Shas72
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at a point with coordinates (x,y), the gradient of the curve is directly propotional to xy^2. At a particular point P with coordinates (1,3) its known that the gradient of this curve is 6. find and simplify an expression for y in terms of x
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mqb2766
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(Original post by Shas72)
at a point with coordinates (x,y), the gradient of the curve is directly propotional to xy^2. At a particular point P with coordinates (1,3) its known that the gradient of this curve is 6. find and simplify an expression for y in terms of x
What are you having difficulties with? Write the "gradient" as a derivative and set proportional to xy^2. Use the two bits of information to determine the integration constant and the proportionality constant.
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Shas72
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(Original post by mqb2766)
What are you having difficulties with? Write the "gradient" as a derivative and set proportional to xy^2. Use the two bits of information to determine the integration constant and the proportionality constant.
I did that i got -1/y=x^2/2+c. I don't know how to go ahead
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mqb2766
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(Original post by Shas72)
I did that i got -1/y=x^2/2+c. I don't know how to go ahead
Assuming its right, use the point (1,3) to get c.
Upload your working if you want it checked.
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Sinnoh
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(Original post by Shas72)
I did that i got -1/y=x^2/2+c. I don't know how to go ahead
Rearrange to make y the subject, and using the initial conditions given that dy/dx = 6 at (1,3), solve for c
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Shas72
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(Original post by mqb2766)
What are you having difficulties with? Write the "gradient" as a derivative and set proportional to xy^2. Use the two bits of information to determine the integration constant and
(Original post by mqb2766)
Assuming its right, use the point (1,3) to get c.
Upload your working if you want it checked.
I got c=-5/6
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mqb2766
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(Original post by Shas72)
I got c=-5/6
Upload your calcs if you want them checked.
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Shas72
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(Original post by mqb2766)
Upload your calcs if you want them checked.
Ok I did dy/dx=xy^2
1/y^2.dy=x.dx
-1/y=x^2/2+c
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mqb2766
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(Original post by Shas72)
Ok I did dy/dx=xy^2
1/y^2.dy=x.dx
-1/y=x^2/2+c
You've not used the proportional to information with the fact that the graident is 6 at (1,3)
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Shas72
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(Original post by mqb2766)
You've not used the proportional to information with the fact that the graident is 6 at (1,3)
I did not understand. Do I just say proportional or equal to and solve
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mqb2766
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(Original post by Shas72)
I did not understand. Do I just say proportional or equal to and solve
the question says dy/dx is directly proportional to xy^2. Use the fact that dy/dx = 6 at (1,3) to determine the proportionality constant. You did proportional stuff at gcse?
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Shas72
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(Original post by mqb2766)
the question says dy/dx is directly proportional to xy^2. Use the fact that dy/dx = 6 at (1,3) to determine the proportionality constant. You did proportional stuff at gcse?
Ok let me try. I seem to have forgotten. Let me try
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Shas72
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(Original post by mqb2766)
the question says dy/dx is directly proportional to xy^2. Use the fact that dy/dx = 6 at (1,3) to determine the proportionality constant. You did proportional stuff at gcse?
So when I do that I get c= 2/3
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mqb2766
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(Original post by Shas72)
So when I do that I get c= 2/3
Again, upload what you've done if you want it checked.
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Shas72
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(Original post by mqb2766)
Again, upload what you've done if you want it checked.
So I did dy/dx=k.xy^2
Substituting the values I got k=2/3
Then I did dy/dx=2/3× xy^2
I got-1/y=x^2/3+c
Substituting (1,3) I get c=-2/3
Hence-1/y=x^2/3-2/3
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mqb2766
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(Original post by Shas72)
So I did dy/dx=k.xy^2
Substituting the values I got k=2/3
Then I did dy/dx=2/3× xy^2
I got-1/y=x^2/3+c
Substituting (1,3) I get c=-2/3
Hence-1/y=x^2/3-2/3
Looks ok.
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Shas72
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(Original post by mqb2766)
Looks ok.
Thanks. Iam stuck with one more problem. Pls help.Name:  15949103606434005661861529311613.jpg
Views: 3
Size:  194.0 KBits the 14th sum
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mqb2766
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(Original post by Shas72)
Thanks. Iam stuck with one more problem. Pls help.Name:  15949103606434005661861529311613.jpg
Views: 3
Size:  194.0 KBits the 14th sum
Have you started by separating the variables? Post where you're up to?
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Shas72
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(Original post by mqb2766)
Have you started by separating the variables? Post where you're up to?
Name:  15949116124677589065051049874043.jpg
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Shas72
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So when I do integration, I get lnx+1/2lnx^2+3=kt+c
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