Multivariate Calculus Tutorial Sheet, #11

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Skill Building Questions

Problem 1.

For each of the functions below, find all the first order partial derivatives:

(a) $f\left(x,y\right)=xy^3+x^2y^2$

$$ \frac{\partial{}f}{\partial{}x}=y^3+2xy^2, \ \ \ \frac{\partial{}f}{\partial{}y}=3xy^2+2x^2y $$








(b) $f\left(x,y\right)=xe^{2x+3y}$

$$ \frac{\partial{}f}{\partial{}x}=e^{2x+3y}+2xe^{2x+3y}, \ \ \ \ \frac{\partial{}f}{\partial{}y}=3xe^{2x+3y} $$








(c) $f\left(x,y\right)=\frac{x-y}{x+y}$

$$ \frac{\partial{}f}{\partial{}x}=\frac{1\left(x+y\right)-1(x-y)}{ {(x+y)}^2}=\frac{2y}{ {(x+y)}^2}, \ \ \ \frac{\partial{}f}{\partial{}y}=\frac{-1\left(x+y\right)-1(x-y)}{ {(x+y)}^2}=\frac{-2x}{ {(x+y)}^2} $$








(d) $f\left(x,y\right)=2x\sin{\left(x^2y\right)}$

$$ \frac{\partial{}f}{\partial{}x}=2\sin{\left(x^2y\right)+4x^2y}\cos{(x^2y)},\ \ \frac{\partial{}f}{\partial{}y}=2x^3\cos{(x^2y)} $$








(e) $f\left(x,y,z\right)=x\cos{z+x^2y^3e^z}$

$$ \frac{\partial{}f}{\partial{}x}=\cos{z+2xy^3e^z, \ \ \ }\frac{\partial{}f}{\partial{}y}=3x^2y^2e^z, \ \ \ \ \frac{\partial{}f}{\partial{}z}=-x\sin{z}+x^2y^3e^z $$








Problem 2.

If $f\left(x,y\right)=\sqrt[ 3 ]{x^3+y^3}$, find $f_x\left(a,0\right)$

Note that by regular differentiation: $$ f'_x(x,y)=\frac{3x^2}{3(x^3+y^3)^{2/3}} $$ $$ \boxed{ f'_x(a,0)=1 } $$








Problem 3.

For the function $u\left(x,y\right)=\ln⁡(1+xy^2)$, show that the second order partial derivative $\frac{\partial^2 u}{\partial x \partial y}$ is equal to $\frac{\partial^2 u}{\partial y \partial x}$ (note the order of differentiation) by calculating both ways.

$$ \frac{\partial{}u}{\partial{}x}=\frac{y^2}{1+xy^2} $$ $$ \frac{\partial^2 u}{\partial y \partial x} = \frac{\partial}{\partial y} \frac{\partial u}{\partial x} = \frac{2y\left(1+xy^2\right) - y^2(2xy)}{(1+xy^2)^2} = \frac{2y+2xy^3-2xy^3}{(1+xy^2)^2} = \frac{2y}{(1+xy^2)^2} $$ $$ \frac{\partial u}{\partial y} = \frac{2 x y}{1+xy^2} $$ $$ \frac{\partial^2 u}{\partial x \partial y} = \frac{\partial}{\partial x} \frac{\partial u}{\partial y} = \frac{(-2 x y)y^2}{((1+xy^2)^2} + \frac{2y}{1+xy^2} = \frac{(-2 x y)y^2 + 2y\left(1+xy^2\right)}{(1+xy^2)^2} = \frac{2y}{(1+xy^2)^2} $$








Problem 4.

Find the turning points, $[x_0, y_0, f(x_0,y_0)]$, for the partial derivatives of the function $f(x,y) = x e^{-x^2 - y^2}$.

$$ f_x(x,y) = \frac{\partial f}{\partial x} = e^{-x^2 - y^2} + -2x^2 e^{-x^2 - y^2} $$ $$ f_y(x,y) = \frac{\partial f}{\partial y} = -2 x y e^{-x^2 - y^2} $$ $\Rightarrow{}$ for $f_x = 0$ $$ f_x = 1 + -2x^2 = 0 \Rightarrow x = \pm\frac{1}{\sqrt{2}} $$ $\Rightarrow{}$ for $f_y = 0$ $$ f_y = -2 x y = 0 \Rightarrow y = 0 $$ $$ \boxed{ f(\pm \frac{1}{\sqrt{2}}, 0) = \pm \frac{1}{\sqrt{2}} e^{-1/2} }$$








Problem 5.

