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Section 9.5 Week 9 Activity

Subsection 9.5.1 Multivariable Limits

Activity 9.5.1.

Calculate this limit or prove that it does not exist. If the limit doesn't exist, describe the behaviour approaching the limit point.

\begin{equation*} \lim_{(x,y) \rightarrow (3,4)} \frac{x^2 + y^2}{x - y + 3} \end{equation*}
Solution.

This limit can be directly evaluated, since there are no division by zero (or similar) problems.

\begin{equation*} \lim_{(x,y) \rightarrow (3,4)} \frac{x^2 + y^2}{x - y + 3} = \frac{3^2 + 4^2}{3-4+3} = \frac{25}{2} \end{equation*}

Activity 9.5.2.

Calculate this limit or prove that it does not exist. If the limit doesn't exist, describe the behaviour approaching the limit point.

\begin{equation*} \lim_{(x,y) \rightarrow (0,0)} \frac{xy + x^2y + xy^2}{x^2y + 3xy} \end{equation*}
Solution.

I can factor \(xy\) out of the numerator and the denominator.

\begin{equation*} \lim_{(x,y) \rightarrow (0,0)} \frac{xy + x^2y + xy^2}{x^2y + 3xy} = \lim_{(x,y) \rightarrow (0,0)} \frac{xy(1 + x + y)}{xy(x + 3)} = \lim_{(x,y) \rightarrow (0,0)} \frac{1 + x + y}{x + 3} \end{equation*}

After this \(xy\) term is cancelled out, the remaining limit can just be evaluted.

\begin{equation*} \lim_{(x,y) \rightarrow (0,0)} \frac{1 + x + y}{x + 3} = \frac{1 + 0 + 0}{0 + 3} = \frac{1}{3} \end{equation*}

Activity 9.5.3.

Calculate this limit or prove that it does not exist. If the limit doesn't exist, describe the behaviour approaching the limit point.

\begin{equation*} \lim_{(x,y) \rightarrow (4,\pi)} \frac{x}{\sin y} \end{equation*}
Solution.

The numerator approaches \(4\) and the denominator approaches \(0\text{,}\) so the limit must diverge to \(\pm \infty\text{.}\) Whether it diverges to positive or negative \(\infty\) will depend on the direction of approach.

Activity 9.5.4.

Calculate this limit or prove that it does not exist. If the limit doesn't exist, describe the behaviour approaching the limit point.

\begin{equation*} \lim_{(x,y) \rightarrow (-3,0)} \frac{x^2 + 1}{y^2 - 4} \end{equation*}
Solution.

This limit can be directly evaluted.

\begin{equation*} \lim_{(x,y) \rightarrow (-3,0)} \frac{x^2 + 1}{y^2 - 4} = \frac{(-3)^1 + 1}{0^2 - 4} = \frac{10}{-4} = \frac{-5}{2} \end{equation*}

Activity 9.5.5.

Calculate this limit or prove that it does not exist. If the limit doesn't exist, describe the behaviour approaching the limit point.

\begin{equation*} \lim_{(x,y) \rightarrow (4,2)} \frac{x^2 + 1}{y^2 - 4} \end{equation*}
Solution.

The numerator approaches \(17\) and the denominator approaches \(0\text{,}\) so the limit must diverge to \(\pm \infty\text{.}\) Whether it diverges to positive or negative \(\infty\) will depend on the direction of approach. Specifically, it depends where the approach falls in relation to the line \(y = 2\text{.}\) The approach cannot cross that line, since the function is undefined all along that line.

Activity 9.5.6.

Calculate this limit or prove that it does not exist. If the limit doesn't exist, describe the behaviour approaching the limit point.

\begin{equation*} \lim_{(x,y) \rightarrow (0,0)} \frac{4x^2 - 3xy - y^2}{2x^2 + xy - 5y^2} \end{equation*}
Solution.

