/*! 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 28 Evaluate the following limits. ... [FREE SOLUTION] | 91Ó°ÊÓ

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Evaluate the following limits. $$\lim _{(u, v) \rightarrow(8,8)} \frac{u^{1 / 3}-v^{1 / 3}}{u^{2 / 3}-v^{2 / 3}}$$

Short Answer

Expert verified
Answer: The value of the multivariable limit is \(\frac{1}{4}\).

Step by step solution

01

Simplify the expression

We have the expression: $$\frac{u^{1 / 3}-v^{1 / 3}}{u^{2 / 3}-v^{2 / 3}}$$ Let \(x = u^{1/3}\) and \(y = v^{1/3}\). Thus, we have: $$\frac{x - y}{x^2 - y^2}$$ Notice that we can factor the denominator using the difference of squares: $$\frac{x - y}{(x - y)(x + y)}$$ Now we can cancel out \((x - y)\) from the numerator and denominator: $$\frac{1}{x + y}$$ Now let's substitute back for \(u\) and \(v\): $$\frac{1}{u^{1/3} + v^{1/3}}$$
02

Convert to polar coordinates

Now let's express \(u\) and \(v\) in polar coordinates. Let \(u = r\cos\theta\) and \(v = r\sin\theta\). Since the limit is approaching \((8,8)\), we let \(r\cos\theta = u-8\) and \(r\sin\theta = v-8\). Now the expression becomes: $$\frac{1}{(r\cos\theta + 8)^{1/3} + (r\sin\theta + 8)^{1/3}}$$
03

Evaluate the limit

Now we need to compute the limit as \((u,v)\rightarrow(8,8)\) or \((r\cos\theta,r\sin\theta) \rightarrow (0,0)\) which is equivalent to \(r \rightarrow 0\): $$\lim_{r\to 0}\frac{1}{(r\cos\theta + 8)^{1/3} + (r\sin\theta + 8)^{1/3}}$$ As \(r \to 0\), both \((r\cos\theta + 8)^{1/3}\) and \((r\sin\theta + 8)^{1/3}\) approach \((8)^{1/3} = 2\). So the limit becomes: $$\frac{1}{2 + 2} = \frac{1}{4}$$ Thus, the limit is \(\frac{1}{4}\).

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Key Concepts

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

Limits and Continuity
Understanding limits and continuity is a fundamental aspect of calculus, including multivariable calculus. At its core, a limit attempts to describe the value that a function approaches as the input (or inputs, in the multi-dimensional case) approaches a certain point. Continuity, on the other hand, implies that at a particular point, the function is defined, and its limit coincides with the function value.

When evaluating limits in multivariable calculus, one must consider all paths leading to the point in question. If limits along different paths yield different results, the limit does not exist. However, if all paths result in the same value and the function is defined at that point with matching value, the function is continuous there. In the given exercise, evaluating the limit as \(u, v\) approaches \(8,8\) tests the functions' behavior in a multi-dimensional space as it nears a specific point.
Polar Coordinates
Polar coordinates provide a way of expressing points in a plane through a radius and an angle relative to a central point (usually the origin). They are especially useful in multivariable calculus for dealing with circular and spiral types of problems where Cartesian coordinates are less practical. The coordinates are given by \(r, \theta\), where \(r\) represents the radius or distance from the origin, and \(\theta\) is the angle from the positive x-axis.

In our exercise, converting to polar coordinates aids in the simplification of the limit evaluation. The coordinate transformation facilitates approaching the limit along a straight line from the origin, which makes the analysis of the limit's behavior more straightforward.
Difference of Squares
The difference of squares is a widely-used algebraic technique for factoring expressions of the form \(a^2 - b^2\). It is based on the identity \(a^2 - b^2 = (a + b)(a - b)\). In the context of the exercise, recognizing the denominator as a difference of squares allows us to factor it and then simplify the expression significantly, turning the complex limit into a more manageable form.

By employing the difference of squares, we can split the denominator and cancel out common factors in both the numerator and the denominator, revealing a simpler relationship between the variables involved, and simplifying further steps in finding the limit.
Factoring Expressions
Factoring expressions is a crucial skill in both single-variable and multivariable calculus. It involves breaking down polynomials into the product of their factors, which can make solving equations, simplifying expressions, and evaluating limits much simpler.

In the problem at hand, after converting \(u\) and \(v\) to third powers of \(x\) and \(y\), respectively, we can factor the denominator as a difference of squares. This transformation exemplifies how factoring can reveal new ways of cancellation, simplification, and evaluation that are less apparent in the original form of an expression.
Multivariable Calculus
Multivariable calculus extends single-variable calculus concepts to functions of several variables. It includes limits, derivatives, and integrals where the functions involve more than one variable. In practice, this field allows us to explore how multi-dimensional spaces behave and interact.

For evaluating limits of functions of multiple variables—as in the given exercise—it's essential to consider the interplay between the variables and the possibility of approaching the target point from different paths in space. The exercise showcases that while the processes are similar to those of single-variable limits, mutivariable limits require a unique approach that often involves creative algebraic manipulations or coordinate transformations to tease out the limiting behavior.

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Most popular questions from this chapter

Suppose you make monthly deposits of \(P\) dollars into an account that earns interest at a monthly rate of \(p \% .\) The balance in the account after \(t\) years is \(B(P, r, t)=P\left(\frac{(1+r)^{12 t}-1}{r}\right),\) where \(r=\frac{p}{100}\) (for example, if the annual interest rate is \(9 \%,\) then \(p=\frac{9}{12}=0.75\) and \(r=0.0075) .\) Let the time of investment be fixed at \(t=20\) years. a. With a target balance of \(\$ 20,000,\) find the set of all points \((P, r)\) that satisfy \(B=20,000 .\) This curve gives all deposits \(P\) and monthly interest rates \(r\) that result in a balance of \(\$ 20,000\) after 20 years. b. Repeat part (a) with \(B=\$ 5000, \$ 10,000, \$ 15,000,\) and \(\$ 25,000,\) and draw the resulting level curves of the balance function.

Challenge domains Find the domain of the following functions. Specify the domain mathematically, and then describe it in words or with a sketch. $$f(x, y, z)=\ln \left(z-x^{2}-y^{2}+2 x+3\right)$$

Heron's formula The area of a triangle with sides of length \(a, b\) and \(c\) is given by a formula from antiquity called Heron's formula: $$A=\sqrt{s(s-a)(s-b)(s-c)}$$ where \(s=\frac{1}{2}(a+b+c)\) is the semiperimeter of the triangle. a. Find the partial derivatives \(A_{\sigma}, A_{b},\) and \(A_{c}\) b. A triangle has sides of length \(a=2, b=4, c=5 .\) Estimate the change in the area when \(a\) increases by \(0.03, b\) decreases by \(0.08,\) and \(c\) increases by 0.6 c. For an equilateral triangle with \(a=b=c,\) estimate the percent change in the area when all sides increase in length by \(p \% .\)

Find the domain of the following functions. If possible, give a description of the domains (for example, all points outside a sphere of radius 1 centered at the origin). $$Q(x, y, z)=\frac{10}{1+x^{2}+y^{2}+4 z^{2}}$$

Traveling waves in general Generalize Exercise 79 by considering a set of waves described by the function \(z=A+\sin (a x-b y),\) where \(a, b,\) and \(A\) are real numbers. a. Find the direction in which the crests and troughs of the waves are aligned. Express your answer as a unit vector in terms of \(a\) and \(b\). b. Find the surfer's direction- that is, the direction of steepest descent from a crest to a trough. Express your answer as a unit vector in terms of \(a\) and \(b\).

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