# Use polar coordinates to find the limit. [Hint: Let x = r cos and y = r sin , and note that (x, y) (0, 0) implies r 0.] lim_((x,y)->(0,0)) (x^2-y^2)/sqrt(x^2+y^2)

Question
Multivariable functions
Use polar coordinates to find the limit. [Hint: Let $$\displaystyle{x}={r}{\cos{{\quad\text{and}\quad}}}{y}={r}{\sin{}}$$ , and note that (x, y) (0, 0) implies r 0.] $$\displaystyle\lim_{{{\left({x},{y}\right)}\to{\left({0},{0}\right)}}}\frac{{{x}^{{2}}-{y}^{{2}}}}{\sqrt{{{x}^{{2}}+{y}^{{2}}}}}$$

2021-02-06
Given the multivariable limit function:
$$\displaystyle\lim_{{{\left({x},{y}\right)}\to{\left({0},{0}\right)}}}\frac{{{x}^{{2}}-{y}^{{2}}}}{\sqrt{{{x}^{{2}}+{y}^{{2}}}}}$$
For polar coordinate system:
$$\displaystyle{x}={r}{\cos{\theta}},{y}={r}{\sin{\theta}}$$
$$\displaystyle\lim_{{{\left({x},{y}\right)}\to{\left({0},{0}\right)}}}\frac{{{x}^{{2}}-{y}^{{2}}}}{\sqrt{{{x}^{{2}}+{y}^{{2}}}}}=\lim_{{{r}\to{0}}}\frac{{{r}^{{2}}{{\cos}^{{2}}\theta}-{r}^{{2}}{{\sin}^{{2}}\theta}}}{\sqrt{{{r}^{{2}}{{\cos}^{{2}}\theta}+{r}^{{2}}{{\sin}^{{2}}\theta}}}}$$
$$\displaystyle\lim_{{{\left({x},{y}\right)}\to{\left({0},{0}\right)}}}\frac{{{x}^{{2}}-{y}^{{2}}}}{\sqrt{{{x}^{{2}}+{y}^{{2}}}}}=\lim_{{{r}\to{0}}}\frac{{{r}^{{2}}{\left({{\cos}^{{2}}\theta}-{r}^{{2}}{{\sin}^{{2}}\theta}\right)}}}{\sqrt{{{r}^{{2}}{\left({{\cos}^{{2}}\theta}+{r}^{{2}}{{\sin}^{{2}}\theta}\right)}}}}$$
We know the trigonometry identities
$$\displaystyle{\left({{\cos}^{{2}}\theta}-{{\sin}^{{2}}\theta}\right)}={\cos{{2}}}\theta$$
$$\displaystyle{\left({{\cos}^{{2}}\theta}{1}{{\sin}^{{2}}\theta}\right)}={1}$$
$$\displaystyle\lim_{{{\left({x},{y}\right)}\to{\left({0},{0}\right)}}}\frac{{{x}^{{2}}-{y}^{{2}}}}{\sqrt{{{x}^{{2}}+{y}^{{2}}}}}=\lim_{{{r}\to{0}}}\frac{{{r}^{{2}}{\cos{{2}}}\theta}}{{r}}$$
$$\displaystyle\lim_{{{\left({x},{y}\right)}\to{\left({0},{0}\right)}}}\frac{{{x}^{{2}}-{y}^{{2}}}}{\sqrt{{{x}^{{2}}+{y}^{{2}}}}}=\lim_{{{r}\to{0}}}{r}{\cos{{2}}}\theta$$
$$\displaystyle\lim_{{{\left({x},{y}\right)}\to{\left({0},{0}\right)}}}\frac{{{x}^{{2}}-{y}^{{2}}}}{\sqrt{{{x}^{{2}}+{y}^{{2}}}}}={0}$$
Hence the limit of the given function is zero.

