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Chapter 12: Problem 66

Use the method of your choice to evaluate the following limits. $$\lim _{(u, v) \rightarrow(-1,0)} \frac{u v e^{-v}}{u^{2}+v^{2}}$$

Short Answer

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Question: Evaluate the limit: $$\lim _{(u, v) \rightarrow(-1,0)} \frac{u v e^{-v}}{u^{2}+v^{2}}$$ Answer: The limit is $$\lim_{(u, v) \rightarrow(-1,0)} \frac{u v e^{-v}}{u^{2}+v^{2}} = 0$$.

Step by step solution

01

Rewrite the function in polar coordinates

To do this, we will utilize the polar coordinate transformations: $$ u = r \cos \theta, \quad v = r \sin \theta $$ Now let's substitute these into our function: $$ f(u,v) = \frac{u v e^{-v}}{u^2+v^2} = \frac{r^2 \cos\theta \sin\theta e^{-r\sin\theta}}{r^2 (\cos^2\theta + \sin^2\theta)} $$ Simplify the function: $$ f(u,v) = \frac{r^2 \cos\theta \sin\theta e^{-r\sin\theta}}{r^2} = \cos\theta \sin\theta e^{-r\sin\theta} $$
02

Rewrite the limit in polar coordinates

Recall that the limit was given as follows: $$ \lim_{(u, v) \rightarrow(-1,0)} \frac{u v e^{-v}}{u^{2}+v^{2}} $$ We need to rewrite this limit in polar coordinates. As \((u,v) \rightarrow(-1,0)\), we have \(u = -1\) and \(v = 0\). Converting this point to polar coordinates, we find \(r = 1\) and \(\theta = \pi\). Our new limit is: $$ \lim_{(r, \theta) \rightarrow(1, \pi)} \cos\theta \sin\theta e^{-r\sin\theta} $$
03

Evaluate the limit

Now, we have to evaluate the limit: $$ \lim_{(r, \theta) \rightarrow(1, \pi)} \cos\theta \sin\theta e^{-r\sin\theta} $$ As \(r \rightarrow 1\) and \(\theta \rightarrow \pi\), we can plug these values into our function: $$ \cos(\pi) \sin(\pi) e^{-1\sin(\pi)} = -1 \cdot 0 \cdot e^0 = -1 \cdot 0 \cdot 1 = 0 $$ Thus, the limit is: $$ \boxed{\lim _{(u, v) \rightarrow(-1,0)} \frac{u v e^{-v}}{u^{2}+v^{2}} = 0} $$

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Most popular questions from this chapter

Potential functions arise frequently in physics and engineering. A potential function has the property that \(a\) field of interest (for example, an electric field, a gravitational field, or a velocity field is the gradient of the potential (or sometimes the negative of the gradient of the potential). (Potential functions are considered in depth in Chapter 14 .) The gravitational potential associated with two objects of mass \(M\) and \(m\) is \(\varphi=-G M m / r,\) where \(G\) is the gravitational constant. If one of the objects is at the origin and the other object is at \(P(x, y, z),\) then \(r^{2}=x^{2}+y^{2}+z^{2}\) is the square of the distance between the objects. The gravitational field at \(P\) is given by \(\mathbf{F}=-\nabla \varphi,\) where \(\nabla \varphi\) is the gradient in three dimensions. Show that the force has a magnitude \(|\mathbf{F}|=G M m / r^{2}\) Explain why this relationship is called an inverse square law.

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