To solve the problem on the gradient function uses the techniques of differential calculus, namely the finding of partial derivatives of the first order in three variables. This assumes that the function and all its partial derivatives have the property of continuity in the function definition.
The gradient is a vector whose direction indicates the direction of fastest increase of the function F. on the graph are chosen at two points M0 and M1, which are the ends of the vector. The gradient magnitude is equal to the rate of increase of the function from point M0 to point M1.
The function is differentiable at all points of this vector, thus the vector's projections on the coordinate axes are all its partial derivatives. Then the formula of the gradient as follows:grad = (∂F/∂x)•i + (∂F/∂y)•j + (∂F/∂z)•k, where i, j, k coordinates of the unit vector. In other words, the gradient function is a vector, whose coordinates are the partial derivatives grad F = (∂F/∂x, ∂F/∂y, ∂F/∂z).
Example1.Let function F = sin(x•z2)/y. You want to find the gradient at the point (π/6, 1/4, 1).
Solution.Determine partial derivatives for each variable: F CH = 1/y•cos(x•z2)•z2;F _y = sin(x•z2)•(-1)•1/(y2);F _z = 1/y•cos(x•z2)•2•x•z.
Substitute the known values of the coordinates of a point:F _x = 4•cos(π/6) = 2•√3; F _y = sin(π/6)•(-1)•16 = -8; F _z = 4•cos(π/6)•2•π/6 = 2•π/√3.
Apply the formula for the gradient function:grаd F = 2•√3•i – 8•j + 2•π/√3•k.
Example 2.Find the coordinates of the gradient of the function F = y•arсtg (z/x) at the point (1, 2, 1).
Solution.H f = 0•аrсtg (z/x) + y•(аrсtg(z/x))’h = y•1/(1 + (z/x)2)•(-z/x2) = -y•z/(x2•(1 + (z/x)2)) = -1;F _y = 1•аrсtg(z/x) = аrсtg 1 = π/4;F _z = 0•аrсtg(z/x) + y•(аrсtg(z/x))’_z = y•1/(1 + (z/x)2)•1/x = y/(x•(1 + (z/x)2)) = 1.grаd = (-1, π/4, 1).
Advice 2: How to find the gradient of the scalar field
The gradient of the scalar field is a vector quantity. Thus, its location is required to determine all the components of the corresponding vector on the basis of knowledge about the distribution of the scalar field.
Read the textbook on higher mathematics, what is the gradient of the scalar field. It is known that this vector quantity has a direction characterized by the maximum velocity of recession of a scalar function. This is the meaning of this vector quantity is justified by the expression to determine its components.
Remember that any vector is determined by the values of its components. The components of the vector are actually the projections of this vector in a particular coordinate axis. Thus, if the three-dimensional space, the vector should have three components.
Make a note of how to define the components of the vector which is the gradient of a field. Each coordinate of this vector is the derivative of the scalar potential on a variable coordinate which is calculated. That is, if you want to calculate "a's" component of the gradient vector field, it is necessary to differentiate a scalar function of the variable "x". Note that the derivative should be private. This means that, while differentiation of other variables, not voting, should be considered constants.
Write the expression for the scalar field. As is known, the term implies only a scalar function of several variables, which are also scalar quantities. The number of variables of a scalar function is limited by the dimension of the space.
Differentiate separate scalar function for each variable. In the end you'll have three new functions. Write each function in the expression for the gradient vector of the scalar field. Each of the obtained functions is actually the coefficient of the unit vector given the coordinates. Thus, the resultant vector of the gradient should look like a polynomial with coefficients in the form of derivatives.