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An unevenly heated metal plate has temperature \(T(x, y)\) in degrees Celsius at a point \((x, y) .\) If \(T(2,1)=119, T_{x}(2,1)=19,\) and \(T_{y}(2,1)=-14,\) estimate the temperature at the point (2.04,0.96) . \(T(2.04,0.96) \approx\) ________.

Short Answer

Expert verified
T(2.04,0.96) \approx 120.32.

Step by step solution

01

Calculate the change in x and change in y

Calculate the difference between the x and y coordinates of the given point (2, 1) and the point we want to estimate the temperature at (2.04, 0.96). That is, calculate Δx and Δy: Δx = 2.04 - 2 = 0.04 Δy = 0.96 - 1 = -0.04
02

Utilize differential approximation

Using the differential approximation, we can estimate the change in temperature (ΔT) using the formula: ΔT ≈ T_x(2, 1) * Δx + T_y(2, 1) * Δy We are given the values for T_x(2, 1) = 19 and T_y(2, 1) = -14. Substitute these values and the values for Δx and Δy found in Step 1 into the formula to get: ΔT ≈ 19 * 0.04 + (-14) * (-0.04)
03

Calculate ΔT

Now, simply calculate the value for ΔT based on the equation found in Step 2: ΔT ≈ 0.76 + 0.56 ΔT ≈ 1.32
04

Estimate the temperature at the point (2.04, 0.96)

We know that T(2, 1) = 119, and we've found the change in temperature, ΔT = 1.32. To find the temperature at the point (2.04, 0.96), simply add the given temperature value (T(2, 1)) and the calculated change in temperature (ΔT): T(2.04, 0.96) ≈ T(2, 1) + ΔT T(2.04, 0.96) ≈ 119 + 1.32 T(2.04, 0.96) ≈ 120.32 So, the estimated temperature at the point (2.04, 0.96) is approximately 120.32 degrees Celsius.

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Key Concepts

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

Multivariable Calculus
Imagine you are on a hiking trip, navigating a mountain terrain. Your goal is to predict the slope at any given point on the trail as it rises and falls in various directions. This is similar to what multivariable calculus allows us to do with mathematical landscapes.

In multivariable calculus, we deal with functions of several variables, such as temperature varies with both x and y coordinates on a metal plate. Just as we analyze how altitude changes with respect to north-south and east-west directions on a mountain, we examine how the function changes in response to its variables in multivariable calculus.

This concept is fundamental when estimating values between known data points, which is essential in many fields such as engineering, physics, and economics. Calculating the temperature at a new point on a metal plate, as presented in our example, leverages the power of multivariable calculus to make predictions about the physical world.
Temperature Gradient
Have you ever felt the gradual change in warmth as you move your hand above a toaster? That is a common experience of a temperature gradient. A temperature gradient represents how temperature changes with distance.

In our exercise, the temperature of the metal plate varies from point to point, creating a temperature gradient. This concept is vital in multivariable calculus as it helps to describe how the temperature changes across different dimensions of the plate. Utilizing temperature gradients allows for a deeper understanding of heat distribution and is a key concept in thermodynamics, meteorology, and material science.

The gradients, represented mathematically by partial derivatives, tell us the rate at which temperature changes in the direction of each variable. For the metal plate example, the horizontal and vertical gradients can predict temperature variations especially when moving from a known point to an adjacent point.
Partial Derivatives
If a recipe offers flexibility in the amount of spices you can add, you might be curious how altering one spice will change the taste without changing others. In the realm of mathematics, this is where partial derivatives come into play.

A partial derivative shows how a function changes as one variable changes while all other variables stay constant. For instance, in our exercise, the function that gives us the temperature at each point of the metal plate depends on both the x-coordinate and y-coordinate. The partial derivatives, denoted as \( T_x \) and \( T_y \), tell us about the temperature rate change with respect to each coordinate.

