Consider the curve a(t) - (2t, t^2, log t) on I: t > 0. Show that this curve passes through the points p = (2,1,0) and q = (4,4,log2) and find its arc length between these points.

Cristofer Graves 2022-07-29 Answered
Consider the curve a ( t ) ( 2 t , t 2 , log t )  on  I : t > 0. Show that this curve passes through the points p = ( 2 , 1 , 0 )  and  q = ( 4 , 4 , log 2 ) and find its arc length between these points.
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Answers (1)

iljovskint
Answered 2022-07-30 Author has 18 answers
Consider the equation of the curve
a ( t ) = 2 t , t 2 . log t , t 0
To show that the curve a ( t ) = 2 t , t 2 . log t is passing
through the points P=(2,1,0) and Q=(4,4, log2)
Compare the equation of the curve with the general curve
equation x ( t ) , y ( t ) , z ( t ) gives
x ( t ) = 2 t , y ( t ) = t 2 , z ( t ) = log t
Now substitute the point P=(2,1,0) in x, y, z gives
2 = 2 t , 1 = t 2 , 0 = log t
Solving these equations gives t=1
Thus, for t=1, the curve passes throug the point P
Now substitute the point Q=(4,4, log 2) in x,y,z gives
4 = 2 t , 4 = t 2 , log 2 = log t
Solving these equations gives t=2
Thus, for t=2, the curve passes throught the point Q
Hence, the curve passes through the point P = (2,1,0) and Q = (4,4, log 2)
Consider the equation of the curve
a ( t ) = 2 t , t 2 . log t , t 0
Find the arc lenght of curve.
From the point P = (2,1,0), we get
( 2 t , t 2 log t ) = ( 2 , 1 , 0 ) 2 t = 2 t = 1
From the point Q = (4,4, log2), we get
( 2 t , t 2 log t ) = ( 4 , 4 , log 2 ) 2 t = 4 t = 2
Thus, the limits of t are 1 t 2
The length of the curve x=f(t), y=g(t), z=h(t)
a t b i s L = a b [ d x d t 2 ] + [ d y d t 2 ] + [ d z d t 2 ] d t
Differentiate x and y with respect to t
d x d t = d d t ( 2 t ) = 2 d y d t = d d t ( t 2 ) = 2 t d z d t = d d t ( log t ) = 1 t
The length of the curve is
a t b i s L = 1 2 [ d x d t 2 ] + [ d y d t 2 ] + [ d z d t 2 ] d t = 1 2 [ 2 ] 2 + [ 2 ] 2 = ( 1 t ) 2 d t = 1 2 4 + 4 t 2 + 1 t 2 d t = 1 2 a t 2 + 4 t 4 + 1 t 2 d t = 1 2 ( 2 t 2 + 1 ) 2 t 2 d t = 1 2 2 t 2 + 1 t d t = 1 2 2 t + 1 t d t = [ t 2 ln t ] 1 2 = [ 2 2 + ln 2 1 2 ln 1 ] = [ 4 + ln 2 1 0 ] = 3 + ln 2
Hence the length of the curve is L=3+ln2
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