Separable Equations help

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  • #1
Slimsta
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Homework Statement


Find the y-intercept of the curve that passes through the point (4/9,1) and whose slope at (x,y) is −2/y3.


Homework Equations





The Attempt at a Solution


woudnt the y-int be y=-2x/y3 + 5.5 ?
i got it by plugging the slope into m in the equation y=mx+b
then plugged in the given points to get b..
but the question wants a numerical solution :S
 

Answers and Replies

  • #2
gabbagabbahey
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i got it by plugging the slope into m in the equation y=mx+b

That's the equation of a straight line, with constant slope, [itex]m[/itex]. The slope you are given isn't constant, it depends on [itex]y[/itex]. Your curve is not a straight line.

What is the slope of any curve [itex]f(x)[/itex] at the point [itex](x,f(x))[/itex]?...Use that.
 
  • #3
Dick
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That makes no sense whatsoever. I think you know that. They want the y intercept of the curve. I think you have to solve for the curve. Whatever happened to the "separable equations" part of your post title? Shouldn't you separate dy/dx=(-2/y^3) and solve it??
 
  • #4
Slimsta
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That makes no sense whatsoever. I think you know that. They want the y intercept of the curve. I think you have to solve for the curve. Whatever happened to the "separable equations" part of your post title? Shouldn't you separate dy/dx=(-2/y^3) and solve it??

okay so i get dy/dx=(-2/y^3)
==> y^3 dy = -2dx
==> (y^4)/4 = -2x + C
==> y= sqrt4(-8x + 4C)

now what do i do?
 
  • #5
rock.freak667
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okay so i get dy/dx=(-2/y^3)
==> y^3 dy = -2dx
==> (y^4)/4 = -2x + C
==> y= sqrt4(-8x + 4C)

now what do i do?

They gave you a point that it passes through, so find, the constant C.
 
  • #6
Slimsta
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They gave you a point that it passes through, so find, the constant C.

well if i plug (9/4, 1) in, i get C=4.75
from there y= sqrt4(-8x + 19)
now what do i do with this?
 
  • #7
Office_Shredder
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So you have a curve passing through the point given, with the slope that they require. What does the question tell you to do with it?
 
  • #8
Slimsta
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So you have a curve passing through the point given, with the slope that they require. What does the question tell you to do with it?

i got 2.087
once i got y = (-8x + 19)^0.25
y-int is when x=0 so y=19^0.25 = 2.087

its weird how the slope they gave us dont even help.. its there to make the question scary i guess.
 
  • #9
Office_Shredder
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You used the slope to figure out what y was as a function of x. How can you plug in x=0 if you don't know what the function is?
 

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