I will try to prove that this limit does not exist by approaches along a variety of lines. I do this by setting \(y = mx\text{,}\) turing this into a single variable limit in \(x\text{.}\)

\begin{equation*} \lim_{(x,y) \rightarrow (0,0)} \frac{4x^2 - 3xy - y^2}{2x^2 + xy - 5y^2} = \lim_{x \rightarrow 0} \frac{4x^2 - 3xmx - m^2x^2}{2x^2 + xmx - 5m^2 x^2} \end{equation*}

Then I can factor \(x^2\) out of numerator and denominator.

\begin{equation*} \lim_{x \rightarrow 0} \frac{4x^2 - 3mx^2 - m^2x^2}{2x^2 + mx^2 - 5m^2 x^2} = \lim_{x \rightarrow 0} \frac{x^2(4 - 3m - m^2)}{x^2 (2 + m - 5m^2)} = \lim_{x \rightarrow 0} \frac{4 - 3m - m^2}{2 + m - 5m^2} \end{equation*}

Then I can just evaluate the limit, since there are no \(x\) variables left at all..

\begin{equation*} \lim_{x \rightarrow 0} \frac{4 - 3m - m^2}{2 + m - 5m^2} = \frac{4 - 3m - m^2}{2 + m - 5m^2} \end{equation*}

Now I can argue that the limit depends on the slope of the line of approach. If \(m=0\text{,}\) the limit is \(\frac{4}{2} = 2\text{.}\) If \(m=1\text{,}\) the limit is \(\frac{4 - 3 - 1}{2 + 1 - 5} = 0\text{.}\) Since the limit depends on the direction of approach, the limit does not exist.

Activity 9.5.7.

Calculate this limit or prove that it does not exist. If the limit doesn't exist, describe the behaviour approaching the limit point.

\begin{equation*} \lim_{(x,y,z) \rightarrow (0,0,0)} \frac{z^2 - xy + y^2 - 4}{5xy^2 - xyz + 3} \end{equation*}
Solution.

This limit can be directly evaluted.

\begin{equation*} \lim_{(x,y,z) \rightarrow (0,0,0)} \frac{z^2 - xy + y^2 - 4}{5xy^2 - xyz + 3} = \frac{0 - 0 + 0 - 4}{0 - 0 + 3} = \frac{-4}{3} \end{equation*}

Activity 9.5.8.

Calculate this limit or prove that it does not exist. If the limit doesn't exist, describe the behaviour approaching the limit point.

\begin{equation*} \lim_{(x,y,z) \rightarrow (0,0,0)} \frac{xy - y^2 + yz}{z^2 - 3xy + x^2} \end{equation*}
Solution.

I want to argue that this limit doesn't exist by approaching along various lines. I can set \(y = mx\) and \(z = mx\) to make this a single-variable limit approach along some lines which depend on the parameter \(m\text{.}\) (There aren't all possible lines of approach, but all I need to show is that there are at least a few directions which lead to different values).

\begin{equation*} \lim_{(x,y,z) \rightarrow (0,0,0)} \frac{xy - y^2 + yz}{z^2 - 3xy + x^2} = \lim_{x \rightarrow 0} \frac{xmx - m^2x^2 + m^2x^2}{m^2x^2 - 3mxx + x^2} \end{equation*}

Then I can simplify, factor out \(x^2\text{,}\) cancel it off, and evlaute the limit.

\begin{align*} \lim_{x \rightarrow 0} \frac{mx^2 - m^2x^2 + m^2x^2}{m^2x^2 - 3mx^2 + x^2} \amp = \lim_{x \rightarrow 0} \frac{mx^2}{x^2(m^2 - 3m + 1)} \\ \amp = \lim_{x \rightarrow 0} \frac{m}{m^2 - 3m + 1} = \frac{m}{m^2 - 3m + 1} \end{align*}

I get a result which does depend on the direction of approach, determined by the parameter \(m\text{.}\) The limit approaches a variety of values based on this parameter, so the limit cannot exist.

Activity 9.5.9.

Calculate this limit or prove that it does not exist. If the limit doesn't exist, describe the behaviour approaching the limit point.

\begin{equation*} \lim_{(x,y,z) \rightarrow (0,0,0)} \arctan \frac{1}{|x+y+z|} \end{equation*}
Solution.

I can work out this limit logically. As we approach the origin, the denoimnator approaches \(0\text{.}\) The absolute value ensure that the denominator, hence the whole limit, is positive. Since the numerator is fixed, the fraction is approaching \(\infty\text{.}\) Then I apply the arctangent function. That function has a horizontal asymptote at \(y = \frac{\pi}{2}\) as the input goes to \(\infty\text{.}\) Therefore, I can conclude that this limit approaches \(\frac{\pi}{2}\text{.}\)

Subsection 9.5.2 Partial Derivatives

Activity 9.5.10.

