find an equation of the curve that passes through the point (1,1) and whose slope at (x,y) is y^2/x^3
Well, I'm not a mathematician, but I do know a thing or two about humor. How about this equation of a curve: "FunnyCurve" = (1/3)x^3 + (1/2)x^2 + (3/2)x - 1.
Why is it funny, you ask? Well, it's because no matter what point you choose on this curve, you'll always end up laughing at the absurdity of its equation!
To find the equation of the curve that passes through the point (1,1) and whose slope at any point (x,y) is y^2/x^3, we can use the method of separation of variables.
First, let's assume the equation of the curve is in the form y = f(x).
Since the slope at any point (x,y) is given by dy/dx = y^2/x^3, we can write it as follows:
dy/dx = y^2/x^3
Let's now separate the variables by multiplying both sides of the equation by x^3/y^2:
x^3/y^2 dy/dx = 1
Integrating both sides with respect to x will give us:
∫ x^3/y^2 dy = ∫ dx
To integrate the left-hand side, we can use the u-substitution method. Let's substitute u = y/x:
d(u) = (u/x - y/x^2) dx
dy = x du
Substituting these into the integral equation, we get:
∫ x^2/u^2 (x du) = ∫ dx
Simplifying the integral equation, we have:
∫ x^3/u^2 du = ∫ dx
Taking the integral of both sides, we get:
x^4/(-u) = x + C
Replacing u with y/x, we have:
y/x = -x^3/(x + C)
Now, to find the value of C, we can substitute the coordinates of the given point (1,1) into the equation:
1/1 = -1^3/(1 + C)
1 = -1/(1 + C)
1 + C = -1
C = -2
Substituting the value of C back into the equation, we have:
y/x = -x^3/(x - 2)
Finally, we can multiply both sides by x to get the equation in terms of y:
y = -x^4/(x - 2)
Therefore, the equation of the curve that passes through the point (1,1) and whose slope at any point (x,y) is y^2/x^3 is y = -x^4/(x - 2).
To find an equation of the curve that passes through the point (1,1) and whose slope at (x,y) is y^2/x^3, we can use the method of separation of variables.
First, let's differentiate the given slope equation implicitly with respect to x:
dy/dx = (d/dx)(y^2/x^3)
= (2y * dy/dx)/x^3 - (3y^2 * 1/x^4)
Now, we can write this equation in a form that separates the variables:
x^3 * dy/dx = 2y * dy - 3y^2 * dx/x
x^3 * dy - 2y * dy = -3y^2 * dx/x
(dy * (x^3 - 2y)) = (-3y^2 * dx/x)
Now, to integrate both sides of this equation:
∫ (dy * (x^3 - 2y)) = ∫ (-3y^2 * dx/x)
Let's solve the left-hand side first:
∫ (x^3 - 2y) dy = ∫ (-3y^2 dx/x)
∫ x^3 * dy - 2∫ y * dy = -3∫ y^2 dx/x
Integrating both sides:
(x^3 * y) - (y^2) = -3∫ y^2 dx/x + C
where C is the constant of integration.
Now, let's substitute the point (1,1) into the equation to find the value of C:
(1^3 * 1) - (1^2) = -3∫ 1^2 dx/1 + C
1 - 1 = -3∫ 1 dx + C
0 = -3x + C
Since the curve passes through the point (1,1), we substitute x = 1 and y = 1:
0 = -3(1) + C
0 = -3 + C
C = 3
Therefore, the equation of the curve is:
x^3 * y - y^2 = -3∫ y^2 dx/x + 3
dy/dx=y^2/x^3
y^-2 dy= x^-3 dx
-y^-1=-x^-2*1/2 + C
check:take the derivative.
y^-2 y'=x^-3
dy/dx=y^2/x^3
so find c at by putting 1,1 into the equation
1/y=1/x^2 + C