What an interesting approach. it made me pause when you made the creative assumption that Y and X are linearly related. That assumption gives a clear solution to the first problem, but does that method exclude other reasonable solutions, too? After all, what if X and Y are instead exponentially related, or related by a quadratic? Is it fair to assume linearity?
P. S. This is a graph defined parametrically - in the form of x(t) and y(t). The graph has two branches - the lower one (t from 0 to 2) and the upper one (t from 2 to infinity).
problem xʸ = y ²ˣ In the real world, y could be a complicated function of x. But for simplicity's sake assume y is a linear multiple of x. y = kx , where k ∈ ℝ. xᵏˣ = (kx) ²ˣ x = 0 gives the solution 1 = 1 since 0⁰ = 1. Therefore x = y = 0 is a solution corresponding to the parametric point x = 0 for any k valu. Assume x ≠ 0 since x = 0 is a solution. Take lns. (k x) ln x = 2x ( ln k + ln x ) x ≠ 0, so divide by x. k ln x = 2( ln k + ln x ) Solve for x. (k -2) ln x = 2 ln k As long as k ≠ 2, ln x = 2 ln k / (k-2) Parametric curve with parameter k not 2: x = k²ᐟ⁽ᵏ⁻²⁾ y = kx = k ᵏᐟ⁽ᵏ⁻²⁾ There is a discontinuity problem at k = 0. Then x = 0, y = 0 by y = kx But we have y = k ᵏᐟ⁽ᵏ⁻²⁾ = 1 = 0⁰. y can't be 1 and 0 simultaneously. Therefore k ≠ 0. answer (x, y) ∈ { ( k²ᐟ⁽ᵏ⁻²⁾, k ᵏᐟ⁽ᵏ⁻²⁾), (k ≠ 0, 2, k ∈ ℝ) }
What an interesting approach. it made me pause when you made the creative assumption that Y and X are linearly related. That assumption gives a clear solution to the first problem, but does that method exclude other reasonable solutions, too? After all, what if X and Y are instead exponentially related, or related by a quadratic? Is it fair to assume linearity?
graphing the equation in Desmos does not give the same graph as the one you have included in your video. why so?
Yes, I also got completely different graphs if I build the functions x(k) and y(k).
P. S. This is a graph defined parametrically - in the form of x(t) and y(t). The graph has two branches - the lower one (t from 0 to 2) and the upper one (t from 2 to infinity).
problem
xʸ = y ²ˣ
In the real world, y could be a complicated function of x. But for simplicity's sake assume y is a linear multiple of x.
y = kx
, where k ∈ ℝ.
xᵏˣ = (kx) ²ˣ
x = 0 gives the solution
1 = 1
since 0⁰ = 1. Therefore x = y = 0 is a solution corresponding to the parametric point x = 0 for any k valu.
Assume x ≠ 0 since x = 0 is a solution.
Take lns.
(k x) ln x = 2x ( ln k + ln x )
x ≠ 0, so divide by x.
k ln x = 2( ln k + ln x )
Solve for x.
(k -2) ln x = 2 ln k
As long as k ≠ 2,
ln x = 2 ln k / (k-2)
Parametric curve with parameter k not 2:
x = k²ᐟ⁽ᵏ⁻²⁾
y = kx = k ᵏᐟ⁽ᵏ⁻²⁾
There is a discontinuity problem at k = 0.
Then
x = 0, y = 0 by y = kx
But we have
y = k ᵏᐟ⁽ᵏ⁻²⁾ = 1 = 0⁰.
y can't be 1 and 0 simultaneously.
Therefore k ≠ 0.
answer
(x, y) ∈ { ( k²ᐟ⁽ᵏ⁻²⁾, k ᵏᐟ⁽ᵏ⁻²⁾), (k ≠ 0, 2, k ∈ ℝ) }
0^0 is undefined and NOT equal to 0
@ i thought it was 1.
@ I didn't say it was 0. I said it is 1.