In mathematics, the directional derivative of a multivariate differentiable function along a given vector V at a given point P intuitively represents the instantaneous rate of change of the function, moving through P, in the direction of V. It therefore generalizes the notion of a partial derivative, in which the direction is always taken parallel to one of the coordinate axes. The directional derivative is a special case of the Gateaux derivative.
The directional derivative of a scalar function f( ) = f ( x1,x2,.......xn)
Source: - Wikipedia
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Examples for the directional derivative:
Example 1:
The directional derivative of the function f = sqrt(x2+y2) in the direction defined through vector 2i + 2j + 2k is known through the scalar product hatn·grad f, where the unit vector hatn is
Solution:-
hatn = (2i+2j+2k) / sqrt (2^2+ 2^2+2^2)
= (2i + 2j + 2k) / sqrt9
= 2/3 i + 2/3 j + 2/3 k
The gradient of the function f is (0-2)/sqrt(0^2+(-2)^2)
grad f = x / sqrt(x^2+y^2)i + y / sqrt(x^2+y^2)j+0 k
= (x i+yj) / sqrt(x^2+y^2)
Therefore the required directional derivative is hatn.gradf =(2/3i+2/3j+2/3k).((x i+yj)/sqrt(x^2+y^2)) = 2/3 (x+y)/sqrt(x^2+y^2)
At the point (0,-2,1) it is equal to 2/3 =(0-2)/sqrt(0^2+(-2)^2)
= 2/3 x -2/2
= -2/3
More Examples for the directional derivative:
Example 1:
The directional derivative of the function
f = 2x2 − 2y2
Solution:-
In the unit vector j direction is known through the scalar product j ·rf.
The gradient of the function
f = 2x2 − 2y2 is
rf = 4xi − 4yj
So the directional derivative in the j direction is
j ·rf = j · (4xi − 4yj) = −4y
And at the point (1, 2, 3) it has the value −4 × 2 = −8.
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Example 2:
The directional derivative of the function
f = 4x2 − 5y2
Solution:-
In the unit vector i direction is known through the scalar product i·rf.
The gradient of the function
f = 4x2 − 5y2 is
rf = 8xi − 10yj
So the directional derivative in the i direction is
i ·rf = i · (8xi − 10yj) = 8x
And at the point (2, 4, 6) it has the value 8 × 2 = 16.
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