What is the arclength of #f(t) = (t^3-t+55,t^2-1)# on #t in [2,3]#?

Answer 1

#L=56/3+2/9sum_(n=1)^oo((1/2),(n))2^nint_6^9(1/(u-1)-1/(u+1))^(2n-1)du# units.

#f(t)=(t^3-t+55,t^2-1)#
#f'(t)=(3t^2-1,2t)#

Arclength is given by:

#L=int_2^3sqrt((3t^2-1)^2+(2t)^2)dt#

Expand the squares:

#L=int_2^3sqrt(9t^4-2t^2+1)dt#

Complete the square in the square root:

#L=1/3int_2^3sqrt((9t^2-1)^2+8)dt#

Rearrange:

#L=1/3int_2^3(9t^2-1)sqrt(1+8/(9t^2-1)^2)dt#
For #t in [2,3]#, #8/(9t^2-1)^2<1#. Take the series expansion of the square root:
#L=1/3int_2^3(9t^2-1){sum_(n=0)^oo((1/2),(n))(8/(9t^2-1)^2)^n}dt#
Isolate the #n=0# term and simplify:
#L=1/3int_2^3(9t^2-1)dt+1/3sum_(n=1)^oo((1/2),(n))8^nint_2^3 1/(9t^2-1)^(2n-1)dt#

Apply partial fraction decomposition:

#L=1/3[3t^3-t]_ 2^3+2/3sum_(n=1)^oo((1/2),(n))2^nint_2^3(1/(3t-1)-1/(3t+1))^(2n-1)dt#
For ease of reading, apply the substitution #3t=u#:
#L=56/3+2/9sum_(n=1)^oo((1/2),(n))2^nint_6^9(1/(u-1)-1/(u+1))^(2n-1)du#
The #n=1# case is trivial.
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Answer 2

To find the arclength of the curve defined by ( f(t) = (t^3 - t + 55, t^2 - 1) ) on the interval ( t ) in ([2,3]), we use the formula for arc length:

[ L = \int_{a}^{b} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} dt ]

where (a = 2) and (b = 3).

Differentiating (x(t) = t^3 - t + 55) and (y(t) = t^2 - 1) with respect to (t), we get:

[ \frac{dx}{dt} = 3t^2 - 1 ] [ \frac{dy}{dt} = 2t ]

Substituting these derivatives into the formula, we get:

[ L = \int_{2}^{3} \sqrt{(3t^2 - 1)^2 + (2t)^2} dt ]

Simplify and compute the integral:

[ L = \int_{2}^{3} \sqrt{9t^4 - 6t^2 + 1 + 4t^2} dt ] [ L = \int_{2}^{3} \sqrt{9t^4 - 2t^2 + 1} dt ]

[ L = \int_{2}^{3} \sqrt{(3t^2 - 1)^2} dt ] [ L = \int_{2}^{3} (3t^2 - 1) dt ] [ L = \left[\frac{t^3}{3} - t\right]_{2}^{3} ] [ L = \left[\frac{3^3}{3} - 3\right] - \left[\frac{2^3}{3} - 2\right] ] [ L = \left[9 - 3\right] - \left[\frac{8}{3} - 2\right] ] [ L = 6 - \left(\frac{8}{3} - 2\right) ] [ L = 6 - \frac{2}{3} ] [ L = \frac{16}{3} ]

So, the arclength of ( f(t) ) on ( t ) in ([2,3]) is ( \frac{16}{3} ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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