1)Graph of the following curve is given. Find its length. r=5cos^2(theta/2) 2)De

Question

1)Graph of the following curve is given. Find its length.
r=5cos^2(theta/2)

2)Determine whether the sequence converges or diverges.
a(n)=n(n-6)

Explanation / Answer

1) Recall that the arc-length of the curve given by r = f(?) on [a, b] is given by: L = ? v[r^2 + (dr/d?)^2] d? (from ?=a to b). With r = 5cos? and dr/dt = -5sin?, we see that the required arc-length is: L = ? v[r^2 + (dr/d?)^2] d? (from ?=a to b) = ? v[(5cos?)^2 + (-5sin?)^2] d? (from ?=0 to 3p/4) = ? v(25cos^2? + 25sin^2?) d? (from ?=0 to 3p/4) = 5 ? d? (from ?=0 to 3p/4), since cos^2? + sin^2? = 1 = 5? (evaluated from ?=0 to 3p/4) = 5(3p/4 - 0) = 15p/4, 2) You didn't state what n converges to, but I'll assume 8. 1) Divide numerator and denominator by n^3, getting: (1 / (9 + 1/n^3)) As n->8, n^3->8, and (1/n^3)->0. So as n->8, (1 / (9 + (1/n^3))-> 1/9. Answer to #1: This sequence converges to 1/9. ------------- 2) sqrt((n+9)/(36n+9)) Divide the numerator and denominator inside of the square root sign by n. sqrt( [(1 + (9/n)) / (36 + (9/n))] ) As n->8, (9/n) -> 0, so the limit of this sequence is sqrt(1/36) = 1/6. Answer to #2: this sequence converges to 1/6. --------------- 3) ln(2n^2+3)-ln(n^2+2) This is a basic theorem about logarithms: Theorem) log(x / y) = log(x) - log(y) for any base and for both x and y positive. The logarithm of z is not defined for z 8, (3/n^2) and (2/n^3)->0

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