The directional derivative of f(x,y,z)=2x2+3y2+z2 at the point P(2,1,3) in the direction of the vector →a=^i−2^k is
A
−2.785
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B
−2.145
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C
−1.789
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D
1.000
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Solution
The correct option is C−1.789 f(x,y,z)=2x2+3y2+z2 ▽f=^i∂f∂x+^j∂f∂y+^k∂f∂z =^i(4x)+^j(6y)+^k(2z) =8^i+6^j+6^k; at P(2,1,3)
Directional derivative in the direction of →a=^i−2^k is =▽f.→a|→a|=(8^i+6^j+6^k).(^i−2^k)√1+4 =8−12√5=−4√5 =−1.789