# Consider the parametric equation #x= 10(cost+tsint)# and #y= 10(sint-tcost)#, What is the length of the curve from #0# to #((3pi)/2)#?

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To find the length of the curve from ( t = 0 ) to ( t = \frac{3\pi}{2} ), you can use the arc length formula for parametric equations:

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

Given ( x = 10(\cos t + t\sin t) ) and ( y = 10(\sin t - t\cos t) ), you need to compute ( \frac{dx}{dt} ) and ( \frac{dy}{dt} ), and then integrate from ( t = 0 ) to ( t = \frac{3\pi}{2} ).

[ \frac{dx}{dt} = -10(t\cos t) ] [ \frac{dy}{dt} = 10(t\sin t) ]

Now, plug these into the arc length formula:

[ L = \int_{0}^{\frac{3\pi}{2}} \sqrt{\left(-10(t\cos t)\right)^2 + \left(10(t\sin t)\right)^2} dt ]

[ L = \int_{0}^{\frac{3\pi}{2}} \sqrt{100t^2(\cos^2 t + \sin^2 t)} dt ]

[ L = \int_{0}^{\frac{3\pi}{2}} \sqrt{100t^2} dt ]

[ L = \int_{0}^{\frac{3\pi}{2}} 10t dt ]

[ L = 10\int_{0}^{\frac{3\pi}{2}} t dt ]

[ L = 10\left[\frac{t^2}{2}\right]_{0}^{\frac{3\pi}{2}} ]

[ L = 10\left(\frac{9\pi^2}{8}\right) ]

[ L = \frac{45\pi^2}{4} ]

So, the length of the curve from ( t = 0 ) to ( t = \frac{3\pi}{2} ) is ( \frac{45\pi^2}{4} ).

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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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