How do you find the length of the curve #x=3t+1, y=2-4t, 0<=t<=1#?
Given that
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To find the length of the curve defined by ( x = 3t + 1 ) and ( y = 2 - 4t ) for ( 0 \leq t \leq 1 ), you can use the formula for the arc length of a parametric curve:
[ s = \int_{a}^{b} \sqrt{\left(\frac{{dx}}{{dt}}\right)^2 + \left(\frac{{dy}}{{dt}}\right)^2} , dt ]
where ( a ) and ( b ) are the limits of the parameter ( t ). In this case, ( a = 0 ) and ( b = 1 ).
Plugging in the given parametric equations:
[ \frac{{dx}}{{dt}} = 3 ] [ \frac{{dy}}{{dt}} = -4 ]
Substitute these derivatives into the formula and integrate from 0 to 1:
[ s = \int_{0}^{1} \sqrt{(3)^2 + (-4)^2} , dt ]
[ s = \int_{0}^{1} \sqrt{9 + 16} , dt ]
[ s = \int_{0}^{1} \sqrt{25} , dt ]
[ s = \int_{0}^{1} 5 , dt ]
[ s = 5t \Bigg|_{0}^{1} ]
[ s = 5(1) - 5(0) ]
[ s = 5 ]
Therefore, the length of the curve is ( 5 ) units.
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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.
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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