Calculate the divergence ($\nabla \cdot$) and the curl ($\nabla \times$) of the following vector fields:

(a) $\mathbf{x} = \begin{bmatrix} x \newline y \newline z \end{bmatrix}$

For any column vector v: $$ \nabla \cdot \mathbf{v} = \frac{\partial v_x}{\partial x} + \frac{\partial v_y}{\partial y} + \frac{\partial v_z}{\partial z}$$ $$\nabla\times\mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \newline \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \newline v_x & v_y & v_z \end{vmatrix} = \begin{bmatrix} \frac{\partial v_z}{\partial y} - \frac{\partial v_y}{\partial z} \newline \frac{\partial v_x}{\partial z} - \frac{\partial v_z}{\partial x} \newline \frac{\partial v_y}{\partial x} - \frac{\partial v_x}{\partial y} \end{bmatrix} $$
$$ \boxed{ \nabla \cdot \mathbf{x} = 3, \ \ \ \nabla \times \mathbf{x} = \begin{bmatrix} 0 \newline 0 \newline 0 \end{bmatrix} } $$








(b) $\mathbf{u} = \begin{bmatrix} -y \newline x \newline 3 z^2 \end{bmatrix}$

$$ \boxed{ \nabla \cdot \mathbf{u} = 6 z, \ \ \ \nabla \times \mathbf{u} = \begin{bmatrix} 0 \newline 0 \newline 2 \end{bmatrix} } $$








(c) $\mathbf{a} = \begin{bmatrix} 0 \newline 0 \newline x z \end{bmatrix}$

$$ \boxed{ \nabla \cdot \mathbf{a} = x, \ \ \ \nabla \times \mathbf{a} = \begin{bmatrix} 0 \newline -z \newline 0 \end{bmatrix} } $$








(d) $\mathbf{p} = \begin{bmatrix} x y z \newline x^2 + y^2 \newline -z \end{bmatrix}$

$$ \boxed{ \nabla \cdot \mathbf{p} = y z + 2 y -1, \ \ \ \nabla \times \mathbf{p} = \begin{bmatrix} 0 \\ x y \\ 2x-xz \end{bmatrix} } $$








(e) $\mathbf{q} = \begin{bmatrix} \sin x \cos y \sin z \newline \cos x \sin y \sin z \newline \cos z \end{bmatrix}$

$$\boxed{ \nabla \cdot \mathbf{q} = -\sin z + 2 \cos x \cos y \sin z, \ \ \ \nabla \times \mathbf{q} = \begin{bmatrix} -\cos x \sin y \cos z \newline \sin x \cos y \cos z \newline 0 \end{bmatrix} }$$








Problem 6.

Calculate the gradient $(\nabla)$ and Laplacian $(\nabla^2)$ of the following functions:
(a) $7xy^2+z^4$

