Find the radius of curvature for the curve y = x² + 3x at the point (1,2)
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Answer:
general equation of sphere,
x
2
+y
2
+2gx+2fy+2hz+c=0
centre (−g,−f.−h) and radius =
g
2
+f
2
+h
2
−c
At z=0, x
2
+y
2
+2gx+2fy+c=0−−−(i)
x
2
+y
2
−4=0−−−(ii)
comparing (i) and (ii), g=0,f=0,c=−4
∴ general equation reduces to,
x
2
+y
2
+z
2
+2hz+4=0 centre (0,0,h), radius=
h
2
−c
from fig., P is perpendicular drawn from centre to plane and r is radius of sphere. ∴p
2
+9=r
2
9
4h
2
+9=h
2
+4
⇒h=±3
∴ equation of spheres are,
x
2
+y
2
+z
2
+6z−4=0 and x
2
+y
2
+z
2
+6z−4=0
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