Calculating the length of a tangent curve

[keeaga]

Considering f(x) = tan(x) * 5 / 8 ...

how can I find the length of the curve, specifically, between (0, 0) and (1, 1) ?

if anyone can help I would be happy.

Thanks
Keeaga

 

[DonAntonio]

Considering f(x) = tan(x) * 5 / 8 ...

how can I find the length of the curve, specifically, between (0, 0) and (1, 1) ?

if anyone can help I would be happy.

Thanks
Keeaga

You can't: the point (1,1) is not on the function's graph.

DonAntonio

 

[keeaga]

Actually, it is... that's what the 5/8 is for. It shrinks the tangent vertically just enough for the curve to cross (-1,-1), (1,1), and (0,0).

I still don't know how to go about finding the length of the curve though.

Keeaga

 

[Mandlebra]

Actually, it is... that's what the 5/8 is for. It shrinks the tangent vertically just enough for the curve to cross (-1,-1), (1,1), and (0,0).

I still don't know how to go about finding the length of the curve though.

Keeaga

Are you implying tan(1)*5/8= 1?

 

[DonAntonio]

Actually, it is... that's what the 5/8 is for. It shrinks the tangent vertically just enough for the curve to cross (-1,-1), (1,1), and (0,0).

I still don't know how to go about finding the length of the curve though.

Keeaga

No, it really doesn't: \,\tan 1=1.55741\Longrightarrow \frac{5}{8}\tan 1 = 0.97338\neq 1\Longrightarrow (1,1)\, is not on the graph of the function, and neither

is the point \,(-1,-1)\,

DonAntonio

 

[keeaga]

Ok, sorry, you're right... Thought it crossed 1,1 but that was based on a graph of it only.

Still, anyone know how generally to find the length of a tangent curve?

KTM

 

[Vargo]

Check out the wikipedia entry on arclength where there are many formulas. Also, any calculus text will have arclength formulas. The key to them all is the Pythagorean theorem

ds = sqrt(dx^2+dy^2). Divide out dx and you get sqrt(1+(dy/dx)^2) dx