Calculate this partial derivative.

\begin{equation*} \frac{\del}{\del x} 4x^2 - 4y^2 + 3xy + 4 \end{equation*}
Solution.

I differetiate in the \(x\) variable, treating other variables as constant.

\begin{align*} \frac{\del}{\del x} 4x^2 - 4y^2 + 3xy + 4 \amp = 8x - 0 + 3y = 8x + 3y \end{align*}

Activity 9.5.11.

Calculate this partial derivative.

\begin{equation*} \frac{\del}{\del y} \cos (3x^2 + 4y^2) \end{equation*}
Solution.

I differetiate in the \(y\) variable, treating other variables as constant.

\begin{align*} \frac{\del}{\del y} \cos (3x^2 + 4y^2) \amp = \frac{d}{du} \cos u \Bigg|_{u=3x^2 + 4y^2} \frac{d}{dy} (3x^2 + 4y^2) \\ \amp = -\sin u \Bigg|_{u = 3x^2 + 4y^2} (8y) = -8y \sin (3x^2 + 4y^2) \end{align*}

Activity 9.5.12.

Calculate this partial derivative.

\begin{equation*} \frac{\del}{\del z} \ln (xyz) \end{equation*}
Solution.

I differetiate in the \(z\) variable, treating other variables as constant.

\begin{align*} \frac{\del}{\del z} \ln (xyz) \amp = \frac{d}{du} \ln u \Bigg|_{u = xyz} \frac{\del}{\del z} xyz = \frac{1}{u} \Bigg|_{u = xyz} xy = \frac{xy}{xyz} = \frac{1}{z} \end{align*}

Activity 9.5.13.

Calculate this partial derivative.

\begin{equation*} \frac{\del^2}{\del x^2} e^{x^2 + y^2} \end{equation*}
Solution.

I differetiate in the \(x\) variable twice, treating other variables as constant. I use the chain rule in the first step, and then the product rule and the chain rule for the second partial derivative.

\begin{align*} \frac{\del^2}{\del x^2} e^{x^2 + y^2} \amp = \frac{\del}{\del x} \left[ \frac{d}{du} e^u \Bigg|_{u = x^2 + y^2} \frac{\del}{\del x} (x^2 + y^2) \right]\\ \amp = \frac{\del}{\del x} \left[ e^u \Bigg|_{u = x^2 + y^2} (2x) \right] = \frac{\del}{\del x} 2xe^{x^2 + y^2}\\ \amp = \frac{\del}{\del x} (2x) (e^{x^2+y^2} + 2x \frac{\del}{\del x} e^{x^2 + y^2} \\ \amp = 2e^{x^2 + y^2} + 2x \frac{d}{du} e^u \Bigg|_{u = x^2 + y^2} \frac{\del}{\del x} (x^2 + y^2) \\ \amp = 2e^{x^2 + y^2} + 2x e^u \Bigg|_{u = x^2 + y^2} (2x) = 2e^{x^2 + y^2} + 4x^2 e^{x^2 + y^2}\\ \amp = (2+4x^2) e^{x^2 + y^2} \end{align*}

Activity 9.5.14.

Calculate this partial derivative.

\begin{equation*} \frac{\del^2}{\del y^2} x^2 + 3y^2 + 4y^2 x^3 - 5x^3 y + 8y^4 \end{equation*}
Solution.

I differetiate in the \(y\) variable twice, treating other variables as constant.

\begin{gather*} \frac{\del^2}{\del y^2} x^2 + 3y^2 + 4y^2 x^3 - 5x^3 y + 8y^4 = \frac{\del}{\del y} \left[ 0 + 6y + 8yx^3 - 5x^3 + 32y^3 \right]\\ = 6 + 8x^3 - 0 + 96y^2 = 6 + 8x^3 + 96y^2 \end{gather*}

Activity 9.5.15.

Calculate this partial derivative.

\begin{equation*} \frac{\del^2}{\del z^2} xyz + x^2yz - xy^2z - xyz^2 \end{equation*}
Solution.

I differetiate in the \(z\) variable twice, treating other variables as constant.

\begin{align*} \frac{\del^2}{\del z^2} xyz + x^2yz - xy^2z - xyz^2 \amp = \frac{\del}{\del z} \left[ xy + x^2y - xy^2 + 2xyz \right] \\ = 0 + 0 - 0 + 2xy = 2xy \end{align*}

Activity 9.5.16.

