Asked by Sara
Please help me with the following 2 questions.
1. Find the area of the curved surface of a right-circular cone of radius 6 and height 4 by rotating the straight line segment from (0,0) to (6,4) about the y-axis.
2. Use the Maclaurin polynomial of degree 4 to approximate sin(0.2). The formula given is the sum between k = 0 and infinity of [-1^k]/(2k+1)!x^(2k+1). I tried using the following but got it wrong
0.2 - (0.2)^3/3! + (0.2)^5/5! - (0.2)^7/7! + (0.2)^9/9!.
Please help.
Thanks
1. Find the area of the curved surface of a right-circular cone of radius 6 and height 4 by rotating the straight line segment from (0,0) to (6,4) about the y-axis.
2. Use the Maclaurin polynomial of degree 4 to approximate sin(0.2). The formula given is the sum between k = 0 and infinity of [-1^k]/(2k+1)!x^(2k+1). I tried using the following but got it wrong
0.2 - (0.2)^3/3! + (0.2)^5/5! - (0.2)^7/7! + (0.2)^9/9!.
Please help.
Thanks
Answers
Answered by
Steve
The area of a cone of radius r and slant height s is πrs
r = 6
s^2 = 6^2 + 4^2
s = √52 = 2√13
area = 12π√13
For MacLaurin polynomials, I may be wrong, but I thought you wanted degree 4. That would just be
0 + 1/1! x + 0/2! x^2 + -1/3! x^3 + 0/4! x^4
= x - x^3/3!
= .2 - .008/6 = 0.1987
r = 6
s^2 = 6^2 + 4^2
s = √52 = 2√13
area = 12π√13
For MacLaurin polynomials, I may be wrong, but I thought you wanted degree 4. That would just be
0 + 1/1! x + 0/2! x^2 + -1/3! x^3 + 0/4! x^4
= x - x^3/3!
= .2 - .008/6 = 0.1987
Answered by
Sara
Thanks Steve.
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