42.) Do the following for the curve y = -4x^2 on [-2, 2] a.) Set up an integral

Question

42.) Do the following for the curve y = -4x^2 on [-2, 2] a.) Set up an integral for the length of the curve b.) Use your grapher's or computer's integral evaluator to find the curve's length numerically (just use wolfram alpha in my opinion)

Explanation / Answer

Since the ellipse is symmetric about both the x and y axes, it suffices to find the length of the ellipse in the first quadrant, and then quadruple that result. For x, y > 0, we have y^2 = b^2 (1 - x^2/a^2) = (b/a)^2 (a^2 - x^2) ==> y = (b/a) v(a^2 - x^2). Since dy/dx = (b/a) * -x/v(a^2 - x^2), the arc length equals 4 * ?(x = 0 to a) v[1 + (dy/dx)^2] dx = 4 * ?(x = 0 to a) v[1 + ((b/a) v(a^2 - x^2))^2] dx = 4 * ?(x = 0 to a) v[1 + (b/a)^2 (a^2 - x^2)] dx = 4 * ?(x = 0 to a) v[1 + b^2 (1 - a^2 x^2)] dx. I hope this helps! You are on the right track, but: dy/dt = (4/3)(3/2)vt = 2vt (not 8vt). So, the integrand should be: v[(dx/dt)^2 + (dy/dt)^2] = v[(1 - 2t)^2 + (2vt)^2] = v(4t^2 - 4t + 1 + 4t) = v(4t^2 + 1). So, use your calculator to calculate ? v(4t^2 + 1) dt (from t=1 to 2) and you'll get the desired answer. Alternatively, this can be integrated by hand with a trigonometric substitution. I hope this helps!

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