Engineering Board Exam Reviewer · Curvature and Radius of Curvature · medium
Find the curvature of y=x3 at x=1.
- 0.1897
- 0.06
- 0.6
- 1.8974
Answer
A. 0.1897
Solution
y'=3x2=3 and y''=6x=6. = |y''|≤ft(1+y'2 )3/2= 6103/2 0.1897.
More Curvature and Radius of Curvature questions
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