Calculating Partial Derivatives

When calculating partial derivatives, we treat one variable as the focus and consider the other variables as constants. By doing so, we can anticipate how a slight shift in one direction affects the outcome without being clouded by changes in other directions. This precision is pivotal when assessing how temperature fluctuates in various parts of the metal plate or predicting the effects of changing input variables in any multivariable function.

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Most popular questions from this chapter

In this exercise we consider how to apply the Method of Lagrange Multipliers to optimize functions of three variable subject to two constraints. Suppose we want to optimize \(f=f(x, y, z)\) subject to the constraints \(g(x, y, z)=c\) and \(h(x, y, z)=k\). Also suppose that the two level surfaces \(g(x, y, z)=c\) and \(h(x, y, z)=k\) intersect at a curve \(C\). The optimum point \(P=\left(x_{0}, y_{0}, z_{0}\right)\) will then lie on \(C\). a. Assume that \(C\) can be represented parametrically by a vector-valued function \(\mathbf{r}=\mathbf{r}(t) .\) Let \(\overrightarrow{O P}=\mathbf{r}\left(t_{0}\right) .\) Use the Chain Rule applied to \(f(\mathbf{r}(t)), g(\mathbf{r}(t)),\) and \(h(\mathbf{r}(t)),\) to explain why $$\begin{array}{l} \nabla f\left(x_{0}, y_{0}, z_{0}\right) \cdot \mathbf{r}^{\prime}\left(t_{0}\right)=0 \\ \nabla g\left(x_{0}, y_{0}, z_{0}\right) \cdot \mathbf{r}^{\prime}\left(t_{0}\right)=0, \text { and } \\ \nabla h\left(x_{0}, y_{0}, z_{0}\right) \cdot \mathbf{r}^{\prime}\left(t_{0}\right)=0\end{array}$$ Explain how this shows that \(\nabla f\left(x_{0}, y_{0}, z_{0}\right), \nabla g\left(x_{0}, y_{0}, z_{0}\right),\) and \(\nabla h\left(x_{0}, y_{0}, z_{0}\right)\) are all orthogonal to \(C\) at \(P\). This shows that \(\nabla f\left(x_{0}, y_{0}, z_{0}\right), \nabla g\left(x_{0}, y_{0}, z_{0}\right),\) and \(\nabla h\left(x_{0}, y_{0}, z_{0}\right)\) all lie in the same plane. b. Assuming that \(\nabla g\left(x_{0}, y_{0}, z_{0}\right)\) and \(\nabla h\left(x_{0}, y_{0}, z_{0}\right)\) are nonzero and not parallel, explain why every point in the plane determined by \(\nabla g\left(x_{0}, y_{0}, z_{0}\right)\) and \(\nabla h\left(x_{0}, y_{0}, z_{0}\right)\) has the form \(s \nabla g\left(x_{0}, y_{0}, z_{0}\right)+t \nabla h\left(x_{0}, y_{0}, z_{0}\right)\) for some scalars \(s\) and \(t\). c. Parts (a.) and (b.) show that there must exist scalars \(\lambda\) and \(\mu\) such that $$\nabla f\left(x_{0}, y_{0}, z_{0}\right)=\lambda \nabla g\left(x_{0}, y_{0}, z_{0}\right)+\mu \nabla h\left(x_{0}, y_{0}, z_{0}\right)$$ So to optimize \(f=f(x, y, z)\) subject to the constraints \(g(x, y, z)=c\) and \(h(x, y, z)=k\) we must solve the system of equations $$\nabla f(x, y, z)=\lambda \nabla g(x, y, z)+\mu \nabla h(x, y, z)$$ $$\begin{array}{l}g(x, y, z)=c, \text { and } \\\h(x, y, z)=k .\end{array}$$ for \(x, y, z, \lambda,\) and \(\mu .\) Use this idea to find the maximum and minimum values of \(f(x, y, z)=\) \(x+2 y\) subject to the constraints \(y^{2}+z^{2}=8\) and \(x+y+z=10\).

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