Solve for the y-Intercept: Curve Through (4/9,1) with Slope of -2/y3

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In summary, to find the y-intercept of the curve that passes through the point (4/9,1) and whose slope at (x,y) is −2/y3, you can use the equation y = sqrt4(-8x + 19) and plug in x=0 to get the y-intercept of 2.087. The given slope was not directly used, but was used to find the function y as a function of x.
  • #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
 
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  • #2
Slimsta said:
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
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
Dick said:
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
Slimsta said:
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
rock.freak667 said:
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
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
Office_Shredder said:
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 don't even help.. its there to make the question scary i guess.
 
  • #9
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?
 

Related to Solve for the y-Intercept: Curve Through (4/9,1) with Slope of -2/y3

1. What are separable equations?

Separable equations are a type of differential equation where the variables can be separated into two different functions that are multiplied together. This allows us to solve the equation by integrating each function separately.

2. How do I solve a separable equation?

To solve a separable equation, you first need to separate the variables into two functions. Then, integrate each function separately with respect to its variable. Finally, combine the two solutions and solve for any constants that may be present.

3. What is the purpose of solving separable equations?

Solving separable equations allows us to find the general solution to a differential equation. This can be useful in many fields of science and engineering where differential equations are used to model real-life phenomena.

4. Are there any special cases or methods for solving separable equations?

Yes, there are some special cases where separable equations can be solved using specific methods. For example, exact equations can be solved using the method of integrating factors. Additionally, some equations may require substitutions or other techniques to be solved.

5. Can separable equations be used to model real-life situations?

Yes, separable equations can be used to model many real-life situations, such as population growth, radioactive decay, and chemical reactions. By solving these equations, we can make predictions and understand the behavior of these systems.

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