Question

True or False? In Exercises 59-62, determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. If \(f(x, y)=\sqrt{1-x^{2}-y^{2}}\), then \(D_{\mathbf{u}} f(0,0)=0\) for any unit vector \(\mathbf{u}\).

Short answer

Yes, the statement is true. For any unit vector \(\mathbf{u}\), the directional derivative of \(f\) at the point \((0,0)\) is indeed 0.

Step-by-step solution

Understand and Define the Directional Derivative

The directional derivative of a function \(f(x,y)\) at a point \((x_0, y_0)\) in the direction of a unit vector \(\mathbf{u} = (u_1, u_2)\) is given by \(D_{\mathbf{u}}f(x_0,y_0) = u_1 \cdot f_x(x_0, y_0) + u_2 \cdot f_y(x_0, y_0)\), where \(f_x\) and \(f_y\) are the partial derivatives of \(f\) with respect to \(x\) and \(y\) respectively.

Compute the Partial Derivatives of \(f\)

We compute the partial derivatives of \(f(x, y) = \sqrt{1-x^{2}-y^{2}}\). Utilizing the chain rule, we get \(f_x(x,y) = -\frac{x}{\sqrt{1-x^{2}-y^{2}}}\) and \(f_y(x,y) = -\frac{y}{\sqrt{1-x^{2}-y^{2}}}\). Then, we evaluate these at the point \((0,0)\), obtaining \(f_x(0,0) = 0\) and \(f_y(0,0) = 0\).

Substitute and Evaluate the Directional Derivative

Substituting the values from Step 2 into the formula from Step 1 gives: \(D_{\mathbf{u}}f(0,0) = u_1 \cdot f_x(0,0) + u_2 \cdot f_y(0,0) = 0\). We see that for any unit vector \(\mathbf{u} = (u_1, u_2)\), the directional derivative of \(f\) at \((0,0)\) is indeed equal to 0, thus confirming the statement provided in the problem.

Key concepts

Partial Derivatives

In calculus, partial derivatives play a central role when we're dealing with multivariable functions. Unlike a derivative of a function with only one variable, the partial derivative tells us how the function changes as only one of the variables changes, keeping others constant. Essentially, it gives us the slope of the function in the direction of the variable we’re differentiating against.

For a given function, say, f(x, y), the notation f_x represents the partial derivative with respect to x, and similarly, f_y represents the partial derivative with respect to y. The process to find a partial derivative is the same as finding a regular derivative, but we only focus on the variable of interest and treat all other variables as constants.

Calculating partial derivatives is essential for understanding surfaces in multivariable calculus and is widely used in optimization problems, physics, engineering, and economics. The exercise presented demonstrates the concept of evaluating partial derivatives at a certain point, which are necessary to determine the rate of change of the function along a specific axis.

Chain Rule

The chain rule is a powerful tool in calculus used to find the derivative of composite functions. When we have a function that is the composition of two or more functions, the chain rule allows us to differentiate the outer function with respect to the inner one.

For example, if a multivariable function f(x, y) is defined in terms of another function, g, such as f(x, y) = h(g(x, y)), and we want to differentiate f with respect to x, we use the chain rule: f_x = h'_g(g(x, y)) * g_x. This rule is not just for functions of two variables—it extends to functions of any number of variables.

Applying the chain rule correctly is crucial to find the rates of change in more complex situations. In our exercise, the chain rule was utilized to find the partial derivatives of a square root function involving multiple variables, simplifying the process of differentiation and contributing significantly to finding the directional derivative.

Vector Calculus

Vector calculus extends the basic ideas of calculus to vector fields. It's a branch of mathematics that deals with functions that have vector outputs and inputs. This concept becomes especially useful in physics and engineering where fields like electromagnetism and fluid dynamics often require an understanding of how vectors change in space.

A unit vector, represented as \(\textbf{u}\) in our exercise, is a vector with a magnitude of one. In calculus, we use these unit vectors to express directions. The directional derivative of a function in the direction of a unit vector shows the rate at which the function changes in that particular direction. In the context of the exercise, D_{\(\textbf{u}\)}f(0,0) should be interpreted as the instantaneous rate of change of f at the point (0,0) in the direction of the unit vector \(\textbf{u}\).

Vector calculus is instrumental in multidimensional optimization problems, as it provides tools to analyze functions over different paths and in various directions within a field, which is exactly what the directional derivative represents.