### Relevant Questions

The bulk density of soil is defined as the mass of dry solidsper unit bulk volume. A high bulk density implies a compact soilwith few pores. Bulk density is an important factor in influencing root development, seedling emergence, and aeration. Let X denotethe bulk density of Pima clay loam. Studies show that X is normally distributed with $$\displaystyle\mu={1.5}$$ and $$\displaystyle\sigma={0.2}\frac{{g}}{{c}}{m}^{{3}}$$.
(a) What is thedensity for X? Sketch a graph of the density function. Indicate onthis graph the probability that X lies between 1.1 and 1.9. Findthis probability.
(b) Find the probability that arandomly selected sample of Pima clay loam will have bulk densityless than $$\displaystyle{0.9}\frac{{g}}{{c}}{m}^{{3}}$$.
(c) Would you be surprised if a randomly selected sample of this type of soil has a bulkdensity in excess of $$\displaystyle{2.0}\frac{{g}}{{c}}{m}^{{3}}$$? Explain, based on theprobability of this occurring.
(d) What point has the property that only 10% of the soil samples have bulk density this high orhigher?
(e) What is the moment generating function for X?
Let $$\displaystyle{z}={e}^{{{x}^{{2}}+{3}{y}}}+{x}^{{3}}{y}^{{2}}$$, where $$\displaystyle{x}={t}{\cos{{r}}}{\quad\text{and}\quad}{y}={r}{t}^{{4}}$$. Use the chain rule fot multivariable functions (Cals III Chain Rule) to find $$\displaystyle\frac{{\partial{z}}}{{\partial{r}}}$$ and $$\displaystyle\frac{{\partial{z}}}{\partial}{t}{)}$$. Give your answers in terms of r and t only. Be sure to show all of your work.
Let $$\displaystyle{f{{\left({x},{y}\right)}}}=-\frac{{{x}{y}}}{{{x}^{{2}}+{y}^{{2}}}}$$.
Find limit of $$\displaystyle{f{{\left({x},{y}\right)}}}{a}{s}{\left({x},{y}\right)}\rightarrow{\left({0},{0}\right)}{i}{)}{A}{l}{o}{n}{g}{y}{a}\xi{s}{\quad\text{and}\quad}{i}{i}{)}{a}{l}{o}{n}{g}{t}{h}{e}{l}\in{e}{y}={x}.{E}{v}{a}{l}{u}{a}{t}{e}\Lim{e}{s}\lim_{{{\left({x},{y}\right)}\rightarrow{\left({0},{0}\right)}}}{y}.{\log{{\left({x}^{{2}}+{y}^{{2}}\right)}}}$$,by converting to polar coordinates.
Nonexistence of a limit Investiage the limit $$\displaystyle\lim_{{{\left({x},{y}\right)}\to{\left({0},{0}\right)}}}\frac{{\left({x}+{y}\right)}^{{2}}}{{{x}^{{2}}+{y}^{{2}}}}$$
What tis the complete domain D and range R of the following multivariable functions: $$\displaystyle{w}{\left({x},{y}\right)}=\sqrt{{{y}-{4}{x}^{{2}}}}$$
Let $$\displaystyle{z}{\left({x},{y}\right)}={e}^{{{3}{x}{y}}},{x}{\left({p},{q}\right)}=\frac{{p}}{{q}}{\quad\text{and}\quad}{y}{\left({p},{q}\right)}=\frac{{q}}{{p}}$$ are functions. Use multivariable chain rule of partial derivatives to find
(i) $$\displaystyle\frac{{\partial{z}}}{{\partial{p}}}$$
(ii) $$\displaystyle\frac{{\partial{z}}}{{\partial{q}}}$$.
