Determine the integral that can be used to find the arc length of the curve defined by the parametric equations [tex]x(t)=6 t^2-4 t+6[/tex] and [tex]y(t)=4 t^2+6 t+3[/tex] from [tex]1 \leq t \leq 9[/tex].
Provide your answer below:
[tex]s=\sqrt{\square} \square dt[/tex]
Answer (2)
Find the derivatives of the parametric equations: d t d x = 12 t − 4 and d t d y = 8 t + 6 .
Square the derivatives: ( d t d x ) 2 = 144 t 2 − 96 t + 16 and ( d t d y ) 2 = 64 t 2 + 96 t + 36 .
Add the squares: ( d t d x ) 2 + ( d t d y ) 2 = 208 t 2 + 52 .
The arc length integral is: s = ∫ 1 9 208 t 2 + 52 d t , so the answer is s = ∫ 1 9 208 t 2 + 52 d t .
Explanation
Problem Setup We are given the parametric equations x ( t ) = 6 t 2 − 4 t + 6 and y ( t ) = 4 t 2 + 6 t + 3 , and we want to find the arc length of the curve from 1 ≤ t ≤ 9 . 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 integration for the parameter t . In this case, a = 1 and b = 9 .
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 ( 6 t 2 − 4 t + 6 ) = 12 t − 4 d t d y = d t d ( 4 t 2 + 6 t + 3 ) = 8 t + 6
Squaring the Derivatives Next, we need to square these derivatives: ( d t d x ) 2 = ( 12 t − 4 ) 2 = 144 t 2 − 96 t + 16 ( d t d y ) 2 = ( 8 t + 6 ) 2 = 64 t 2 + 96 t + 36
Adding the Squares Now, we add the squares of the derivatives: ( d t d x ) 2 + ( d t d y ) 2 = ( 144 t 2 − 96 t + 16 ) + ( 64 t 2 + 96 t + 36 ) = 208 t 2 + 52
Taking the Square Root Then, we take the square root of the sum: ( d t d x ) 2 + ( d t d y ) 2 = 208 t 2 + 52 = 52 ( 4 t 2 + 1 ) = 2 13 ( 4 t 2 + 1 ) = 2 52 t 2 + 13
Setting up the Arc Length Integral Finally, we set up the arc length integral: s = ∫ 1 9 208 t 2 + 52 d t = ∫ 1 9 2 52 t 2 + 13 d t Thus, the integral that can be used to find the arc length is s = ∫ 1 9 208 t 2 + 52 d t = ∫ 1 9 2 52 t 2 + 13 d t
Final Answer The integral to find the arc length is: s = ∫ 1 9 208 t 2 + 52 d t We can also write it as: s = ∫ 1 9 2 52 t 2 + 13 d t So, the answer is: s = ∫ 1 9 208 t 2 + 52 d t
Examples
Imagine you are designing a roller coaster. The track's length is crucial for determining the ride's duration and speed. By using parametric equations to define the track's curves and then applying the arc length integral, you can precisely calculate the length of the track. This ensures the roller coaster provides the desired thrill and safety for its riders. The arc length calculation helps engineers to accurately plan and construct complex curved paths in various real-world applications.
To find the arc length of the curve defined by the parametric equations x ( t ) = 6 t 2 − 4 t + 6 and y ( t ) = 4 t 2 + 6 t + 3 from t = 1 to t = 9 , the integral is expressed as s = ∫ 1 9 208 t 2 + 52 d t . This integral represents the arc length of the curve over the specified interval.
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