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Chapter 9: Problem 17

Consider the function \(h\) defined by \(h(x, y)=8-\sqrt{4-x^{2}-y^{2}}\). a. What is the domain of \(h ?\) (Hint: describe a set of ordered pairs in the plane by explaining their relationship relative to a key circle.) b. The range of a function is the set of all outputs the function generates. Given that the range of the square root function \(g(t)=\sqrt{t}\) is the set of all nonnegative real numbers, what do you think is the range of \(h ?\) Why? c. Choose 4 different values from the range of \(h\) and plot the corresponding level curves in the plane. What is the shape of a typical level curve? d. Choose 5 different values of \(x\) (including at least one negative value and zero), and sketch the corresponding traces of the function \(h\). e. Choose 5 different values of \(y\) (including at least one negative value and zero), and sketch the corresponding traces of the function \(h\). f. Sketch an overall picture of the surface generated by \(h\) and write at least one sentence to describe how the surface appears visually. Does the surface remind you of a familiar physical structure in nature?

Short Answer

Expert verified
The domain of h is the set of all ordered pairs (x, y) that lie inside or on the circle with radius 2 centered at the origin. The range of h is [6, 8]. The level curves of h are circles of varying radii. The x-traces and y-traces are sections of a circle. The overall surface generated by h resembles a circular crater with a peak at the origin and gradually descending towards the boundary circle, which can resemble a volcano-like structure in nature.

Step by step solution

01

1. Determine the domain of h

To determine the domain of h, we need to find the set of all possible input values (x, y) for which the function is defined. The function h is defined as \(h(x, y) = 8 - \sqrt{4 - x^2 - y^2}\). The only restriction here is the square root: we need to ensure the term inside the square root is nonnegative. Therefore: \(4 - x^2 - y^2 \geq 0\) Add x虏 and y虏 to both sides: \(x^2 + y^2 \leq 4\) This inequality represents a filled circle with a radius of 2, centered at the origin (0, 0). Therefore, the domain of h is the set of all ordered pairs (x, y) such that they lie inside or on the circle with radius 2 centered at the origin.
02

2. Determine the range of h

The range of h is the set of all possible output values it can produce. Note that the term inside the square root, \(4 - x^2 - y^2\), ranges from 0 (when x虏 + y虏 = 4) to 4 (when x = 0 and y = 0). Thus, the range of the square root part, \(\sqrt{4 - x^2 - y^2}\), is [0, 2]. Since h(x, y) is defined as 8 minus the square root part, the range of h is given by 8 minus the values in [0, 2], resulting in [6, 8].
03

3. Plot level curves

To plot the level curves, we need to choose 4 values from the range of h (which we found to be [6, 8]) and equate the function h(x, y) to these values. Let's choose 6, 7, 7.5, and 8 as the values. We then have: 1. \(h(x, y) = 6 \Rightarrow 8 - \sqrt{4 - x^2 - y^2} = 6\) 2. \(h(x, y) = 7 \Rightarrow 8 - \sqrt{4 - x^2 - y^2} = 7\) 3. \(h(x, y) = 7.5 \Rightarrow 8 - \sqrt{4 - x^2 - y^2} = 7.5\) 4. \(h(x, y) = 8 \Rightarrow 8 - \sqrt{4 - x^2 - y^2} = 8\) Solving these equations for the corresponding values of x and y will yield the level curves, which will turn out to be circles of varying radii.
04

4. Sketch x-traces of h

To sketch the x-traces, we need to choose 5 values of x and find the corresponding values of y for which h(x, y) is defined. Let's choose x = -2, -1, 0, 1, and 2. For each value of x, we can plug it into the domain inequality: \(x^2 + y^2 \leq 4\) Solve for y and sketch the corresponding values of y for each x value. These will be sections of a circle.
05

5. Sketch y-traces of h

Similarly, to sketch the y-traces, we need to choose 5 values of y and find the corresponding values of x for which h(x, y) is defined. Let's choose y = -2, -1, 0, 1, and 2. For each value of y, plug it into the domain inequality: \(x^2 + y^2 \leq 4\) Solve for x and sketch the corresponding values of x for each y value. These will also be sections of a circle.
06

6. Describe the surface generated by h

When we combine the level curves and traces, we obtain the overall surface generated by the function h(x, y). The surface appears like a circular crater with a peak at the origin and gradually descending towards the boundary circle (x虏 + y虏 = 4) in the shape of a paraboloid. This surface can resemble a volcano-like structure in nature.