The gradient is found by,
$$ \nabla(7xy^2+z^4) = \begin{bmatrix} \frac{\partial{}f}{\partial{}x} \newline \frac{\partial{}f}{\partial{}y} \newline \frac{\partial{}f}{\partial{}z} \end{bmatrix}(7xy^2+z^4) \Rightarrow\quad\boxed{\begin{bmatrix} 7y^2 \newline 14xy \newline 4z^3 \end{bmatrix}}$$
The Laplacian is effectively applying $\nabla$ twice, hence there are 2 methods,
Method 1: Using the answer for the gradient, we can apply $\nabla$ again,
$$ \nabla\begin{bmatrix} 7y^2 \newline 14xy \newline 4z^3 \end{bmatrix}\Rightarrow\quad\begin{bmatrix} \frac{\partial{}f}{\partial{}x} \newline \frac{\partial{}f}{\partial{}y} \newline \frac{\partial{}f}{\partial{}z} \end{bmatrix}\begin{bmatrix} 7y^2 \newline 14xy \newline 4z^3 \end{bmatrix} \Rightarrow 0 + 14x + 12z^2\Rightarrow\quad\boxed{14x + 12z^2}$$
Method 1 is a two-step process, applying $\begin{bmatrix} \frac{\partial{}f}{\partial{}x} \newline \frac{\partial{}f}{\partial{}y} \newline \frac{\partial{}f}{\partial{}z} \end{bmatrix}$ twice.
Method 2:
$$\nabla^2(7xy^2+z^4) = (\frac{\partial{}^2}{\partial{}x^2}+\frac{\partial{}^2}{\partial{}y^2}+\frac{\partial{}^2}{\partial{}z^2})(7xy^2+z^4)$$
$$\Rightarrow\quad 0 + 14x + 12z^2\Rightarrow\quad\boxed{14x + 12z^2}$$
Method 2 is a one step process, applying $(\frac{\partial{}^2}{\partial{}x^2}+\frac{\partial{}^2}{\partial{}y^2}+\frac{\partial{}^2}{\partial{}z^2})$ once.








(b) $\sin (xy) + 2z^2$

$$ \nabla(sin (xy) + 2z^2) = \begin{bmatrix} \frac{\partial{}f}{\partial{}x} \newline \frac{\partial{}f}{\partial{}y} \newline \frac{\partial{}f}{\partial{}z} \end{bmatrix}(\sin (xy) + 2z^2)\Rightarrow\quad\boxed{\begin{bmatrix} y\cos(xy)\newline x\cos(xy) \newline 4z \end{bmatrix}}$$
$$ \nabla^2(sin (xy) + 2z^2) = (\frac{\partial{}^2}{\partial{}x^2}+\frac{\partial{}^2}{\partial{}y^2}+\frac{\partial{}^2}{\partial{}z^2})(sin (xy) + 2z^2) $$
$$ \Rightarrow\quad -y^2\sin(xy)-x^2\sin(xy) + 4 \Rightarrow\quad\boxed{4 - (x^2+y^2)\sin (xy)} $$








Problem 7.

Show that the function $u\left(x,y\right)=\ln⁡(1+xy^2)$ satisfies the partial differential equation: \(2\frac{ {\partial{}}^2u}{ {\partial{}x}^2}+y^3\frac{ {\partial{}}^2u}{\partial{}y\partial{}x}=0\)

$$ \frac{\partial{}u}{\partial{}x}=\frac{y^2}{1+xy^2}\ ;\ \ \ \frac{ {\partial{}}^2u}{ {\partial{}x}^2}=-\frac{y^2}{ {\left(1+{xy}^2\right)}^2}.y^2=\frac{-y^4}{ {\left(1+{xy}^2\right)}^2} $$
$$ \frac{ {\partial{}}^2u}{\partial{}y\partial{}x}=\frac{\partial{}}{\partial{}y}\left(\frac{\partial{}u}{\partial{}x}\right)=\frac{2y\left(1+xy^2\right)-y^2(2xy)}{ {(1+xy^2)}^2}=\frac{2y+2xy^3-2xy^3}{ { {(1+xy}^2)}^2}=\frac{2y}{ { {(1+xy}^2)}^2} $$
$$ \boxed{ 2\frac{ {\partial{}}^2u}{\partial{}x^2}+y^3\frac{ {\partial{}}^2u}{\partial{}y\partial{}x}=-\frac{2y^4}{ { {(1+xy}^2)}^2}+\frac{ {2y}^4}{ { {(1+xy}^2)}^2}=0 } $$








Exam Style Questions

Problem 8.

