Problem 1 - Active Calculus Multivariable

Question

Consider the function \(f(x, y, z)=x y+y z^{2}+x z^{3}\). Find the gradient of \(f\) : ( ____________, ____________) Find the gradient of \(f\) at the point (2,3,5) . ( ____________, ____________) Find the rate of change of the function \(f\) at the point (2,3,5) in the direction \(\mathbf{u}=\langle 4 / \sqrt{66},-5 / \sqrt{66},-5 / \sqrt{66}\rangle .\)

Step-by-Step Solution

Answer

The gradient of f is \(\nabla f = (y+z^3, x+z^2, 2yz+3xz^2)\). At the point (2, 3, 5), the gradient is (128, 27, 160). The rate of change of the function at this point in the given direction is \(\frac{-423}{\sqrt{66}}\).

Step 1: Write down the given function

We have a function in three variables: \( f(x, y, z) = xy + yz^2 + xz^3 \).

Step 2: Find the gradient of the function

The gradient of a function is a vector containing the partial derivatives with respect to each variable. Calculate the partial derivatives: \(\frac{\partial f}{\partial x} = y + z^3\) \(\frac{\partial f}{\partial y} = x + z^2\) \(\frac{\partial f}{\partial z} = 2yz + 3xz^2\) The gradient of function f is therefore: \(\nabla f = (y+z^3, x+z^2, 2yz+3xz^2)\)

Step 3: Find the gradient at the point (2, 3, 5)

Substitute the values of the point (2, 3, 5) into the gradient to find its value at this specific point: \(\nabla f(2, 3, 5) = (3+5^3, 2+5^2, 2(3)(5)+3(2)(5^2)) = (128, 27, 160)\) Thus, the gradient of f at the point (2, 3, 5) is (128, 27, 160).

Step 4: Find the rate of change in the given direction

The rate of change of f in the direction of the vector \(\mathbf{u} = \langle 4/\sqrt{66}, -5/\sqrt{66}, -5/\sqrt{66} \rangle\) can be found by taking the dot product of the gradient and the direction vector: \(Rate\:of\:change = \nabla f \cdot \mathbf{u}\) \(= (128, 27, 160) \cdot \left(\frac{4}{\sqrt{66}},\frac{-5}{\sqrt{66}},\frac{-5}{\sqrt{66}}\right)\) \(= \frac{128 \cdot 4 - 27 \cdot 5 - 160 \cdot 5}{\sqrt{66}}\) \(= \frac{512 - 135 - 800}{\sqrt{66}}\) \(= \frac{-423}{\sqrt{66}}\) The rate of change of the function f at the point (2, 3, 5) in the direction of vector \(\mathbf{u}\) is \(\frac{-423}{\sqrt{66}}\).

Key Concepts

Partial Derivatives

Partial derivatives are taken for functions with multiple variables. Instead of differentiating the whole function with respect to one variable while ignoring others, they allow us to focus on the effect of change in one variable at a time. In our exercise, the function \(f(x, y, z) = xy + yz^2 + xz^3\) comprises three variables: \(x\), \(y\), and \(z\). The partial derivative of \(f\) with respect to \(x\) is found by treating \(y\) and \(z\) as constants. This yields \(\frac{\partial f}{\partial x} = y + z^3\). Likewise, \(\frac{\partial f}{\partial y} = x + z^2\) and \(\frac{\partial f}{\partial z} = 2yz + 3xz^2\) are calculated similarly. Partial derivatives are helpful to understand how \(f\) changes as each individual variable changes, keeping others fixed.

Rate of Change

The concept of rate of change is vital in understanding how a quantity shifts over time or concerning another variable. In multivariable calculus, it's not just about changing in time, but also in spatial dimensions. One way to measure this is through the gradient, as it provides the direction and magnitude of the steepest ascent. Specifically, in our example, we looked for the rate of change in a specific direction given by the vector \(\mathbf{u}\). The formula involves computing the dot product of the gradient (calculated as \(abla f(x, y, z)\)) with the direction vector \(\mathbf{u}\). This resulted in the expression for the rate of change: \(\frac{-423}{\sqrt{66}}\). It tells us how fast and in what manner \(f\) changes as we move in the direction of \(\mathbf{u}\).

Directional Derivative

Directional derivatives give us insight into the rate at which a function changes as we move in a specified direction. To calculate a directional derivative, we use a unit vector \(\mathbf{u}\) that signifies the direction we're interested in. In our exercise, this unit vector was \(\langle 4/\sqrt{66}, -5/\sqrt{66}, -5/\sqrt{66} \rangle\). Once the gradient, \(abla f\), is known, the directional derivative is the dot product \(abla f \cdot \mathbf{u}\). This computation measures the change of \(f\) following \(\mathbf{u}\) and indicates both the slope and sign (positive for increase, negative for decrease). In our context, the directional derivative showed the rate of descent as the function decreased in the chosen direction.

Multivariable Functions

Multivariable functions, such as \(f(x, y, z) = xy + yz^2 + xz^3\), are important in modeling scenarios with more than a single independent variable. They allow for the exploration of how changes in variables can affect outcomes differently in different dimensions. Analyzing such functions is crucial in fields like physics or economics, where the real-world phenomena depend on various inputs. For these functions, partial derivatives, gradients, and directional derivatives are tools that help understand how slight variations in one or more variables influence the behavior of the function, enabling predictions and optimizations. They are foundational concepts to tackle complex systems where each variable plays a unique role.