# Examine what is length of curve given by equations x=4cost, y=4sint, 0<=t<=360 degrees?

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### 1 Answer

You need to use the following formula that gives the arclength of a curve given by parametric equations, such that:

`L = int_a^b sqrt(((dx)/(dt))^2 + ((dy)/(dt))^2) dt`

The problem provides the following parametric equations, such that:

`x = 4cos t => dx = -4sin t dt => (dx)/(dt) = -4sin t => ((dx)/(dt))^2 = 16sin^2 t`

`y = 4sin t => dy = 4cos t dt => (dy)/(dt) = 4cos t => ((dy)/(dt))^2 = 16cos^2 t`

Replacing `16sin^2 t` and `16cos^2 t` for `((dx)/(dt))^2 + ((dy)/(dt))^2 ` yields:

`L = int_0^(2pi) sqrt(16sin^2 t + 16cos^2 t) dt`

Factoring out 16 yields:

`L = int_0^(2pi) sqrt(16(sin^2 t + cos^2 t)) dt`

`L = int_0^(2pi) 4sqrt(sin^2 t + cos^2 t) dt`

Using Pythagorean trigonometric identity `sin^2 t + cos^2 t = 1 ` yields:

`L = int_0^(2pi) 4sqrt 1 dt`

Taking out the constant yields:

`L = 4int_0^(2pi) dt => L = 4t|_0^(2pi)`

Using the fundamental theorem of calculus, yields:

`L = 4(2pi - 0) => L = 8pi`

**Hence, evaluating the arc length of the given curve, under the given conditions, yields **`L = 8pi.`

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