/sci/ - Science & Math

[Anonymous No.2698629]

If one have a tangent line that goes through the X-axis at one location, and the Y-axis at a location (described with a variable), how can one go forth to find a value of the variable that makes the distance between the two points as small as possible?

[Anonymous No.2698740]

hmm. mind blown.

[Anonymous No.2698766]

Right triangles

[Anonymous No.2699089]

>>2698766
explain.

[Anonymous No.2699159]

>>2698766
Do you mean a^2+b^2=c^2 ?
Does that solve anyhing here?

[Anonymous No.2699174]

You mean you're given an x-intercept and have to find the y-intercept that minimizes distance? That's just x = 0.

[Anonymous No.2699202]

>>2699174
Shit, I meant y = 0.

[Anonymous No.2699278]

>>2699174
let's say the tangen line is: 2/x^(1/2)
And you know where the line intersects with the axises.

How to change x so that the distance between the two points are as small as possible?

the distance between the lines would be the tangent line

[Anonymous No.2699299]

>>2699278
>And you know where the line intersects with the axises.
You're given both intercepts? Then you can't change x without changing the intercepts. I'm not sure I completely understand the question.

[Anonymous No.2699338]

>>2699299
Yes, given both intercepts.
And can change the x in the equation.

for what value of x is the line shortest?
not just in this case, but generally

[Anonymous No.2699413]

>>2699338
Oh, I get what you're saying.

Let f(x) be the function you're given. The tangent line to that function at (x, f(x)) is y = f'(x) * x + f(x).

x-intercept:
0 = f'(x) * x + f(x), and solve for x.

y-intercept:
y = f'(x) * 0 + f(0) = f(0)

The distance between the two points is [f(0)^2 + (x-intercept)^2]^(1/2), and just find the minimum of that function.

[Anonymous No.2699438]

>>2699413
Never mind, that's not right. The x-intercept should be this:
0 = f'(c) * x + f(c)
f'(c) * x = -f(c)
x = -f(c) / f'(c)

And then put that into the distance formula.