Question

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.

Answer 1

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.

Answer 2

(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

Answer 3

Won't it be times two, since it's

(1 over 2/3)(1+2x)^3/2

Answer 4

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 .

Answer 5

How do you integrate?

Answer 6

@order the coefficient of X is 2 the divide by 2 here.

Answer 7

Why is the final answer not \[{(1+2x)^{3/2} \over {3\over2}}?\]

Answer 8

@order do you have any problem..

Answer 9

Yes... How do you integrate it?

I did it like this:

\[\int\limits 2 + \sqrt 2x^{1/2}\]\[y= x +{ \sqrt2x^{3/2} \over {3\over2}}\]\[y=x+ {2\sqrt2x^{3/2} \over 3}\]

Answer 10

\[\int\limits 1~~~ sorry\]

Answer 11

Would that be correct?

Answer 12

@Taufique

Answer 13

@FoolForMath

Answer 14

@kropot72

Answer 15

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.

Answer 16

I expanded the bracket though.. Would that be wrong?

Answer 17

no, it is not wrong.There is many way to integrate any function with respect to variable

Answer 18

where is your confusion??

Answer 19

Is my integration correct, or wrong?

Answer 20

there is mathematical processing error ..so i am unable to read it.

Answer 21

x + {2{sqrt 2}x^3/2 over 3}

Answer 22

When you split the square root, it's totally WRONG!

Answer 23

this is incorect..

Answer 24

Simple put sqrt ( 4 + 4) not equal to squrt ( 4 ) + sqrt ( 4)

Answer 25

you cant expand it in this way ..this is wrong way..

Answer 26

If you expand the brackets, isn't it

int sqrt 1 + sqrt 2 sqrtx dx?

Answer 27

Ok... So, that doesn't work...

Answer 28

Technically, exponential doesn't work!

Answer 29

OK. So, can you show me once again, how to integrate it?

Answer 30

All you do is applying traditional method U substitution!

Answer 31

How?

Answer 32

u = √ ( 1 + 2x) -> u² = 1 + 2x
=> 2udu = 2dx
Plug them in!

Answer 33

Hello order,
Your method of integration is not valid.
One method to integrate the function is as follows:
z = 1+2x
dz = 2dx
\[\int\limits z ^{1/2}.dz/2 = 1/2\int\limits z ^{1/2}.dz = 1/3(1+2x)^{3/2} + C\]

Answer 34

As long as you obtain the result (1/3) u^(3/2) + C. That's the correct result!

Answer 35

hmmm. ok

Answer 36

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

Answer 37

Thanks, I understand now :)

Answer 38

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.

Answer 39

Hello Shahan03,

Sorry but your integration is invalid. The basic method for integration cannot be directly used on the particular function to be integrated.

Answer 40

THE VALUE OF C MUST BE 2..

Answer 41

kropot72: Why is that so? I have an exam in few days and so your reply will be very helpful.

Answer 42

Thanks a lot for pointing that out. I was indeed wrong. Apologies for the inconvenience. :)

Answer 43

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.