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Key Concepts

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

Domain of a Function
In multivariable calculus, the domain of a function consists of all the possible input values that will output real numbers. For the given function h, it is defined as h(x, y)=8- \(\sqrt{4-x^{2}-y^{2}}\). The domain must be such that no undefined mathematical operations occur, like taking the square root of a negative number.

For function h, we ensure that the expression under the square root, 4 - x虏 - y虏, remains non-negative for it to be real. This condition creates a filled-in circle, or a 'disk', as the domain where any point (x, y) inside or on the edge of that circle is valid. This means the function h is defined for all points within the circle of radius 2 centered at the origin in the xy-plane.
Range of a Function
The range of a function is the set of all possible output values a function can produce when evaluating the domain. For the function h(x, y) under consideration, it becomes crucial to determine the range of the square root function within it to ensure the output is real. The function g(t)= \(\sqrt{t}\) has a range of all nonnegative real numbers.

As for h, we adjust the range by the transformation 8 - \(\sqrt{4 - x^2 - y^2}\), resulting in the range from 6 to 8, including both endpoints. This tells us h will only yield real numbers between 6 and 8 for any (x, y) within the domain.
Level Curves
Level curves are a powerful tool for visualizing functions of two variables. A level curve represents all points where the function has the same output value. In graphing the set of points for various fixed outputs of h(x, y), we get circles with various radii on the xy-plane.

By selecting specific output values, like 6, 7, 7.5, and 8 from the range of h, we look at the inverse operation of 8 - \(\sqrt{4 - x^2 - y^2}\) to find the corresponding x and y. These values produce circles that decrease in radius as the output value increases, creating concentric circles with the smallest corresponding to the highest value, 8.
Function Traces
Function traces offer a slice of the graph of the function by fixing one variable and letting the other vary. When we sketch the traces of h(x, y) by fixing x and varying y, and vice versa, we essentially look at cross-sections of the surface created by the function.

By choosing five different values for x (such as -2, -1, 0, 1, 2) and for y, we determine the corresponding y and x values that satisfy the domain condition. These traces reflect vertical and horizontal cross-sections through the surface, which in this case, are segments of circles centered at the origin.
Surface Sketching
Surface sketching synthesizes what we understand from level curves and traces into a three-dimensional perspective. For the function h(x, y), when we draw the level curves and connect them with the function traces, a 3D image begins to form.

This visualization helps us see not just individual points or curves but the entire shape that the function model; our surface resembles a downward opening paraboloid. It's a shape with a high point at the center, tapering off smoothly and symmetrically to the edges, akin to a three-dimensional representation of a topographic map with elevations.
Paraboloid
A paraboloid is a specific type of surface commonly found within multivariable calculus. It can be construed as the shape generated by a parabola that is rotated around its axis of symmetry. Depending on the direction of the opening of this parabola, a paraboloid can be upward or downward opening.

In the case of function h(x, y)=8- \(\sqrt{4-x^{2}-y^{2}}\), the surface created is a downward opening paraboloid. This shape is reminiscent of natural structures, such as volcanic craters, which have a peak at the center鈥攖he origin鈥攄escending steeply to the edges. The level curves and function traces of h substantiate this by showing circular curves and circular cross-sections decreasing in size towards the peak of the 'volcano.'

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Most popular questions from this chapter

Consider the concentration, \(\mathrm{C},\) (in \(\mathrm{mg} /\) liter \()\) of a drug in the blood as a function of the amount of drug given, \(\mathrm{x},\) and the time since injection, \(\mathrm{t} .\) For \(0 \leq x \leq 5 \mathrm{mg}\) and \(t \geq 0\) hours, we have $$ C=f(x, t)=22 t e^{-(5-x) t} $$ f(1,4)= __________ Give a practical interpretation of your answer: \(f(1,4)\) is \(\odot\) the change in concentration of a \(4 \mathrm{mg}\) dose in the blood 1 hours after injection. \(\odot\) the concentration of a \(4 \mathrm{mg}\) dose in the blood 1 hours after injection. \(\odot\) the amount of a \(1 \mathrm{mg}\) dose in the blood 4 hours after injection. \(\odot\) the change in concentration of a \(1 \mathrm{mg}\) dose in the blood 4 hours after injection. \(\odot\) the concentration of a \(1 \mathrm{mg}\) dose in the blood 4 hours after injection. \(\odot\) the amount of a \(4 \mathrm{mg}\) dose in the blood 1 hours after injection.

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