Determine the integral that can be used to find the arc length of the curve defined by the parametric equations [tex]$x(t)=-9 t^2-6 t-5$[/tex] and [tex]$y(t)=7 t^3-2 t^2+8 t+4$[/tex] from [tex]$1 \leq t \leq 2$[/tex].
Answer (2)
Find the derivatives of the parametric equations: d t d x = − 18 t − 6 and d t d y = 21 t 2 − 4 t + 8 .
Square the derivatives: ( d t d x ) 2 = 324 t 2 + 216 t + 36 and ( d t d y ) 2 = 441 t 4 − 168 t 3 + 352 t 2 − 64 t + 64 .
Add the squares: ( d t d x ) 2 + ( d t d y ) 2 = 441 t 4 − 168 t 3 + 676 t 2 + 152 t + 100 .
Set up the arc length integral: s = ∫ 1 2 441 t 4 − 168 t 3 + 676 t 2 + 152 t + 100 d t .
Explanation
Problem Setup We are given the parametric equations x ( t ) = − 9 t 2 − 6 t − 5 and y ( t ) = 7 t 3 − 2 t 2 + 8 t + 4 , and we want to find the arc length of the curve from $1
\le t \le 2$. The formula for the arc length of a parametric curve is given by
s = ∫ a b ( d t d x ) 2 + ( d t d y ) 2 d t
where a and b are the limits of the parameter t . In this case, a = 1 and b = 2 .
Finding Derivatives First, we need to find the derivatives of x ( t ) and y ( t ) with respect to t .
d t d x = d t d ( − 9 t 2 − 6 t − 5 ) = − 18 t − 6
d t d y = d t d ( 7 t 3 − 2 t 2 + 8 t + 4 ) = 21 t 2 − 4 t + 8
Squaring Derivatives Next, we need to square these derivatives:
( d t d x ) 2 = ( − 18 t − 6 ) 2 = ( 18 t + 6 ) 2 = 324 t 2 + 216 t + 36
( d t d y ) 2 = ( 21 t 2 − 4 t + 8 ) 2 = 441 t 4 − 168 t 3 + 352 t 2 − 64 t + 64
Sum of Squares Now, we add the squares of the derivatives:
( d t d x ) 2 + ( d t d y ) 2 = ( 324 t 2 + 216 t + 36 ) + ( 441 t 4 − 168 t 3 + 352 t 2 − 64 t + 64 )
( d t d x ) 2 + ( d t d y ) 2 = 441 t 4 − 168 t 3 + 676 t 2 + 152 t + 100
Square Root of Sum Then, we take the square root of the sum:
( d t d x ) 2 + ( d t d y ) 2 = 441 t 4 − 168 t 3 + 676 t 2 + 152 t + 100
Arc Length Integral Finally, we set up the arc length integral:
s = ∫ 1 2 441 t 4 − 168 t 3 + 676 t 2 + 152 t + 100 d t
Final Answer Therefore, the integral that can be used to find the arc length of the curve is
s = ∫ 1 2 441 t 4 − 168 t 3 + 676 t 2 + 152 t + 100 d t
Examples
Arc length calculations are crucial in various fields. For instance, in manufacturing, determining the length of a curved component, like a car's exhaust pipe, ensures accurate material usage and fit. In robotics, calculating the path length of a robot arm helps optimize movements and conserve energy. Even in architecture, finding the length of a curved facade is essential for precise material ordering and construction planning. Understanding arc length allows engineers and designers to create efficient and accurate designs.
The integral for finding the arc length of the curve defined by the given parametric equations from t = 1 to t = 2 is s = ∫ 1 2 441 t 4 − 168 t 3 + 676 t 2 + 152 t + 100 d t .
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