To determine:
a) Whether the statement, " The point with Cartesian coordinates $$\displaystyle{\left[\begin{array}{cc} -{2}&\ {2}\end{array}\right]}$$ has polar coordinates $$\displaystyle{\left[{b}{f}{\left({2}\sqrt{{{2}}},\ {\frac{{{3}\pi}}{{{4}}}}\right)}\ {\left({2}\sqrt{{{2}}},{\frac{{{11}\pi}}{{{4}}}}\right)}\ {\left({2}\sqrt{{{2}}},\ -{\frac{{{5}\pi}}{{{4}}}}\right)}\ {\quad\text{and}\quad}\ {\left(-{2}\sqrt{{2}},\ -{\frac{{\pi}}{{{4}}}}\right)}\right]}$$ " is true or false.
b) Whether the statement, " the graphs of $$\displaystyle{\left[{r}{\cos{\theta}}={4}\ {\quad\text{and}\quad}\ {r}{\sin{\theta}}=\ -{2}\right]}$$ intersect exactly once " is true or false.
c) Whether the statement, " the graphs of $$\displaystyle{\left[{r}={4}\ {\quad\text{and}\quad}\ \theta={\frac{{\pi}}{{{4}}}}\right]}$$ intersect exactly once ", is true or false.
d) Whether the statement, " the point $$\displaystyle{\left[\begin{array}{cc} {3}&{\frac{{\pi}}{{{2}}}}\end{array}\right]}{l}{i}{e}{s}{o}{n}{t}{h}{e}{g}{r}{a}{p}{h}{o}{f}{\left[{r}={3}{\cos{\ }}{2}\ \theta\right]}$$ " is true or false.
e) Whether the statement, " the graphs of $$\displaystyle{\left[{r}={2}{\sec{\theta}}\ {\quad\text{and}\quad}\ {r}={3}{\csc{\theta}}\right]}$$ are lines " is true or false.
A surface is represented by the following multivariable function,
$$\displaystyle{f{{\left({x},{y}\right)}}}=\frac{{1}}{{3}}{x}^{{3}}+{y}^{{2}}-{2}{x}{y}-{6}{x}-{3}{y}+{4}$$
a) Calculate $$\displaystyle{f}_{{\times}},{f}_{{{y}{x}}},{f}_{{{x}{y}}}{\quad\text{and}\quad}{f}_{{{y}{y}}}$$
b) Calculate coordinates of stationary points.
c) Classify all stationary points.
COnsider the multivariable function $$\displaystyle{g{{\left({x},{y}\right)}}}={x}^{{2}}-{3}{y}^{{4}}{x}^{{2}}+{\sin{{\left({x}{y}\right)}}}$$. Find the following partial derivatives: $$\displaystyle{g}_{{x}}.{g}_{{y}},{g}_{{{x}{y}}},{g{{\left(\times\right)}}},{g{{\left({y}{y}\right)}}}$$.
To find: The equivalent polar equation for the given rectangular-coordinate equation.
Given:
$$\displaystyle\ {x}=\ {r}{\cos{\theta}}$$
$$\displaystyle\ {y}=\ {r}{\sin{\theta}}$$
b. From rectangular to polar:
$$\displaystyle{r}=\pm\sqrt{{{x}^{{{2}}}\ +\ {y}^{{{2}}}}}$$
$$\displaystyle{\cos{\theta}}={\frac{{{x}}}{{{r}}}},{\sin{\theta}}={\frac{{{y}}}{{{r}}}},{\tan{\theta}}={\frac{{{x}}}{{{y}}}}$$
Calculation:
Given: equation in rectangular-coordinate is $$\displaystyle{y}={x}$$.
Converting into equivalent polar equation -
$$\displaystyle{y}={x}$$
Put $$\displaystyle{x}={r}{\cos{\theta}},\ {y}={r}{\sin{\theta}},$$
$$\displaystyle\Rightarrow\ {r}{\sin{\theta}}={r}{\cos{\theta}}$$
$$\displaystyle\Rightarrow\ {\frac{{{\sin{\theta}}}}{{{\cos{\theta}}}}}={1}$$
$$\displaystyle\Rightarrow\ {\tan{\theta}}={1}$$
Thus, desired equivalent polar equation would be $$\displaystyle\theta={1}$$
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