Find Length of Spiral Given by r(t) Equation

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To find the length of the spiral defined by r(t) = (e^-t cos t, e^-t sin t), a line integral approach is used. The arc length formula for parametric curves is applied, leading to the expression L = ∫√(x'(t)^2 + y'(t)^2) dt. After calculating the derivatives and simplifying, the integral reduces to L = ∫√2 e^-t dt. Evaluating this from t = 0 to infinity results in a total length of √2 units for the spiral.
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a spiral given by r(t) = (e^-t cos t, e^-t sin t) , t is greater than or equal to 0, how would i find the length of the spiral?
thanks anyone!
 
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do a line integral. length = S(dr) from t = 0 on out to infinity in this case. dr is sqrt(dx^2+dy^2) (assuming this is two and not three dimensions) where dx is the derivative of the i component of r(t) and dy is the derivative of the j component of r(t). Oh sorry S also means integral in my bastardized way of writing math with basic text.
 
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To find the length of the spiral given by the equation r(t) = (e^-t cos t, e^-t sin t), we can use the arc length formula for parametric curves. This formula is given by:

L = ∫√(x'(t)^2 + y'(t)^2) dt

Where x'(t) and y'(t) are the derivatives of the x and y components of the curve, respectively.

In this case, we have x(t) = e^-t cos t and y(t) = e^-t sin t. Taking the derivatives, we get:

x'(t) = -e^-t cos t - e^-t sin t
y'(t) = -e^-t sin t + e^-t cos t

Plugging these into the arc length formula, we get:

L = ∫√((-e^-t cos t - e^-t sin t)^2 + (-e^-t sin t + e^-t cos t)^2) dt

Simplifying and combining like terms, we get:

L = ∫√(2e^-2t) dt

Integrating, we get:

L = ∫√2 e^-t dt = -√2 e^-t + C

Evaluating this from t = 0 to t = ∞, we get:

L = -√2 e^-∞ + √2 e^0 = √2

Therefore, the length of the spiral is √2 units.
 
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