Question

What function f(x) meets the requirements below and gives the smallest arc length possible?
f(0)=0, f(1)=0
f(x)> or = 0
0< or = x 11 years ago

Answer 1

is that: f(x) is 0 or positive, when x is betweeen 0 and 1?

Answer 2

if so, then what is the shortest distance between 2 points?

Answer 3

I forgot part of the problem. \[\int\limits_{0}^{1} f(x) dx = 1\]

Answer 4

can you verify this criteria?

f(x)> or = 0
0< or = x <or = 1 (x is between 0 and 1)

Answer 5

if im thinking right, f(x) is 0 or positive, then an absolute value will always be shorter than a quadratic or arcing function

Answer 6

but then im prolly over thinking something since the area of 1 would alter what im thinking of

Answer 7

\[\int ds=\int_{0}^{1}\sqrt{1+(f'(x))^2}~dx\]
is a minimum

Answer 8

f(x) is greater than or equal to zero (it's a positive function).
Zero is less than or equal to x which is less than or equal to one. (The only x values that matter are between zero and one).

Answer 9

Yeah, an absolute value would be shorter but I think I need an arc length because that's what we're doing in this chapter. Your graph is what I was thinking, but I don't know how to minimize the arc while keeping the area equal to one.

Answer 10

given 2 roots 0 and 1:

f(x) = -kx(x-1) forms a quadratic

f'(x) = -k(x-1)-kx = -k(2x+1)

Answer 11

I recognize your integral too, that's also this chapter. I just don't know what to do with all this parts that I have? They aren't fitting together in my head?

For your equation, assuming k is a constant?

Answer 12

yea, k is some constant
\[-k\int_{0}^{1}x(x-1)=1\]
\[-k[\frac13-\frac12]=1\]
\[-k[\frac{2-3}{6}]=1\]
\[k[\frac{1}{6}]=1\]

\[k = 6\] is the only quadratic that will fit an area equal to 1

Answer 13

a circle or ellipse might work but they maybe problematic

Answer 14

Yeah I think I need to use an ellipse. The quadratic is too long of an arc. I understand everything, I just don't know how to fit the equation of an ellipse to the minimization integral while keeping the area under the curve equal to one.

Another idea I have was to someone manipulate sin(x). Possibly make the period extremely large so the arc is minimally small? But again, how do I connect that idea and keep the area equal to one?

Answer 15

hmm, one idea. minimum arclength means we want our derivative of arclength at 0
\[a(x)=\int_{0}^{x}\sqrt{1+(f'(t))^2}~dt\]
\[a'(x)=\sqrt{1+(f'(x))^2}=0\]
well, i dont think that helped us out any

Answer 16

a circle wont fit