Q. 22 - Calculus

Question

Given a vector-valued function r(t) with domain $\mathrm{ℝ},$ what is the relationship between the graph of r(t) and the graph of kr(t), where k > 1 is a scalar?

Step-by-Step Solution

Answer

The relationship between the graph of r(t) and the graph of kr(t), where k > 1 is a scalar is that the arc length of the function will become k times.

Step 1. Given Information.

The given vector-valued function r(t) has a domain $\mathrm{ℝ}.$

Step 2. Explanation.

Let a vector-valued function r(t) in $\mathrm{ℝ}$ is $r\left(t\right)=\left(\cos t, \sin t, t\right), t\in \left[0,2\mathrm{\pi }\right].$

So, the graph of r(t) is

The arc length of the curve we get by the graph is $l=2\sqrt{2}\mathrm{\pi }.$

Step 3. Explanation.

Now, multiply by scalar k, so $kr\left(t\right)=\left(k\cos t, k\sin t, kt\right), t\in \left[0,2\mathrm{\pi }\right].$

Take $k=7, \mathrm{so} kr\left(t\right)=\left(7\cos t,7\sin t, 7t\right).$

So, the graph is

Now, the arc length of the curve by the graph is $l=14\sqrt{2}\mathrm{\pi }.$

So, we can depict by the graphs and arc lengths that the arc length of the function will become k times.