Problem 63 - Calculus with Concepts in Calculus

Question

Suppose \(T(x, y, z)=x^{3} y+3 x^{2} y^{2} z\). Find the directional derivative of \(T\) at \((1,1,-1)\) in the direction of the gradient.

Step-by-Step Solution

Answer

The directional derivative is \( \sqrt{43} \).

Step 1: Find the Gradient Vector of Function

To find the directional derivative in the direction of the gradient, first determine the gradient vector of the function.The function given is:\[ T(x, y, z) = x^3 y + 3x^2 y^2 z \]Compute the partial derivatives:- \( \frac{\partial T}{\partial x} = 3x^2 y + 6xy^2z \)- \( \frac{\partial T}{\partial y} = x^3 + 6x^2 yz \)- \( \frac{\partial T}{\partial z} = 3x^2 y^2 \)The gradient of \( T \) is:\[ abla T = \left(3x^2 y + 6xy^2z, x^3 + 6x^2 yz, 3x^2 y^2\right) \]

Step 2: Evaluate the Gradient at the Given Point

We need to evaluate the gradient vector at the point \((1,1,-1)\).Substitute \(x = 1\), \(y = 1\), and \(z = -1\) into the gradient:- \( \frac{\partial T}{\partial x} = 3\cdot1^2\cdot1 + 6\cdot1\cdot1^2\cdot(-1) = 3 - 6 = -3 \)- \( \frac{\partial T}{\partial y} = 1^3 + 6\cdot1^2\cdot1\cdot(-1) = 1 - 6 = -5 \)- \( \frac{\partial T}{\partial z} = 3\cdot1^2\cdot1^2 = 3 \)Thus, \( abla T (1, 1, -1) = (-3, -5, 3) \).

Step 3: Compute Magnitude of the Gradient Vector

The directional derivative in the direction of the gradient involves computing the magnitude of the gradient vector we found.Calculate the magnitude of \( abla T (1, 1, -1) \):\[ \lVert abla T (1, 1, -1) \rVert = \sqrt{(-3)^2 + (-5)^2 + (3)^2} = \sqrt{9 + 25 + 9} = \sqrt{43} \]

Step 4: Directional Derivative in the Gradient Direction

The directional derivative of a function at a point in the direction of the gradient vector is simply the magnitude of the gradient vector at that point.Thus, the directional derivative of \( T \) at \((1,1,-1)\) in the direction of its gradient is:\[ \sqrt{43} \]

Key Concepts

Gradient Vector

In the realm of multivariable calculus, the gradient vector is a fundamental concept used to find the directional derivative. It's like a navigational arrow pointing us in the direction of greatest increase of a function at a given point. For a function of several variables, the gradient is a vector of its partial derivatives.

For example, consider a function with three variables:
- If we have the function \(T(x, y, z) = x^3 y + 3x^2 y^2 z\), the gradient vector, denoted by \( abla T \), is composed of its partial derivatives with respect to each variable.
- Here are the partial derivatives:

• \( \frac{\partial T}{\partial x} = 3x^2 y + 6xy^2z \)
• \( \frac{\partial T}{\partial y} = x^3 + 6x^2 yz \)
• \( \frac{\partial T}{\partial z} = 3x^2 y^2 \)

Thus, the gradient vector is given by: \[ abla T = \left(3x^2 y + 6xy^2z, x^3 + 6x^2 yz, 3x^2 y^2\right) \]
This vector is crucial for determining how a multi-variable function changes at a particular point.

Partial Derivatives

Partial derivatives are the building blocks of the gradient vector. They measure how a function changes as one of its variables changes, while holding the other variables constant. This is especially useful when analyzing functions with multiple variables.

Let's look at an example where we find partial derivatives for a function \(T(x, y, z)\).
- Consider the function \(T(x, y, z) = x^3 y + 3x^2 y^2 z\).

• The partial derivative with respect to \(x\) is \( \frac{\partial T}{\partial x} = 3x^2 y + 6xy^2z \).
• The partial derivative with respect to \(y\) is \( \frac{\partial T}{\partial y} = x^3 + 6x^2 yz \).
• The partial derivative with respect to \(z\) is \( \frac{\partial T}{\partial z} = 3x^2 y^2 \).

Each of these derivatives provides information on how the function behaves as one variable changes while keeping the others fixed. This understanding helps in constructing the gradient and analyzing the direction of change of the function.

Magnitude of Vector

The magnitude of a vector often referred to as the length or norm, gives us a scalar quantity that describes how long the vector is. When dealing with gradient vectors, its magnitude is key for finding the directional derivative in the function's steepest ascent.To understand it, let's take the gradient vector we found earlier at the point \((1,1,-1)\): \(abla T (1, 1, -1) = (-3, -5, 3)\).
To calculate the magnitude of this vector, we use the following formula:
\[\lVert abla T (1, 1, -1) \rVert = \sqrt{(-3)^2 + (-5)^2 + (3)^2} \]
- This simplifies to \(\sqrt{9 + 25 + 9} = \sqrt{43}\).
This shows the magnitude of the vector at that point, allowing us to determine the directional derivative as the function's most rapid increase.

Knowing how to calculate and interpret the magnitude of a vector is crucial, especially when solving problems involving the directional derivative in multivariable calculus.