anonymous
  • anonymous
Is my integration correct, or wrong?
anonymous
  • anonymous
there is mathematical processing error ..so i am unable to read it.
anonymous
  • anonymous
x + {2{sqrt 2}x^3/2 over 3}
anonymous
  • anonymous
When you split the square root, it's totally WRONG!
anonymous
  • anonymous
this is incorect..
anonymous
  • anonymous
Simple put sqrt ( 4 + 4) not equal to squrt ( 4 ) + sqrt ( 4)
anonymous
  • anonymous
you cant expand it in this way ..this is wrong way..
anonymous
  • anonymous
If you expand the brackets, isn't it int sqrt 1 + sqrt 2 sqrtx dx?
anonymous
  • anonymous
Ok... So, that doesn't work...
anonymous
  • anonymous
Technically, exponential doesn't work!
anonymous
  • anonymous
OK. So, can you show me once again, how to integrate it?
anonymous
  • anonymous
All you do is applying traditional method U substitution!
anonymous
  • anonymous
How?
anonymous
  • anonymous
u = √ ( 1 + 2x) -> u² = 1 + 2x => 2udu = 2dx Plug them in!
kropot72
  • kropot72
Hello order, Your method of integration is not valid. One method to integrate the function is as follows: z = 1+2x dz = 2dx \[\int\limits z ^{1/2}.dz/2 = 1/2\int\limits z ^{1/2}.dz = 1/3(1+2x)^{3/2} + C\]
anonymous
  • anonymous
As long as you obtain the result (1/3) u^(3/2) + C. That's the correct result!
anonymous
  • anonymous
hmmm. ok
anonymous
  • anonymous
you can do direct method: dy=1/2*sqrt(1+2x)d(2x).....2X part is variable integrate both sides y=1/2*((1+2x)^3/2)/(3/2))+c hence y=1/3(1+2x)^3/2+c
anonymous
  • anonymous
Thanks, I understand now :)
anonymous
  • anonymous
What's with so much confusion. :O \[\int\limits_{}^{} (1+2x)^{0.5}dx = (1+2x)^{0.5+1}\div0.5+1 + constant\]\[(1+2x)^{1.5} \div 1.5 + constant\] Now use the information given to find out the constant. \[11 = (1+2(4)) \div 1.5 + constant\] The constant comes out to be -7. Therefore, the equation of the curve will be \[y = ((1+2x)^{1.5} \div 1.5) - 7\] To find out the point on the y-axis, put x as zero. The answer comes out to be 3.4641.
kropot72
  • kropot72
Hello Shahan03, Sorry but your integration is invalid. The basic method for integration cannot be directly used on the particular function to be integrated.
anonymous
  • anonymous
THE VALUE OF C MUST BE 2..
anonymous
  • anonymous
kropot72: Why is that so? I have an exam in few days and so your reply will be very helpful.
anonymous
  • anonymous
Thanks a lot for pointing that out. I was indeed wrong. Apologies for the inconvenience. :)
kropot72
  • kropot72
To check whether or not your method of integration is valid, differentiate your result with respect to the variable. If your method is correct you will obtain the original function.

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