A LEVEL: Rate of change questionWatch

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#1
Rate of change of y with respect to x, when x =0, y = 0; when x=2, y =6

dy/dx = klnx

find y when x=8 in the form:

(Aln2 -B)/(ln2-C)

I first separated the variables and got lnk = xlnx -x +c, but now I'm confused since there is no y in the equation. Do I have to e everything to work out k ?

I then tried to integrate by parts but it didn't look right. I got y=kxlnx -kx +c
Last edited by noorgal; 5 months ago
0
5 months ago
#2
(Original post by noorgal)
Rate of change of y with respect to x, when x =0, y = 0; when x=2, y =6

dy/dx = klnx

find y when x=8 in the form:

(Aln2 -B)/(ln2-C)

I first separated the variables and got lnk = xlnx -x +c, but now I'm confused since there is no y in the equation. Do I have to e everything to work out k ?

I then tried to integrate by parts but it didn't look right. I got y=kxlnx -kx +c
Don't know how you arrived at "lnk = xlnx -x +c", as separating variables would give you a "y".

is correct.

However, this is not defined for x=0.

Can you put up a link for, or image of, the original question?
Last edited by ghostwalker; 5 months ago
0
#3
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#4
(Original post by ghostwalker)
Don't know how you arrived at "lnk = xlnx -x +c", as separating variables would give you a "y".

is correct.

However, this is not defined for x=0.

Can you put up a link for, or image of, the original question?
I’ve posted the question
0
5 months ago
#5
(Original post by noorgal)
dy/dx = klnx seemingly is right

timesing by dx and dividing by k gives:

1/k *dy = lnx dx
y/k = ... use integration by parts. u=lnx, dv=1. du=1/x, v = x
y/k = xln(x) - (integral of(1)dx))
y/k = xln(x) - x + c
y= kxln(x) - kx + c

when x=0, y=0
Hence c=0

6= 2kln(2) - 2k
3= kln(2) - k
3/k = ln(2) - 1
k = 3/(ln(2) - 1) -- leaving it in this form because we'll need the fraction later on.

sub into: y/k = xln(x) - x
(y(ln(2) - 1))/3 = xln(x) - x
y* ((ln(2)-1)/3) = 8ln(8) -8
divide by ((ln(2)-1)/3)
y= 24ln(8) - 24/(ln(2) - 1)

Now simply use your rules of logs and you've got it.

Edit:

I was gonna say this is wrong actually because ln0 is undefined however the x being 0 makes it ln0^0 = ln1 = 0.
Last edited by username3154862; 5 months ago
0
5 months ago
#6
OK, you can get away with xlnx being 0 at x=0, since the limit as x goes to 0, of xlnx is 0.
0
#7
Thanks for the help!
(Original post by Relentas)
dy/dx = klnx seemingly is right

timesing by dx and dividing by k gives:

1/k *dy = lnx dx
y/k = ... use integration by parts. u=lnx, dv=1. du=1/x, v = x
y/k = xln(x) - (integral of(1)dx))
y/k = xln(x) - x + c
y= kxln(x) - kx + c

when x=0, y=0
Hence c=0

6= 2kln(2) - 2k
3= kln(2) - k
3/k = ln(2) - 1
k = 3/(ln(2) - 1) -- leaving it in this form because we'll need the fraction later on.

sub into: y/k = xln(x) - x
(y(ln(2) - 1))/3 = xln(x) - x
y* ((ln(2)-1)/3) = 8ln(8) -8
divide by ((ln(2)-1)/3)
y= 24ln(8) - 24/(ln(2) - 1)

Now simply use your rules of logs and you've got it.

Edit:

I was gonna say this is wrong actually because ln0 is undefined however the x being 0 makes it ln0^0 = ln1 = 0.
0
#8
thanks for the help!
(Original post by ghostwalker)
OK, you can get away with xlnx being 0 at x=0, since the limit as x goes to 0, of xlnx is 0.
0
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