Find the curve’s unit tangent vector. Also, find the length of the indicated portion of the curve. r(t)=(2cos⁡t)i+(2sin⁡t)j+5tk,0≤t≤π

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Find the curve’s unit tangent vector. Also, find the length of the indicated portion of the curve.  r(t)=(2cos⁡t)i+(2sin⁡t)j+5tk,0≤t≤π

2k1enyvp · Accepted Answer

r=(2cos⁡t)i+(2sin⁡t)j+5tk  ⇒v=(−2sin⁡t)i+(2cos⁡t)j+5  ⇒|v|=(−2sin⁡t)2+(2cos⁡t)2+(5)2  =4sin2⁡t+4cos2⁡t+5=3  T=v|v|=(−23sin⁡t)i+(23cos⁡t)j+53k  and Length =∫0π|v|dt=∫0π3dt=|3t|0π=3π  Result:  T=&amp;lt;−23sin⁡t,23cos⁡t,53&amp;gt;,L=3π

Nick Camelot · Accepted Answer

Answer: 3πExplanation: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&#039;(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&#039;(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&#039;(t) is 3 for all values of t.To find the unit tangent vector T(t), we divide the vector r&#039;(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=∫0π‖r′(t)‖dt=∫0π3dt=3t|0π=3π−0=3π.Therefore, the length of the indicated portion of the curve is 3π units.

madeleinejames20 · Accepted Answer

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)𝐤Step 2: Simplify the derivative expressions:d𝐫(t)dt=−2sint𝐢+2cost𝐣+5𝐤Step 3: Find the magnitude of the derivative vector:‖d𝐫(t)dt‖=(−2sint)2+(2cost)2+(5)2Step 4: Simplify the magnitude expression:‖d𝐫(t)dt‖=4sin2t+4cos2t+5Step 5: Find the unit tangent vector by dividing the derivative vector by its magnitude:𝐓(t)=d𝐫(t)dt‖d𝐫(t)dt‖=−2sint𝐢+2cost𝐣+5𝐤4sin2t+4cos2t+5Step 6: Calculate the length of the indicated portion of the curve by integrating the magnitude of the derivative vector over the given interval:Length=∫0π‖d𝐫(t)dt‖dt=∫0π4sin2t+4cos2t+5dt

Eliza Beth13 · Accepted Answer

First, let&#039;s find the derivative of the position vector r(t) with respect to t:r(t)=(2cost)i+(2sint)j+(5t)kDifferentiating 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+5kThe unit tangent vector, T(t), is obtained by normalizing the derivative vector, r&#039;(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}=∫0π‖r′(t)‖dtNow let&#039;s compute these values:First, we calculate the magnitude of the derivative vector:‖r′(t)‖=(−2sint)2+(2cost)2+(5)2Simplifying, we get:‖r′(t)‖=4sin2t+4cos2t+5Next, we find the unit tangent vector:T(t)=(−2sint)i+(2cost)j+5k4sin2t+4cos2t+5Finally, we calculate the length of the indicated portion of the curve:{Length}=∫0π4sin2t+4cos2t+5dtWe can solve this integral to find the length of the curve.

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

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