Question

137. Find a curve through the origin in the xy-plane whose length from x = 0 to x = 1 is L = ∫ from 0 to 1 of sqrt(1 + (1/4)e^x) dx.

Accepted Answer

Recognize that the problem asks for a curve $$ y = f(x) $$ passing through the origin $$ (0,0) $$ whose arc length from $$ x=0  to$$ $$ x=1  is $$given by $$ L = \int_0^1 \sqrt{1 + \frac{1}{4} e^x} \, dx . $$Recall the formula for the arc length of a curve $$ y = f(x) $$ from $$ x=a  to$$ $$ x=b $$:

$$
L = \int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx
$$ Compare the given integral's integrand $$ \sqrt{1 + \frac{1}{4} e^x} $$ with the arc length formula's integrand $$ \sqrt{1 + (y')^2} . $$This implies:

$$
1 + \left(\frac{dy}{dx}\right)^2 = 1 + \frac{1}{4} e^x
$$

which simplifies to:

$$
\left(\frac{dy}{dx}\right)^2 = \frac{1}{4} e^x
$$ Take the square root of both sides to find $$ \frac{dy}{dx} $$:

$$
\frac{dy}{dx} = \pm \frac{1}{2} e^{x/2}
$$

Choose the positive root to find an increasing function (or negative if decreasing is desired). Integrate $$ \frac{dy}{dx} = \frac{1}{2} e^{x/2} $$ with respect to $$ x  to $$find $$ y(x) $$:

$$
y = \int \frac{1}{2} e^{x/2} \, dx + C
$$

Use the initial condition $$ y(0) = 0  to $$solve for $$ C .$$

137. Find a curve through the origin in the xy-plane whose length from x = 0 to x = 1 is L = ∫ from 0 to 1 of sqrt(1 + (1/4)e^x) dx.

Verified video answer for a similar problem:

Key Concepts

Arc Length of a Curve

Differential Equation from Arc Length

Initial Condition and Integration