Given the expressions,
\(f(u,v)=2u^3-7uv+v^2,\quad u(x,y)=\frac{x}{y}, \quad v(x,y)=\frac{y^2}{x}\)
Use the multivariate chain rule to calculate $\frac{\partial{}f}{\partial{x}}$ of $f(u(x,y),v(x,y))$.
(Hint: The final expression should be in terms of $x$ and $y$)

We need to use the multivariate chain rule:
$$\frac{\partial{}f}{\partial{}x} = \frac{\partial{}f}{\partial{}u}\frac{\partial{}u}{\partial{}x} + \frac{\partial{}f}{\partial{}v}\frac{\partial{}v}{\partial{}x}$$
Calculate each of the needed partial derivatives,
$$\frac{\partial{}f}{\partial{}u} = 6u^2 - 7v, \frac{\partial{}f}{\partial{}v} = -7u +2v$$
$$\frac{\partial{}u}{\partial{}x} = \frac{1}{y}, \frac{\partial{}v}{\partial{}x} = -\frac{y^2}{x^2}$$
Substitute all of these derivatives into the multivariate chain rule equation,
$$\frac{\partial{}f}{\partial{}x} = (6u^2-7v)(\frac{1}{y})+(-7u+2v)(-\frac{y^2}{x^2})$$
Substitute $u$ and $v$,
$$\frac{\partial{}f}{\partial{}x}=(\frac{6x^2}{y^2}-\frac{7y^2}{x})(\frac{1}{y})+(-\frac{7x}{y}+\frac{2y^2}{x})(-\frac{y^2}{x})$$
$$\Rightarrow \frac{6x^2}{y^3} - \frac{7y}{x} + \frac{7y}{x} - \frac{2y^4}{x^3}$$
Simplify,
$$\frac{\partial{}f}{\partial{}x} = \frac{6x^2}{y^3} - \frac{2y^4}{x^3}$$








Problem 9.

The total differential is defined as, $ df = (\frac{\partial{}f}{\partial{}x})_y dx + (\frac{\partial{}f}{\partial{}y})_x dy $.
Using this expression, find an expression for the partial derivative, $(\frac{\partial{}\gamma}{\partial{}\beta}) _ \alpha $.
Use the identities, $(\frac{\partial{}a}{\partial{}b})_a = 0$ and $(\frac{\partial{}a}{\partial{}a})_b = 1$.

Start with: $ d\alpha = (\frac{\partial{}\alpha}{\partial{}\beta}) _ \gamma d\beta + (\frac{\partial{}\alpha}{\partial{}\gamma}) _ \beta d\gamma $
Divide through by $d\beta$ holding $\alpha$ constant,
$$ (\frac{\partial{}\alpha}{\partial{}\beta}) _ \alpha = (\frac{\partial{}\alpha}{\partial{}\beta}) _ \gamma (\frac{\partial{}\beta}{\partial{}\beta}) _ \alpha + (\frac{\partial{}\alpha}{\partial{}\gamma}) _ \beta (\frac{\partial{}\gamma}{\partial{}\beta}) _ \alpha$$
Apply the identities,
$$ \Rightarrow\quad 0 = (\frac{\partial{}\alpha}{\partial{}\beta}) _ \gamma + (\frac{\partial{}\alpha}{\partial{}\gamma}) _ \beta (\frac{\partial{}\gamma}{\partial{}\beta}) _ \alpha$$
Rearrange,
$$ (\frac{\partial{}\gamma}{\partial{}\beta}) _ \alpha =\boxed{ -\frac{(\frac{\partial{}\alpha}{\partial{}\beta}) _ \gamma}{(\frac{\partial{}\alpha}{\partial{}\gamma}) _ \beta}} $$








Problem 10.

The ellipsoid $4x^2+2y^2+z^2=16$ intersects the plane $y=2$ in an ellipse. Find parametric equations for the tangent line to this ellipse at the point (1, 2, 2).

