Question

Determine whether the following statements are true and give an explanation or counterexample. a. If \(f(x, y)=x^{2}+y^{2}-10,\) then \(\nabla f(x, y)=2 x+2 y\) b. Because the gradient gives the direction of maximum increase of a function, the gradient is always positive. c. The gradient of \(f(x, y, z)=1+x y z\) has four components. d. If \(f(x, y, z)=4,\) then \(\nabla f=\mathbf{0}\)

Short answer

Question: Determine whether the following statements are true or false. If a statement is false, provide a counterexample. a. The gradient of the function \(f(x, y) = x^{2} + y^{2} - 10\) is equal to \((2x + 2y)\). b. The gradient of any function of two or three variables is always positive. c. The gradient of the function \(f(x, y, z) = 1 + xyz\) has four components. d. The gradient of the function \(f(x, y, z) = 4\) is equal to the zero vector. Answer: a. False. The gradient of the function \(f(x, y) = x^{2} + y^{2} - 10\) is \((2x, 2y)\), not \((2x + 2y)\). b. False. The gradient is not always positive. Consider the function \(f(x, y, z) = -x^2 - y^2\), whose gradient is negative. c. False. The gradient of the function \(f(x, y, z) = 1 + xyz\) has three components, not four: \((yz, xz, xy)\). d. True. The gradient of the constant function \(f(x, y, z) = 4\) is indeed equal to the zero vector, \(\mathbf{0}\).

Step-by-step solution

Compute the gradient of the given function

Calculate the gradient of \(f(x, y)=x^{2}+y^{2}-10\). The gradient is the vector of partial derivatives with respect to each variable, so we have \(\nabla f(x,y) = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right)\).

Compare the computed gradient with the given vector

Calculate the partial derivatives as follows: \(\frac{\partial f}{\partial x} = 2x\) and \(\frac{\partial f}{\partial y} = 2y\). So, the computed gradient is \(\nabla f(x, y) = (2x, 2y)\). This is not equal to the given vector \((2x+2y)\). Therefore, statement a is *false*. For statement b: The gradient gives the direction of maximum increase in the function, not its magnitude. Therefore, it is not always positive. For instance, if we have a function \(f(x, y, z) = -x^2 - y^2\), then its gradient is negative. Hence, statement b is *false*. For statement c:

Compute the gradient of the given function

Calculate the gradient of \(f(x, y, z) = 1 + xyz\). The gradient is the vector of partial derivatives with respect to each variable, so we have \(\nabla f(x,y,z)=\left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}\right)\).

Compare the number of components with the claimed number

Calculate the partial derivatives as follows: \(\frac{\partial f}{\partial x} = yz\), \(\frac{\partial f}{\partial y} = xz\), and \(\frac{\partial f}{\partial z} = xy\). So, the computed gradient is \(\nabla f(x, y, z)=(yz, xz, xy)\). This gradient has *three components*, not four. Therefore, statement c is *false*. For statement d: If \(f(x,y,z)=4\), it means \(f\) is a constant function. The gradient of a constant function is always zero because partial derivatives with respect to each variable are zero. So, \(\nabla f = \mathbf{0}\). In this case, statement d is *true*.