Consider the vector function given below.
1 answer:
The tangent vector is by definition the derivative of r(t) with respect to t: <span>T' = dr/dt = <6t, tsin(t), tcos(t)> </span> <span>The unit vector T = T'/|T'| = <6t, tsin(t), tcos(t)>/sqrt(36t^2 + t^tsin(t)^2 +t^2cos(t^2)) </span> <span>T = <6t, tsin(t), tcos(t)>/(t*sqrt(37)) = <6, sin(t), cos(t)>/sqrt(37) </span> <span>Now the normal unit vector N is perpendicular to r/|r| and T. It is the second derivative of r/|r| with repsect to time </span> <span>N' = d^2r/dt^2 = <6, sin(t) + tcos(t), cos(t) - tsin(t)> </span> <span>N= N'/|N'| = <6, sin(t) + tcos(t), cos(t) - tsin(t)>/sqrt(36 + sin^2t +2tsin(t)cos(t)+t^2cos^2t + cos^2(t) -2tcos(t) sin(t) +t^2sin^2t) </span> <span>N = <6, sin(t) + tcos(t), cos(t) - tsin(t)>/sqrt(37 +t^2)</span>
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It depends on what you draw. So I'd like to say C, because you could pull up all the white marbles the first few draws, or none the first few and every other marble. It really depends.
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Answer:
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Step-by-step explanation:
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Step-by-step explanation:
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Answer:
5a + 6b + 7
step-by-step explanation:
16 + 8a -3a + 6a -9: subtract 9 from 16
7 + 8a -3a + 6b Now subtract -3a from 8a
5a + 6b + 7 is the final answer.