Calculate this partial derivative.

\begin{equation*} \frac{\del^2}{\del x \del y} \sin (xy) \end{equation*}
Solution.

I differetiate in \(y\) and then in \(x\text{,}\) in each step treating other variables as constant. I use the chain rule in the first derivative, then product rule and chain rule in the second derivative.

\begin{align*} \frac{\del^2}{\del x \del y} \sin (xy) \amp = \frac{\del}{\del x} \left[ \frac{d}{du} \sin u \Bigg|_{u = xy} \frac{\del}{\del y} xy \right] \\ \amp = \frac{\del}{\del x} \left[ \cos u \Bigg|_{u = xy} (x) = \frac{\del}{\del x} x \cos (xy) \right] \\ \amp = \left( \frac{\del}{\del x} x \right) \cos (xy) + x \frac{\del}{\del x} \cos xy \\ \amp = 1 \cos (xy) + x \frac{d}{du} \cos u \Bigg|_{u = xy} \frac{\del}{\del x} xy \\ \amp = \cos (xy) - x \sin u \Bigg|_{u = xy} (y) = \cos (xy) - xy \sin (xy) \end{align*}

Activity 9.5.17.

Calculate this partial derivative.

\begin{equation*} \frac{\del^2}{\del x \del y} \frac{1}{x^2 + y^2 + 3} \end{equation*}
Solution.

I differetiate in \(y\) and then in \(x\text{,}\) in each step treating other variables as constant. I use the chain rule in the first step, and then the quotient rule in the second.

\begin{align*} \frac{\del^2}{\del x \del y} \frac{1}{x^2 + y^2 + 3} \amp = \frac{\del}{\del x} \left[ \frac{d}{du} \frac{1}{u} \Bigg|_{u = x^2 + y^2 + 3} \frac{\del}{\del y} (x^2 + y^2 + 3) \right] \\ \amp = \frac{\del}{\del x} \left[ \frac{-1}{u^2} \Bigg|_{u = x^2 + y^2 +3} (2y) \right] = \frac{\del}{\del x} \left[ \frac{-2y}{(x^2 + y^2 + 3)^2} \right] \\ \amp = \frac{(x^2 + y^2 + 3)\frac{\del}{\del x} (-2y) + 2y \frac{\del}{\del x} (x^2 +y^2 +3)}{(x^2 + y^2 + 3)^4} \\ \amp = \frac{0 + (2y)(2x)}{(x^2 + y^2 + 3)^4} = \frac{4xy}{(x^2 + y^3 + 3)^4} \end{align*}

Activity 9.5.18.

Calculate this partial derivative.

\begin{equation*} \frac{\del^2}{\del x \del y} x^2 y^3 - 7x^4y^3 - x^5 + 3y^7 \end{equation*}
Solution.

I differetiate in \(y\) and then in \(x\text{,}\) in each step treating other variables as constant.

\begin{align*} \frac{\del^2}{\del x \del y} x^2 y^3 - 7x^4y^3 - x^5 + 3y^7 \amp = \frac{\del}{\del x} \left[ 3x^2y^2 - 21x^4y^2 - 0 + 21y^6 \right] \\ \amp = 6xy^2 - 84x^3y^2 - 0 + 0 = 6xy^2 - 84x^3y^2 \end{align*}

Activity 9.5.19.

Calculate this partial derivative.

\begin{equation*} \frac{\del^3}{\del x \del y \del z} 3xyz - xy^2 - xz^2 + yz^2 \end{equation*}
Solution.

I differetiate in \(z\text{,}\) then in \(y\text{,}\) and then in \(z\text{,}\) in each step treating other variables as constant.

\begin{align*} \frac{\del^3}{\del x \del y \del z} 3xyz - xy^2 - xz^2 + yz^2 \amp = \frac{\del^2}{\del x \del y} \left[ 3xy - 0 - 2xy + 2yz \right] \\ \amp = \frac{\del}{\del x} \left[ 3x - 2x + 2z \right] = 3 - 2 = 0 = 1 \end{align*}

Activity 9.5.20.

