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The temperature on an unevenly heated metal plate positioned in the first quadrant of the \(x y\) -plane is given by $$C(x, y)=\frac{25 x y+25}{(x-1)^{2}+(y-1)^{2}+1}$$ Assume that temperature is measured in degrees Celsius and that \(x\) and \(y\) are each measured in inches. (Note: At no point in the following questions should you expand the denominator of \(C(x, y) .)\) a. Determine \(\left.\frac{\partial C}{\partial x}\right|_{(x, y)}\) and \(\left.\frac{\partial C}{\partial y}\right|_{(x, y)}\). b. If an ant is on the metal plate, standing at the point (2,3) , and starts walking in the direction parallel to the positive \(y\) axis, at what rate will the temperature the ant is experiencing change? Explain, and include appropriate units. c. If an ant is walking along the line \(y=3\) in the positive \(x\) direction, at what instantaneous rate will the temperature the ant is experiencing change when the ant passes the point (1,3)\(?\) d. Now suppose the ant is stationed at the point (6,3) and walks in a straight line towards the point (2,0) . Determine the average rate of change in temperature (per unit distance traveled) the ant encounters in moving between these two points. Explain your reasoning carefully. What are the units on your answer?

Step by step solution

01

Computing the partial derivatives

The given temperature function is \(C(x, y)=\frac{25 x y+25}{(x-1)^{2}+(y-1)^{2}+1}\). To find \(\frac{\partial C}{\partial x}\) (the derivative with respect to \(x\)), use the quotient rule: \[ \frac{\partial C}{\partial x} = \frac{25y((x-1)^2 +(y-1)^2 +1) - ((25xy+25)2(x-1))}{((x-1)^2 +(y-1)^2 +1)^2} \] Simplify this to get: \[ \frac{\partial C}{\partial x} =\frac{25y(x-1)^2 +25y(y-1)^2 +25y -50x(x-1)}{(x-1)^2 +(y-1)^2 +1}}. \] For \(\frac{\partial C}{\partial y}\) (the derivative with respect to \(y\)), use the same quotient rule: \[ \frac{\partial C}{\partial y} = \frac{25x((x-1)^2 +(y-1)^2 +1) - ((25xy+25)2(y-1))}{((x-1)^2 +(y-1)^2 +1)^2} \] Simplify this to get: \[ \frac{\partial C}{\partial y} = \frac{25x(x-1)^2 +25x(y-1)^2 +25x -50y(y-1)}{(x-1)^2 +(y-1)^2 +1}}. \]

02

Evaluating the \(y\) derivative at a given point

To find the rate of temperature change as the ant moves along the positive \(y\) axis from the point (2,3), plug \(x = 2\) and \(y = 3\) into the expression derived for \(\frac{\partial C}{\partial y}\).

03

Evaluating the \(x\) derivative at a given point

The ant is moving along a line of constant \(y\) hence we need to evaluate the partial derivative with respect to \(x\) at the point (1,3). Plug \(x = 1\) and \(y = 3\) into the expression derived for \(\frac{\partial C}{\partial x}\).

04

Finding the average rate of change in temperature

The average rate of change in temperature is the difference in temperature between the points divided by the distance between these two points. Calculate the distance between (2,0) and (6,3) using the distance formula \(\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\), then find the difference in temperature by subtracting \(C(2,0)\) from \(C(6,3)\), and divide the difference in temperature by the distance to find the average rate of change.

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Key Concepts

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

Multivariable Calculus

Multivariable calculus extends the concepts of single-variable calculus to functions with several variables. These functions can describe a wide range of phenomena, from the changing temperatures across a surface to the economics of supply and demand. A fundamental tool in multivariable calculus is the partial derivative, which measures how a function changes as one of its input variables is varied while the others are held constant.

For example, given the temperature function of a metal plate, such as in the exercise, finding the partial derivative with respect to x tells us how temperature changes in the east-west direction at a particular point, while doing so with respect to y tells us about the north-south temperature change at that point. In problems involving multivariable functions, it's crucial to grasp which variable is changing and how it affects the overall function. Understanding partial derivatives allows one to compute rates of change specific to each variable, leading to better insight into the behavior of complex systems.

Temperature Gradient

The temperature gradient is a vector that points in the direction of the greatest rate of increase of the temperature and whose magnitude is the rate of increase per unit distance. It is given by the gradient of the temperature function, which is a vector of all its first partial derivatives. In the context of the exercise, the temperature gradient would help the ant understand not merely how temperature changes in a particular direction, but where to go if it wishes to move towards higher or lower temperatures.

The concept of the gradient is particularly useful in scenarios where the direction in which a variable changes most rapidly is of interest. For instance, in meteorology, understanding the temperature gradient can be crucial for predicting weather patterns. In the case of our metal plate, the gradient would point towards the hottest part of the plate and the ant could follow this to find the warmest spot.

Rate of Change

The rate of change, in a mathematical context, refers to how a quantity changes with respect to variation in another quantity. For a function, this concept is quantified by its derivative. In the context of our exercise, we are concerned with how temperature changes as position changes; thus, we look at the partial derivatives of the temperature function.

The rate of temperature change, which the ant experiences as it moves, is directly given by the partial derivatives calculated. The derivatives we calculate are known as the instantaneous rates of change at a particular point because they describe the temperature change at exactly that point and in a specific direction. On the other hand, if we wanted to know the average rate of change of temperature over a distance, we would compare the temperatures at two distinct points and divide by the distance traveled, as seen in parts (b) and (d) of the exercise. Understanding these rates of change is vital for making predictions and understanding how different variables affect one another in both theoretical and real-world applications.