Find the length of the curve. r(t)=&lt;t,3cos⁡t,3sin⁡t&gt;,−5≤t≤5

Step 1We know that the arc length L of a curve r(t) from -a to a is,L=∫−aa|r′(t)|⋅dtTherefore, let us find the derivative of given vector function component-wise, we get,r(t)=&lt;t,3cos⁡t,3sin⁡t&gt;,−5≤t≤5r′(t)=&lt;1,−3sin⁡t,3cos⁡t&gt;Step 2 Plug into the arc length formulaL=∫{−5}5|r′(t)|dt=∫{−5}512+(−3sin⁡t)2+(3cos⁡t)2dt=∫{−5}51+9sin2t+9cos2tdt=∫{−5}51+9(sin2t+cos2t)dt=∫{−5}51+9(1)dt=∫{−5}510dt=[10t]{−5}5=510−(−510)=1010The length of curve r(t)=⟨t,3cos⁡t,sin⁡t⟩ from -5 to 5 is 1010

Resolved: "Find the length of the curve. r(t)=<t,3cos⁡t,3sin⁡t>,−5≤t≤5"

High School
Geometry
Trigonometry
Trigonometric Functions

postillan4

Answered question

2021-10-28

Find the length of the curve.

r (t) = < t, 3 \cos t, 3 \sin t >, - 5 \leq t \leq 5

Answer & Explanation

opsadnojD

Skilled2021-10-29Added 95 answers

Step 1
We know that the arc length L of a curve r(t) from -a to a is,

L = \int_{- a}^{a} | r^{'} (t) | \cdot d t

Therefore, let us find the derivative of given vector function component-wise, we get,

r (t) = < t, 3 \cos t, 3 \sin t >, - 5 \leq t \leq 5

r^{'} (t) = < 1, - 3 \sin t, 3 \cos t >

Step 2
Plug into the arc length formula

L = \int_{{- 5}}^{5} | r^{'} (t) | d t

= \int_{{- 5}}^{5} \sqrt{1^{2} + {(- 3 \sin t)}^{2} + {(3 \cos t)}^{2}} d t

= \int_{{- 5}}^{5} \sqrt{1 + 9 \sin^{2} t + 9 \cos^{2} t} d t

= \int_{{- 5}}^{5} \sqrt{1 + 9 (\sin^{2} t + \cos^{2} t)} d t

= \int_{{- 5}}^{5} \sqrt{1 + 9 (1)} d t

= \int_{{- 5}}^{5} \sqrt{10} d t

= {[\sqrt{10 t}]}_{{- 5}}^{5}

= 5 \sqrt{10} - (- 5 \sqrt{10})

= 10 \sqrt{10}

The length of curve

r (t) = ⟨ t, 3 \cos t, \sin t ⟩

from -5 to 5 is

10 \sqrt{10}

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Trigonometry

Didn't find what you were looking for?