/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 33 The temperature \(T\) at the poi... [FREE SOLUTION] | 91影视

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Chapter 16: Problem 33

The temperature \(T\) at the point \(P(x, y, z)\) in an \(x y z-\) coordinate system is given by \(T=8\left(2 x^{2}+4 y^{2}+9 z^{2}\right)^{1 / 2}\). Use differentials to approximate the temperature difference between the points (6,3,2) and (6.1,3.3,1.98) .

Short Answer

Expert verified
The approximate temperature difference is 2.96 units.

Step by step solution

01

Identifying the Function and Initial Values

The temperature function is given by \( T = 8\sqrt{2x^2 + 4y^2 + 9z^2} \). We need to find the temperature difference between the points \((6,3,2)\) and \((6.1,3.3,1.98)\). The initial point is \((x_0, y_0, z_0) = (6, 3, 2)\), and the change in variables are \(dx = 0.1\), \(dy = 0.3\), and \(dz = -0.02\).
02

Calculating the Partial Derivatives

We first find the partial derivatives of \(T\) with respect to \(x\), \(y\), and \(z\). These are:\[ \frac{\partial T}{\partial x} = \frac{8 \cdot (2x)}{\sqrt{2x^2 + 4y^2 + 9z^2}} \]\[ \frac{\partial T}{\partial y} = \frac{8 \cdot (4y)}{\sqrt{2x^2 + 4y^2 + 9z^2}} \]\[ \frac{\partial T}{\partial z} = \frac{8 \cdot (9z)}{\sqrt{2x^2 + 4y^2 + 9z^2}} \]
03

Evaluating Partial Derivatives at Initial Point

Substitute \(x = 6\), \(y = 3\), and \(z = 2\) into the partial derivatives:\[ \frac{\partial T}{\partial x}(6, 3, 2) = \frac{8 \cdot (12)}{\sqrt{72 + 36 + 36}} = \frac{96}{12} = 8 \]\[ \frac{\partial T}{\partial y}(6, 3, 2) = \frac{8 \cdot (12)}{\sqrt{144}} = 8 \]\[ \frac{\partial T}{\partial z}(6, 3, 2) = \frac{8 \cdot (18)}{\sqrt{144}} = 12 \]
04

Calculating the Differential \(dT\)

The differential \(dT\) is an approximation of the temperature change:\[ dT = \frac{\partial T}{\partial x} dx + \frac{\partial T}{\partial y} dy + \frac{\partial T}{\partial z} dz \]\[ dT = 8 \cdot 0.1 + 8 \cdot 0.3 + 12 \cdot (-0.02) \]\[ dT = 0.8 + 2.4 - 0.24 = 2.96 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Partial Derivatives
Partial derivatives help us understand how a multivariable function changes when we modify one variable at a time, while keeping the others constant. In the temperature function given as \( T = 8 \sqrt{2x^2 + 4y^2 + 9z^2} \), we have three variables: \(x\), \(y\), and \(z\). To find the partial derivatives, we differentiate this function with respect to each variable individually, treating the other variables as constants, which gives us:
  • \( \frac{\partial T}{\partial x} = \frac{8 \cdot (2x)}{\sqrt{2x^2 + 4y^2 + 9z^2}} \)
  • \( \frac{\partial T}{\partial y} = \frac{8 \cdot (4y)}{\sqrt{2x^2 + 4y^2 + 9z^2}} \)
  • \( \frac{\partial T}{\partial z} = \frac{8 \cdot (9z)}{\sqrt{2x^2 + 4y^2 + 9z^2}} \)
These derivatives tell us how the temperature at a point changes as we slightly adjust its \(x\), \(y\), or \(z\) coordinate.
Temperature Change
In this problem, we want to approximate how the temperature changes between two points, using differentials. Differentials are a useful tool in mathematics for estimating small changes in functions based on their derivatives.

Here, the differential \(dT\) represents the estimated change in temperature, as calculated by the sum of the products of each partial derivative with the change in the respective coordinate:
  • \( dT = \frac{\partial T}{\partial x} dx + \frac{\partial T}{\partial y} dy + \frac{\partial T}{\partial z} dz \)
This gives us an informative approximation of how much the temperature shifts when moving from point \((6,3,2)\) to \((6.1,3.3,1.98)\). Evaluating it provides the value \(dT = 2.96\), suggesting a temperature increase by approximately 2.96 units.
Coordinate System
An \(x y z\) coordinate system is essential for understanding how functions behave in three-dimensional space.

In this context, the coordinate system allows us to define a location using three values: \(x\), \(y\), and \(z\).
  • \(x\) represents the horizontal direction,
  • \(y\) the vertical direction,
  • and \(z\) the depth or altitude.
Using Cartesian coordinates helps calculate the temperature at any point in this 3D space. The given expression \( T = 8 \sqrt{2x^2 + 4y^2 + 9z^2} \) indicates how temperature depends on a point's position in this space, making it crucial to consider all three dimensions when evaluating changes.
Approximation
Approximation is a powerful mathematical technique that provides us close estimates of quantities without precise computations. In this case, we use approximation to estimate the temperature change between two nearby points.

By employing differentials, a simplified version of calculus, we assume that small changes in variables result in small proportional changes in the function's value.
  • This results in the differential formula \(dT = 8 \cdot 0.1 + 8 \cdot 0.3 + 12 \cdot (-0.02)\).
The calculated value, 2.96, reflects the temperature change, approximating real-world conditions where exact calculation may be complex or infeasible.

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