Find the curve’s unit tangent vector. Also, find the length of the indicated portion of the curve. r(t)=(2\cos t)i+(2\sin t)j+\sqrt{5}tk, 0\leq t\leq\pi

Yasmin

Yasmin

Answered question

2021-06-07

Find the curve’s unit tangent vector. Also, find the length of the indicated portion of the curve.
r(t)=(2cost)i+(2sint)j+5tk,0tπ

Answer & Explanation

2k1enyvp

2k1enyvp

Skilled2021-06-08Added 94 answers

r=(2cost)i+(2sint)j+5tk
v=(2sint)i+(2cost)j+5
|v|=(2sint)2+(2cost)2+(5)2
=4sin2t+4cos2t+5=3
T=v|v|=(23sint)i+(23cost)j+53k
and Length =0π|v|dt=0π3dt=|3t|0π=3π
Result:
T=<23sint,23cost,53>,L=3π
Nick Camelot

Nick Camelot

Skilled2023-06-13Added 164 answers

Answer: 3π
Explanation:
To find the unit tangent vector, we first need to differentiate the vector function r(t) with respect to t. The derivative of r(t) is obtained by differentiating each component separately:
r(t)=(ddt(2cost))i+(ddt(2sint))j+(ddt(5t))k=(2sint)i+(2cost)j+5k.
The length of the vector r'(t) represents the magnitude of the velocity vector, which is the rate of change of position with respect to time. To find the length, we calculate the magnitude of r'(t):
r(t)=(2sint)2+(2cost)2+(5)2=4sin2t+4cos2t+5=4(sin2t+cos2t)+5=4+5=9=3.
Thus, the length of the vector r'(t) is 3 for all values of t.
To find the unit tangent vector T(t), we divide the vector r'(t) by its magnitude:
T(t)=r(t)r(t)=(2sint)i+(2cost)j+5k3.
Therefore, the unit tangent vector T(t) is 2sint3i+2cost3j+53k.
To find the length of the indicated portion of the curve, we integrate the magnitude of the velocity vector from t = 0 to t = π:
Length=0πr(t)dt=0π3dt=3t|0π=3π0=3π.
Therefore, the length of the indicated portion of the curve is 3π units.
madeleinejames20

madeleinejames20

Skilled2023-06-13Added 165 answers

Step 1: Find the derivative of the position vector 𝐫(t) with respect to t:
d𝐫(t)dt=(d(2cost)dt)𝐢+(d(2sint)dt)𝐣+(d(5t)dt)𝐤
Step 2: Simplify the derivative expressions:
d𝐫(t)dt=2sint𝐢+2cost𝐣+5𝐤
Step 3: Find the magnitude of the derivative vector:
d𝐫(t)dt=(2sint)2+(2cost)2+(5)2
Step 4: Simplify the magnitude expression:
d𝐫(t)dt=4sin2t+4cos2t+5
Step 5: Find the unit tangent vector by dividing the derivative vector by its magnitude:
𝐓(t)=d𝐫(t)dtd𝐫(t)dt=2sint𝐢+2cost𝐣+5𝐤4sin2t+4cos2t+5
Step 6: Calculate the length of the indicated portion of the curve by integrating the magnitude of the derivative vector over the given interval:
Length=0πd𝐫(t)dtdt=0π4sin2t+4cos2t+5dt
Eliza Beth13

Eliza Beth13

Skilled2023-06-13Added 130 answers

First, let's find the derivative of the position vector r(t) with respect to t:
r(t)=(2cost)i+(2sint)j+(5t)k
Differentiating each component with respect to t, we have:
r(t)=ddt[(2cost)i+(2sint)j+(5t)k]
Taking the derivatives of each term separately, we get:
r(t)=(2sint)i+(2cost)j+5k
The unit tangent vector, T(t), is obtained by normalizing the derivative vector, r'(t), which is given by:
T(t)=r(t)r(t)
To find the length of the indicated portion of the curve, we need to integrate the magnitude of the derivative vector over the interval [0, π]:
{Length}=0πr(t)dt
Now let's compute these values:
First, we calculate the magnitude of the derivative vector:
r(t)=(2sint)2+(2cost)2+(5)2
Simplifying, we get:
r(t)=4sin2t+4cos2t+5
Next, we find the unit tangent vector:
T(t)=(2sint)i+(2cost)j+5k4sin2t+4cos2t+5
Finally, we calculate the length of the indicated portion of the curve:
{Length}=0π4sin2t+4cos2t+5dt
We can solve this integral to find the length of the curve.

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