The key is that since we are looking at the plane $y=2$, the slope is actually just $\frac{\partial{}z}{\partial{}x}$. So, we implicitly differentiate: First, we need to find a function for $z$: \begin{align*} z^2 &= -4x^2-2y^2+16 \newline z &= \sqrt{-4x^2-2y^2+16} \end{align*} So we get: $$ \frac{\partial{}z}{\partial{}x}=\frac{\partial{}}{\partial{}x} \sqrt{-4x^2-2y^2+16} $$ $$ \frac{df(u)}{dx}=\frac{df}{du}.\frac{du}{dx} $$ Let $u=-4x^2-2y^2+16$ $$ \frac{df}{du}=\frac{1}{2\sqrt{u}}\ \ \ \frac{du}{dx}=-8x $$ $$ \frac{\partial{}z}{\partial{}x}=\frac{1}{2\sqrt{u}} (-8x)=\frac{-8x}{2\sqrt{-4x^2-2y^2+16}} $$ Adding the values of x, y; we have: $$ \frac{\partial{}z}{\partial{}x}=\frac{-8}{2\sqrt{4}}=-2 $$ The tangent line through that point, in the y = 2 plane, is: $$ z = -2x + 4 $$ To get the parametric equation of the line, take $x = t$: $$ \boxed{ x = t, \quad y=2, \quad z = -2t + 4 } $$








Extension Questions

Problem 11.

For functions $f(u, v)$, $u(x, y)$, and $v(x, y)$, calculate the partial derivative $\frac{\partial f}{\partial y}$, by direct substitution then the total derivative chain rule. Check that the approaches match. This question only tests your hand at manual working out, so if you are confident in your understanding, feel free to only do a couple.

(a) $ f(u, v) = \sin(u)\exp(-v), \quad u(x, y) = x / y, \quad v (x, y) = x y $

Direct substitution: $f(x, y) = \sin(x / y)\exp(-x y)$ $$ \left(\frac{\partial f}{\partial y}\right)_x = -\frac{x e^{-x y} \cos(x/y)}{y^2} - x e^{-xy} \sin(x/y) $$
$\Rightarrow{}$ Chain rule: $ \left(\frac{\partial f}{\partial y}\middle)_x = \middle(\frac{\partial f}{\partial u}\middle)_v \middle(\frac{\partial u}{\partial y}\middle)_x + \middle(\frac{\partial f}{\partial v}\middle)_u \middle(\frac{\partial v}{\partial y}\right)_x $ \begin{align*} \left(\frac{\partial f}{\partial u}\right)_v &= e^{-v} \cos u & \left(\frac{\partial f}{\partial v}\right)_u &= -e^{-v} \sin u \newline \left(\frac{\partial u}{\partial y}\right)_x &= -\frac{x}{y^2} & \left(\frac{\partial v}{\partial y}\right)_x &= x \end{align*}
$$ \Rightarrow{} \boxed{ \left(\frac{\partial f}{\partial y}\right)_x = -\frac{x e^{-x y} \cos(x/y)}{y^2} - x e^{-xy} \sin(x/y) } $$








(b) $f(u, v) = u^2 + 2 u v + v^2, \quad u(x, y) = \sin(x + 5 y), \quad v(x, y) = \cos(x + 5 y)$

$\Rightarrow$ Direct substitution: $f(x, y) = 1 + 2 \sin(x + 5 y) \cos(x + 5 y)$ $$ \left(\frac{\partial f}{\partial y}\right)_x = 10 \cos^2(x + 5 y) - 10 \sin^2(x + 5 y) $$ $\Rightarrow$ Chain rule \begin{align*} \left(\frac{\partial f}{\partial u}\right)_v &= 2 u + 2 v & \left(\frac{\partial f}{\partial v}\right)_u &= 2 u + 2 v \newline \left(\frac{\partial u}{\partial y}\right)_x &= 5 \cos(x + 5y) & \left(\frac{\partial v}{\partial y}\right)_x &= -5 \sin(x + 5 y) \end{align*}
$$ \Rightarrow{} \boxed{ \left(\frac{\partial f}{\partial y}\right)_x = 10 \cos^2(x + 5 y) - 10 \sin^2(x + 5 y) } $$








(c) $f(u, v) = \frac{\arctan(u)}{1 + v^2}, \quad u(x, y) = \sqrt{x y}, \quad v(x, y) = x \ln(3 y) $

