3. Consider the curve7 (t) = (1, t², „t³)Find the arc length from t = 0 to t = V5. Note: you might need "u-substitution"to finish evaluating the integral. 2. Consider the following vector-valued function7(t) = (t sint + cos t, sin t – t cos t, v6t?)|Compute the arc length from t = 0 to t = T.

Answered: 3. Consider the curve 7 (t) = (1, t²,…

Calculus: Early Transcendentals

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Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

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3. Consider the curve
7 (t) = (1, t², „t³)
Find the arc length from t = 0 to t = V5. Note: you might need "u-substitution"
to finish evaluating the integral.

2. Consider the following vector-valued function
7(t) = (t sint + cos t, sin t – t cos t, v6t?)
|
Compute the arc length from t = 0 to t = T.

Expert Solution

Step 1

As we know that the expression of the arc length for the parametric curve is:

$L = \int_{t_{1}}^{t_{2}} \sqrt{{(\frac{d x}{d t})}^{2} + {(\frac{d y}{d t})}^{2} + {(\frac{d z}{d t})}^{2}} d t$

Given:

$\vec{r} (t) = <1, t^{2}, \frac{1}{3} t^{3}>$ & $t_{1} = 0, t_{2} = \sqrt{5}$

It means that:

$x = 1, y = t^{2}, z = \frac{1}{3} t^{2} N o w, \frac{d x}{d t} = \frac{d}{d t} (1) = 0 \frac{d y}{d t} = \frac{d}{d t} (t^{2}) = 2 t \frac{d z}{d t} = \frac{d}{d t} (\frac{t^{3}}{3}) = \frac{3 t^{2}}{3} = t^{2}$

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3. Consider the curve 7 (t) = (1, t², *) Find the arc length from t = 0 to t = V5. Note: you might need "u-substitution" to finish evaluating the integral.