Understanding Directional Derivatives in Multivariable Calculus

Directional derivatives are a fundamental concept in multivariable calculus, providing a way to measure the rate of change of a function in a specified direction. This article will guide you through the process of calculating the directional derivative of the function f(x,y,z) 5x^2y^5y^2z^{5/2}z^2x at the point (1,1,1) in the direction of the line x-1/2 y-3/-2 z/1. We will break down the steps involved in finding the solution, including the calculation of partial derivatives and the use of a unit vector.

Finding the Unit Vector in the Direction of the Line

First, to calculate the directional derivative, we need to determine a unit vector in the direction of the specified line. The direction ratios of the line are given as (2, -2, 1). The unit vector in this direction is calculated as:

( mathbf{u} left(frac{2}{sqrt{2^2 (-2)^2 1^2}}, frac{-2}{sqrt{2^2 (-2)^2 1^2}}, frac{1}{sqrt{2^2 (-2)^2 1^2}}right) )

Let's simplify the components of the unit vector:

( mathbf{u} left(frac{2}{3}, frac{-2}{3}, frac{1}{3}right) )

Evaluating the Function and its Partial Derivatives

Next, we need to compute the first order partial derivatives of the function f(x,y,z) at the point (1,1,1).

The partial derivative with respect to x is:

( f_x(x,y,z) 1y frac{5}{2}z^2 )

At the point (1,1,1), this becomes:

( f_x(1,1,1) 10(1)(1) frac{5}{2}(1)^2 frac{25}{2} )

The partial derivative with respect to y is:

( f_y(x,y,z) 5x^2 10yz )

At the point (1,1,1), this becomes:

( f_y(1,1,1) 5(1)^2 10(1)(1) 15 )

The partial derivative with respect to z is:

( f_z(x,y,z) 5y^2 5xz )

At the point (1,1,1), this becomes:

( f_z(1,1,1) 5(1)^2 5(1)(1) 10 )

Calculating the Directional Derivative

The directional derivative in the direction of the unit vector u is given by the dot product of the gradient of the function and the unit vector:

( D_u f left( f_x(1,1,1), f_y(1,1,1), f_z(1,1,1) right) cdot mathbf{u} )

Substituting the values, we get:

( D_u f left( frac{25}{2}, 15, 10 right) cdot left( frac{2}{3}, frac{-2}{3}, frac{1}{3} right) )

Performing the dot product:

( D_u f frac{25}{2} cdot frac{2}{3} 15 cdot frac{-2}{3} 10 cdot frac{1}{3} )

Simplifying:

( D_u f frac{50}{6} - frac{30}{3} frac{10}{3} frac {50}{6} - frac{60}{6} frac {20}{6} frac{10}{6} frac{5}{3} )

Therefore, the directional derivative of the function at the point (1,1,1) in the direction of the specified line is:

( D_u f frac{5}{3} )

Conclusion

This article outlines the step-by-step methodology for calculating the directional derivative of a multivariable function. Understanding these concepts is crucial in various applications of calculus, from optimization problems to physics and engineering. By following these steps, you can determine how a function changes along a specific direction within the multi-dimensional space.

Related Keywords

directional derivative multivariable function unit vector