Calculate this partial derivative.

\begin{equation*} \frac{\del^3}{\del x \del y^2} x^2 e^y - xy^2 \end{equation*}
Solution.

I differetiate in \(y\) twice and the in \(x\text{,}\) in each step treating other variables as constant.

\begin{align*} \frac{\del^3}{\del x \del y^2} x^2 e^y - xy^2 \amp = \frac{\del^2}{\del x \del y} x^2 e^y - 2xy \\ \amp = \frac{\del}{\del x} x^2e^y - 2x = 2xe^y - 2 \end{align*}

Subsection 9.5.3 Gradients

Activity 9.5.21.

Calculate the gradient of this function. Use the countour plots from the activity from Week 8 and draw some of the gradient directions, showing that the gradients are indeed perpendicular to the coutour plots.

\begin{equation*} f(x,y) = 3y - 4x^2 \end{equation*}
Solution.

I calculate the two partial derivative.

\begin{gather*} \frac{\del}{\del x} f(x,y) = -8x \\ \frac{\del}{\del y} f(x,y) = 3 \end{gather*}

The gradient has these two as components.

\begin{equation*} \nabla f(x,y) = (-8x,3) \end{equation*}

I've chosen some points and drawn the gradients in Figure 9.5.1

Figure 9.5.1. \(\nabla 3y - 4x^2\)

Activity 9.5.22.

Calculate the gradient of this function. Use the countour plots from the activity from Week 8 and draw some of the gradient directions, showing that the gradients are indeed perpendicular to the coutour plots.

\begin{equation*} f(x,y) = e^{x^2 + y^2} + 4 \end{equation*}
Solution.

I calculate the two partial derivative.

\begin{gather*} \frac{\del}{\del x} f(x,y) = 2xe^{x^2 + y^2}\\ \frac{\del}{\del y} f(x,y) = 2ye^{x^2 + y^2} \end{gather*}

The gradient has these two as components.

\begin{equation*} \nabla f(x,y) = (2xe^{x^2 + y^2}, 2ye^{x^2 + y^2}) \end{equation*}

I've chosen some points and drawn the gradients in Figure 9.5.2

Figure 9.5.2. \(\nabla e^{x^2+y^2} + 4 \)

Activity 9.5.23.

Calculate the gradient of this function. Use the countour plots from the activity from Week 8 and draw some of the gradient directions, showing that the gradients are indeed perpendicular to the coutour plots.

\begin{equation*} f(x,y) = \frac{x}{\sin y} \end{equation*}
Solution.

I calculate the two partial derivative.

\begin{gather*} \frac{\del}{\del x} f(x,y) = \frac{1}{\sin y}\\ \frac{\del}{\del y} f(x,y) = \frac{-x \cos y}{\sin^2 y} \end{gather*}

The gradient has these two as components.

\begin{equation*} \nabla f(x,y) = \left( \frac{1}{\sin y}, \frac{-x \cos y}{\sin^2 y} \right) \end{equation*}

I've chosen some points and drawn the gradients in Figure 9.5.3

Figure 9.5.3. \(\nabla \frac{x}{\sin y}\)

Activity 9.5.24.

Calculate the gradient of this function. Use the countour plots from the activity from Week 8 and draw some of the gradient directions, showing that the gradients are indeed perpendicular to the coutour plots.

\begin{equation*} f(x,y) = \frac{x^2 + 1}{y^2 - 4} \end{equation*}
Solution.

I calculate the two partial derivative.

\begin{gather*} \frac{\del}{\del x} f(x,y) = \frac{2x + 1}{y^2-4}\\ \frac{\del}{\del y} f(x,y) = \frac{-(x^2+1)(2y)}{(y^2-4)^2} \end{gather*}

The gradient has these two as components.

\begin{equation*} \nabla f(x,y) = \left( \frac{2x + 1}{y^2-4}, \frac{-2y(x^2+1)}{(y^2-4)^2} \right) \end{equation*}

I've chosen some points and drawn the gradients in Figure 9.5.4

Figure 9.5.4. \(\nabla \frac{x^2+1}{y^2-4}\)

Subsection 9.5.4 Conceptual Review Questions

  • How do multivariable limits differ from single variable limits?

  • What is a partial derivative?

  • What is a gradient?

  • Why is the generalization of the derivative so complicated?

  • What is a partial differential equation? What does it mean to relate the time and space derivatives of a model?