RaySix

Post

Parametric Curve Length Calculation

Created by @nathanedwards

at November 1st 2023, 5:20:32 pm.

AP_Calculus_AB-Questions

#tangent-line

#arc-length

#parametric-equations

Question:

Let $C$ be the curve defined by the parametric equations:

x = e^t, \quad y = \int_0^t \frac{1}{\sqrt{1+e^{2u}}} \, du, \quad 0 \leq t \leq 1.

(a)

(b) Find the length of the curve $C$ .

Answer:

(a) To find the equation of the tangent line at the point where $t = 0$ , we need to find the slope of the tangent line, which is given by $\frac{dy}{dx}$ .

Since $x = e^t$ , we can write $t = \ln(x)$ .

Now, using the Fundamental Theorem of Calculus, we can find $\frac{dy}{dt}$ as follows:

\frac{dy}{dt} = \frac{d}{dt} \left( \int_0^t \frac{1}{\sqrt{1+e^{2u}}} \, du \right)

By the Leibniz rule for differentiating under the integral sign, we have:

\frac{dy}{dt} = \frac{1}{\sqrt{1+e^{2t}}}

Substituting $t = 0$ , we get:

\frac{dy}{dt}\bigg|_{t=0} = \frac{1}{\sqrt{1+e^{0}}} = 1

Therefore, the slope of the tangent line at the point where $t = 0$ is $1$ .

Now, to find the $y$ -intercept of the tangent line, we substitute $t = 0$ into the parametric equation for $y$ :

y = \int_0^0 \frac{1}{\sqrt{1+e^{2u}}} \, du = 0

This gives us the $y$ -intercept of the tangent line as $0$ .

Hence, the equation of the tangent line to the curve at the point where $t = 0$ is $y = x$ .

(b) To find the length of the curve $C$ , we will use the arc length formula:

\text{{Length}} = \int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt

where $a$ and $b$ are the corresponding values of $t$ for the range of the curve.

Given that $x = e^t$ , we have $\frac{dx}{dt} = e^t$ .

Also, we found earlier that $\frac{dy}{dt} = \frac{1}{\sqrt{1+e^{2t}}}$ . Squaring both these expressions, we get:

\left(\frac{dx}{dt}\right)^2 = e^{2t}, \quad \left(\frac{dy}{dt}\right)^2 = \frac{1}{1+e^{2t}}

Substituting these values into the arc length formula, we have:

\text{{Length}} = \int_0^1 \sqrt{e^{2t} + \frac{1}{1+e^{2t}}} \, dt

To simplify the integral, let's make a substitution:

u = e^t \quad \Rightarrow \quad du = e^t \, dt

When $t = 0$ , we have $u = e^0 = 1$ , and when $t = 1$ , we have $u = e^1 = e$ .

Therefore, the limits of integration change accordingly:

\text{{Length}} = \int_1^e \sqrt{u^2 + \frac{1}{1+u^2}} \, \frac{du}{u}

To combine the terms inside the square root, we multiply the numerator and denominator of the fraction by $(1+u^2)$ :

\text{{Length}} = \int_1^e \sqrt{\frac{u^4 + 2u^2 + 1}{u^2(1+u^2)}} \, \frac{du}{u}

Simplifying further, we get:

\text{{Length}} = \int_1^e \sqrt{\frac{(u^2+1)^2}{u^2(1+u^2)}} \, \frac{du}{u}

Canceling out the common terms in the numerator and denominator, we have:

\text{{Length}} = \int_1^e \frac{u^2+1}{u} \, \frac{du}{\sqrt{u^2(1+u^2)}}

Simplifying the integrand, we get:

\text{{Length}} = \int_1^e \frac{u^2+1}{u\sqrt{1+u^2}} \, du

Using partial fraction decomposition, we can write the integrand as:

\frac{u^2+1}{u\sqrt{1+u^2}} = \frac{1}{u} + \frac{1}{\sqrt{1+u^2}}

The integral then becomes:

\text{{Length}} = \int_1^e \left( \frac{1}{u} + \frac{1}{\sqrt{1+u^2}} \right) \, du

To evaluate this integral, we have:

\text{{Length}} = \ln|u| + \sinh^{-1}(u) \bigg|_1^e

Substituting the limits of integration, we get:

\text{{Length}} = \ln|e| + \sinh^{-1}(e) - \left( \ln|1| + \sinh^{-1}(1) \right)

Simplifying further, we find:

\text{{Length}} = 1 + \sinh^{-1}(e) - \sinh^{-1}(1)

Hence, the length of the curve $C$ is approximately $1.3119$ units.