$\Rightarrow{}$ Direct substitution: $f(x, y) = \frac{\arctan(\sqrt{x y})}{1 + (x \ln(3 y))^2}$ $$ \left(\frac{\partial f}{\partial y}\right)_x = \frac{1}{(1+xy)(1+x^2 \ln(3 y)^2)}\frac{x}{2\sqrt{xy}} -\frac{2 x \ln(3 y) \arctan(\sqrt{x y})}{(1+x^2 \ln(3 y)^2)^2} x / y $$ $\Rightarrow{}$ Chain rule: \begin{align*} \left(\frac{\partial f}{\partial u}\right)_v &= \frac{1}{(1+u^2)(1+v^2)} & \left(\frac{\partial f}{\partial v}\right)_u &= -\frac{2 v \arctan(u)}{(1+v^2)^2} \newline \left(\frac{\partial u}{\partial y}\right)_x &= \frac{x}{2\sqrt{xy}} & \left(\frac{\partial v}{\partial y}\right)_x &= x / y \end{align*}
$$\Rightarrow{} \boxed{ \left(\frac{\partial f}{\partial y}\right)_x = \frac{1}{(1+xy)(1+x^2 \ln(3 y)^2)}\frac{x}{2\sqrt{xy}} -\frac{2 x \ln(3 y) \arctan(\sqrt{x y})}{(1+x^2 \ln(3 y)^2)^2} x / y } $$








(d) $f(u, v) = \tanh(w u + v), \quad u(x, y) = \tanh(x a + y), \quad v(x, y) = y$

$f(x, y) = \tanh(w \tanh(x a + y) + y)$ $$ \left(\frac{\partial f}{\partial y}\right)_x = w \operatorname{sech}(w \tanh(x a + y) + y)^2 \operatorname{sech}(a x + y)^2 + \operatorname{sech}(w \tanh(x a + y) + y)^2 $$ $\Rightarrow{}$ Chain rule: \begin{align*} \left(\frac{\partial f}{\partial u}\right)_v &= w \operatorname{sech}(w u + v)^2 & \left(\frac{\partial f}{\partial v}\right)_u &= \operatorname{sech}(w u + v)^2 \newline \left(\frac{\partial u}{\partial y}\right)_x &= \operatorname{sech}(a x + y)^2 & \left(\frac{\partial v}{\partial y}\right)_x = 1 \end{align*} $$ \boxed{ \left(\frac{\partial f}{\partial y}\right)_x = w \operatorname{sech}(w \tanh(x a + y) + y)^2 \operatorname{sech}(a x + y)^2 + \operatorname{sech}(w \tanh(x a + y) + y)^2 } $$








Problem 12.

If $ g\left(s,t\right)=f(s^2-t^2,\ t^2-s^2) $ and $f$ is differentiable, show that $g$ satisfies the equation: \(t\frac{\partial{}g}{\partial{}s}+s\frac{\partial{}g}{\partial{}t}=0\)

Let $x=s^2-t^2$ and $y=t^2-s^2$. Then $g(s,t)=f(x,y)$, and by the chain rule:
$$ \frac{\partial{}g}{\partial{}s}=\frac{\partial{}f}{\partial{}x}\frac{\partial{}x}{\partial{}s}+\frac{\partial{}f}{\partial{}y}\frac{\partial{}y}{\partial{}s}=(\frac{\partial{}f}{\partial{}x})(2s)+(\frac{\partial{}f}{\partial{}y})(-2s) $$
$$ \frac{\partial{}g}{\partial{}t}=\frac{\partial{}f}{\partial{}x}\frac{\partial{}x}{\partial{}t}+\frac{\partial{}f}{\partial{}y}\frac{\partial{}y}{\partial{}t}=(\frac{\partial{}f}{\partial{}x})(-2t)+(\frac{\partial{}f}{\partial{}y})(2t) $$
$$ \boxed{ t\frac{\partial{}g}{\partial{}s}+s\frac{\partial{}g}{\partial{}t}=(\frac{\partial{}f}{\partial{}x})(2st)+(\frac{\partial{}f}{\partial{}y})(-2st)+(\frac{\partial{}f}{\partial{}x})(-2st)+(\frac{\partial{}f}{\partial{}y})(2st)=0 }$$








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