Q 71. - Calculus

Question

Let k > 0 be a constant, and let f (t) and g(t) be differentiable functions of t with continuous first derivatives for every t ∈ [a, b]. Prove that the arc length of the curve defined by the parametric equations  $x = kf \left(t\right), y = kg\left(t\right), t \in [a, b]$

is k times as long as the arc length of the curve defined by the parametric equations $x = f \left(t\right), y = g\left(t\right), t \in [a, b]$

What is the arc length of the curve defined by the equations

$x = f \left(kt\right), y = g\left(kt\right), t \in [a/k, b/k]?$

Step-by-Step Solution

Answer

$k{\int }_a^b\sqrt{{\left(f^{\text{'}}\left(kt\right)\right)}^2+{\left(g^{\text{'}}\left(kt\right)\right)}^2}dt$

Step 1: Given information

Allow $k > 0$ to be a constant, and $f \left(t\right)$ and $g\left(t\right)$ to be differentiable functions of $t$ with continuous first derivatives for all $t [a, b]$

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

Consider the two parametric curves $x=kf\left(t\right),y=kg\left(t\right),t\in [a,b]$

The objective is to find the arc length of the curves $x=k f\left(t\right), y=k g\left(t\right)$ is $k$ times the arc length of the curves defined by $x=f\left(t\right), y=g\left(t\right)$

If a curve is expressed by parametric equations $x=f\left(t\right), y=g\left(t\right)$ on the interval $[a, b]$ then the arc length is given by the formula,

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

Now consider the parametric equations, 

$x=k f\left(t\right), y=k g\left(t\right)$

Differentiate $x=k f\left(t\right)$ with respect to $t$ then,

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

Thus, 

$\frac{dx}{dt}=k{f}^{\text{'}}\left(t\right)$

Now take the parametric equation $y=\mathrm{kg}\left(t\right)$

Calculate the difference in the equation with respect to $t$

Then,

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

Substitute these values for the numbers in the equation $\left(1\right)$

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

Thus the arc length of the parametric curve defined by $x=k f\left(t\right), y=k g\left(t\right)$ for $t\in [a,b]$ is $k$ times the arc length of the parametric equation defined by $x=f\left(t\right), y=g\left(t\right)$ for $t\in [a,b]$

Consider the curve defined by $x=f(k t), y=g(k t)$ for $t\in [a,b]$

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 $[a, b]$

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

Take the parametric $x=f(k t)$ equation.

Step 4: Calculation

After that, differentiate with respect to $t$

Differentiate $x=f(k t)$ with respect to $t$ then

$$\begin{gathered} \frac{dx}{dt}=\frac{d}{dt}\left(f\right(kt\left)\right) \\ \frac{dx}{dt}=k{f}^{\text{'}}\left(kt\right) \end{gathered}$$

Take the parametric equation $y=g(k t)$ as an example.

Calculate the difference in the equation with respect to $t$

$$\begin{gathered} \frac{dy}{dt}=\frac{d}{dt}\left(g\right(kt\left)\right) \\ \frac{dy}{dt}=k\frac{d}{dt}\left(g\right(kt\left)\right) \\ \frac{dy}{dt}=k{g}^{\text{'}}\left(kt\right) \end{gathered}$$

Substitute the values in (1) 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 $={\int }_a^b\sqrt{{\left(k{f}^{\text{'}}\left(kt\right)\right)}^2+{\left(k{g}^{\text{'}}\left(kt\right)\right)}^2}dt \left[since\frac{dx}{dt}=k{f}^{\text{'}}\left(kt\right),\frac{dy}{dt}=k{g}^{\text{'}}\left(kt\right)\right]$

$={\int }_a^b\sqrt{k^2{\left(f^{\text{'}}\left(kt\right)\right)}^2+k^2{\left(g^{\text{'}}\left(kt\right)\right)}^2}dt$

Thus,

Length of the curve $=k{\int }_a^b\sqrt{{\left(f^{\text{'}}\left(kt\right)\right)}^2+{\left(g^{\text{'}}\left(kt\right)\right)}^2}dt$

Therefore the arc length of the parametric curves is $x=f(k t), y=g(k t)$ on the interval is $[a, b]$ is $k{\int }_a^b\sqrt{{\left(f^{\text{'}}\left(kt\right)\right)}^2+{\left(g^{\text{'}}\left(kt\right)\right)}^2}dt$

Therefore, the answer is $k$ times the arc length, $k{\int }_a^b\sqrt{{\left(f^{\text{'}}\left(kt\right)\right)}^2+{\left(g^{\text{'}}\left(kt\right)\right)}^2}dt$