Calc: Problem Solving

**09Ashaw** · December 14th 2008, 04:12 PM

THe graph of y = f (x) passes through the origin. the arc length of the curve from (0,0) to (x, f(x)) is given by s(x) = (integral from a to x) of the square roote of 1 + e ^ t dt... Identify the function f.

**mr fantastic** · December 14th 2008, 07:26 PM

Originally Posted by 09Ashaw

THe graph of y = f (x) passes through the origin. the arc length of the curve from (0,0) to (x, f(x)) is given by s(x) = (integral from a to x) of the square roote of 1 + e ^ t dt... Identify the function f.

Formula for arclength: $L = \int_{a}^{b} \sqrt{1 + \left( \frac{dy}{dx}\right)^2} \, dx$ .

So it looks like you need to solve $\left( \frac{dy}{dx} \right)^2 = e^x$ subject to the boundary condition that y = 0 when x = 0 ....

**Soroban** · December 14th 2008, 07:44 PM

Hello, 09Ashaw!

The graph of $y \,=\, f(x)$ passes through the origin.
The arc length of the curve from $(0,0)$ to $(x, f(x))$ is given by: . $s(x) \:=\:\int^x_0\sqrt{1 + e^t}\,dt$
Identify the function $f(x)$

The arc length formula is: . $s(x) \;=\;\int^b_a\sqrt{1 + \left(\tfrac{dy}{dx}\right)^2}\,dx$

So we know that: . $\left(\frac{dy}{dx}\right)^2 \:=\:e^t \quad\Rightarrow\quad \frac{dy}{dx} \:=\:e^{\frac{x}{2}} \quad\Rightarrow\quad dy \:=\:e^{\frac{x}{2}}\,dx$

Integrate: . $\int dy \;=\;\int e^{\frac{x}{2}}\,dx \quad\Rightarrow\quad y \;=\;2e^{\frac{x}{2}} + C$

Since the graph contains the origin $(0,0)$ , we have: . $0 \;=\;2e^0 +C \quad\Rightarrow\quad C \:=\:-2$

. . Therefore: . $\boxed{f(x) \;=\;2e^{\frac{x}{2}} - 2}$

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