Problem 22
Question
Suppose that the temperature \(T\) at the point \((x, y, z)\) depends only on the distance from the origin. Show that the direction of greatest increase in \(T\) is either directly toward the origin or directly away from the origin.
Step-by-Step Solution
Verified Answer
The direction of greatest increase in temperature is either toward or away from the origin, depending on the sign of \( \frac{df}{dr} \).
1Step 1: Define the Temperature Function
Since the temperature \( T \) depends only on the distance from the origin, let's express it as a function of the distance \( r \), where \( r = \sqrt{x^2 + y^2 + z^2} \). We can write \( T = f(r) \), meaning that the temperature is a function of \( r \).
2Step 2: Express in Gradient Terms
The gradient of a function \( T(x, y, z) \) gives the direction of the greatest increase. The gradient is expressed as \( abla T = \left( \frac{\partial T}{\partial x}, \frac{\partial T}{\partial y}, \frac{\partial T}{\partial z} \right) \).
3Step 3: Chain Rule for Partial Derivatives
Since \( T = f(r) \) where \( r = \sqrt{x^2 + y^2 + z^2} \), apply the chain rule. Compute \( \frac{\partial T}{\partial x} = \frac{df}{dr} \cdot \frac{\partial r}{\partial x} \), and similarly for \( y \) and \( z \).
4Step 4: Calculate Partial Derivatives of r
The partial derivatives of \( r \) are \( \frac{\partial r}{\partial x} = \frac{x}{r} \), \( \frac{\partial r}{\partial y} = \frac{y}{r} \), \( \frac{\partial r}{\partial z} = \frac{z}{r} \).
5Step 5: Formulate Gradient of T
Substitute these into the chain rule expression: \(abla T = \frac{df}{dr} \left( \frac{x}{r}, \frac{y}{r}, \frac{z}{r} \right) = \frac{df}{dr} \hat{r} \), where \( \hat{r} = \left( \frac{x}{r}, \frac{y}{r}, \frac{z}{r} \right) \) is the unit vector in the direction of \( \vec{r} \).
6Step 6: Analyze the Direction of the Gradient
The gradient \( abla T = \frac{df}{dr} \hat{r} \) indicates the direction of greatest change. If \( \frac{df}{dr} > 0 \), the direction of greatest increase in \( T \) is the same as \( \hat{r} \) (i.e., away from the origin). If \( \frac{df}{dr} < 0 \), the direction of increase is the opposite of \( \hat{r} \) (i.e., toward the origin). Thus, the direction of greatest increase in \( T \) is either directly toward or away from the origin, depending on the sign of \( \frac{df}{dr} \).
Key Concepts
GradientDirectional DerivativePartial DerivativesVector Calculus
Gradient
The concept of the gradient is central in multivariable calculus, as it helps determine the direction of greatest increase of a scalar field, like temperature. The gradient of a function is represented as a vector made up of partial derivatives, providing both direction and rate of change in a multidimensional space.
For a function of several variables, say, temperature \( T(x, y, z) \), the gradient \( abla T \) is given by:
Understanding how to navigate these gradients allows us to predict and describe changes within spatial fields, making it essential in physics and engineering.
For a function of several variables, say, temperature \( T(x, y, z) \), the gradient \( abla T \) is given by:
- \( abla T = \left( \frac{\partial T}{\partial x}, \frac{\partial T}{\partial y}, \frac{\partial T}{\partial z} \right) \)
Understanding how to navigate these gradients allows us to predict and describe changes within spatial fields, making it essential in physics and engineering.
Directional Derivative
The directional derivative extends the concept of a gradient by specifying a particular direction for the rate of change. While a gradient finds the direction of the steepest ascent, the directional derivative measures how a function changes as we move in any specified direction, meaning it can be applied in diverse situations.
To compute the directional derivative of \( T \) at point \((x, y, z)\) in the direction of a unit vector \( \mathbf{u} = (u_x, u_y, u_z)\), the formula is:
To compute the directional derivative of \( T \) at point \((x, y, z)\) in the direction of a unit vector \( \mathbf{u} = (u_x, u_y, u_z)\), the formula is:
- \( D_u T = abla T \cdot \mathbf{u} = \frac{\partial T}{\partial x} u_x + \frac{\partial T}{\partial y} u_y + \frac{\partial T}{\partial z} u_z \)
Partial Derivatives
Partial derivatives are the building blocks of the gradient and are crucial for understanding changes in multivariable functions. They represent the rate of change of a function with respect to one of its variables, while keeping the other variables constant.
For a function \( T(x, y, z) \), the partial derivatives are:
For a function \( T(x, y, z) \), the partial derivatives are:
- \( \frac{\partial T}{\partial x} \)
- \( \frac{\partial T}{\partial y} \)
- \( \frac{\partial T}{\partial z} \)
Vector Calculus
Vector calculus involves understanding and working with functions that have directional attributes and are represented as vectors, crucial for describing physical phenomena in physics and engineering. When dealing with functions like our temperature \( T(x, y, z) \), vector calculus helps us describe the function's behavior as it varies in space.
Key operations in vector calculus include gradients, divergences, and curls. For the exercise provided, the focus is on evaluating the gradient, a foundational vector calculus operation that provides insights into how scalar fields like temperature vary spatially.
Key operations in vector calculus include gradients, divergences, and curls. For the exercise provided, the focus is on evaluating the gradient, a foundational vector calculus operation that provides insights into how scalar fields like temperature vary spatially.
- Gradient: Tells us directions and rates of change.
- Position Vectors: Communicate locations in space, associated with each point \((x, y, z)\).
Other exercises in this chapter
Problem 21
Find parametric equations of the line tangent to the surface \(z=y^{2}+x^{3} y\) at the point \((2,1,9)\) whose projection on the \(x y\)-plane is (a) parallel
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Use differentials to find the approximate amount of copper in the four sides and bottom of a rectangular copper tank that is 6 feet long, 4 feet wide, and 3 fee
View solution Problem 22
In Problems 17-22, sketch the level curve \(z=k\) for the indicated values of \(k\). $$ z=y-\sin x, k=-2,-1,0,1,2 $$
View solution Problem 22
Find the point on the plane \(2 x+4 y+3 z=12\) that is closest to the origin. What is the minimum distance?
View solution