Q 70. - Calculus

Question

Let c and d be constants, and for t ∈ [a, b] let f (t) and g(t) be differentiable functions with continuous first derivatives. Prove that the arc length of the parametric curve given by x = f (t), y(t) = g(t) for t ∈ [a, b] is equal to

the arc length of the parametric curve defined by x = c + f (t), y(t) = d + g(t) for t ∈ [a, b] for every c and d in R.

Step-by-Step Solution

Answer

The arc lengths are Equal.

Step 1: Given information

$x=c+f\left(t\right),y=d+g\left(t\right),t\in [a,b]$

Step 2: Concept

The formula used: Arc length ${\int }_a^b\sqrt{{\left(\frac{dx}{dt}\right)}^2+{\left(\frac{dy}{dt}\right)}^2}dt\text{ or }{\int }_a^b\sqrt{{\left(f^{\text{'}}\left(t\right)\right)}^2+{\left(g^{\text{'}}\left(t\right)\right)}^2}dt$

Step 3: Calculation

$x=c+f\left(t\right),y=d+g\left(t\right),t\in [a,b]$

The goal is to determine whether the curves' arc lengths are equivalent to the arc lengths indicated by $x=f\left(t\right), y=g\left(t\right)$

The arc length is provided by the formula if a curve is described by parametric equations $x=f\left(t\right), y=g\left(t\right)$ on the interval $[a, b]$

${\int }_a^b\sqrt{{\left(\frac{dx}{dt}\right)}^2+{\left(\frac{dy}{dt}\right)}^2}dt\text{ or }{\int }_a^b\sqrt{{\left(f^{\text{'}}\left(t\right)\right)}^2+{\left(g^{\text{'}}\left(t\right)\right)}^2}dt\dots \left(1\right)$

Now consider the parametric equations,

$x=c+f\left(t\right), y=d+g\left(t\right)$

Differentiate $x=c+f\left(t\right)$ with respect to $t$ then

$$\begin{gathered} \frac{dx}{dt}=\frac{d}{dt}(c+f(t\left)\right) \\ \frac{dx}{dt}=\frac{d}{dt}c+\frac{d}{dt}f\left(t\right) \end{gathered}$$

Thus,

$$\begin{gathered} \frac{dx}{dt}=0+\frac{d}{dt}f\left(t\right) \\ \frac{dx}{dt}=f^{\text{'}}\left(t\right) \end{gathered}$$

Now take the parametric equation $y=d+g\left(t\right)$

Differentiate the equation with respect to $t$

Then,

$$\begin{gathered} \frac{dy}{dt}=\frac{d}{dt}(d+g(t\left)\right) \\ \frac{dy}{dt}=\frac{d}{dt}d+\frac{d}{dt}g\left(t\right) \\ \frac{dy}{dt}=0+\frac{d}{dt}g\left(t\right) \\ \frac{dy}{dt}=g^{\text{'}}\left(t\right) \end{gathered}$$

Step 4: Calculation

Substitute the values in $\left(1\right)$ as follows:

Length of the curve $={\int }_a^b\sqrt{{\left(\frac{dx}{dt}\right)}^2+{\left(\frac{dy}{dt}\right)}^2}dt$

Length of the curve $={\int }_a^b\sqrt{{\left(f^{\text{'}}\left(t\right)\right)}^2+{\left(g^{\text{'}}\left(t\right)\right)}^2}dt$ since $\left.f^{\text{'}}\left(t\right)=\frac{dx}{dt},g^{\text{'}}\left(t\right)=\frac{dy}{dt}\right]$

Thus the arc length of the parametric curve defined by $x=f\left(t\right), y=g\left(t\right)$ for $t\in [a,b]$ is equal to the arc length of the parametric equation defined by $x=c+f\left(t\right), y=d+g\left(t\right)$ for $t\in [a,b]$ Therefore, the arc lengths are Equal.