8 Problems
8.1 Week 2
The following question can be turned in for formative feedback. Please submit this question on paper at the next problems class.
The equation \(3y = z^3+3xz\) defines \(z\) implicitly as a function of \(x\) and \(y\). Evaluate all three second partial derivatives of \(z\) with respect to \(x\) and/or \(y\). Verify that \(z\) is a solution of \[ x\frac{\partial^2 z}{\partial y^2} + \frac{\partial^2 z}{\partial x^2} = 0. \]
First derivatives: \[\begin{eqnarray} \frac{\partial}{\partial x}[3y = z^3 + 3xz]\\ 0 = 3z^2\frac{\partial z}{\partial x} + 3z + 3x\frac{\partial z}{\partial x}\\ -3z = \frac{\partial z}{\partial x}(3z^2 + 3x)\\ \frac{\partial z}{\partial x} = \frac{-z}{x+z^2} \end{eqnarray}\] \[\begin{eqnarray} \frac{\partial}{\partial y}[3y = z^3 +3xz]\\ 3 = 3z^2\frac{\partial z}{\partial y} + 3x \frac{\partial z}{\partial y}\\ \frac{\partial z}{\partial y} = \frac{1}{x+z^2} \end{eqnarray}\] Second derivatives: \[\begin{eqnarray} \frac{\partial^2z}{\partial x^2}= \frac{\partial}{\partial x}[-z(x+z^2)^{-1}]\\ = \frac{2xz}{(x+z^2)^3} \end{eqnarray}\] \[\begin{eqnarray} \frac{\partial z}{\partial x} = \frac{-z}{x+z^2}\\ \frac{\partial^2z}{\partial y \partial x} = \frac{\partial}{\partial y} [-z(x+z^2)^{-1}]\\ = \frac{z^2 - x}{(x+z^2)^3} \end{eqnarray}\] \[\begin{eqnarray} \frac{\partial^2 z}{\partial y^2} = \frac{\partial}{\partial y}(x+z^2)^{-1} = \frac{-2z}{(x+z^2)^3} \end{eqnarray}\]
Find the first partial derivatives of the following functions\(f(x,y)\): (i) \(x^2y\), (ii) \(x^2+y^2 + 4\), (iii) \(\sin (x/y)\)
For (i) and (ii) find the the second partial derivatives.
Show that \[ df = x^2\,dy -(y^2+xy)\,dx \] is not an exact differential.
The function \(f(x,y)\) satisfies the differential equation \[ y\frac{\partial f}{\partial x} + x \frac{\partial f}{\partial y} = 0. \] By changing to new variables \(u = x^2 - y^2\) and \(v = 2xy\), show that \(f\) is, in fact, a functions of \(x^2 - y^2\) only.
## Answer In order to change variables we need to understand that \(u\) and \(v\) depend on both \(x\) and \(y\) and vice versa so our chain rule for partial derivatives becomes \[
\frac{\partial f}{\partial x_j} = \sum_{i=1}^n \frac{\partial f}{\partial u_i}\frac{\partial u_i}{\partial x_j}
\] where we want to change the variables of the partial derivative. Each \(x_j\) depends on more than one \(u_i\). In this case we want the partial of \(f\) with respect to \(x\) and the partial of \(f\) with respect to \(y\) in terms of \(u\) and \(v\). \[
\begin{aligned}
&\frac{\partial f}{\partial x} = \frac{\partial f}{\partial u}\frac{\partial u}{\partial x} + \frac{\partial f}{\partial v}\frac{\partial v}{\partial x} \\
&\frac{\partial f}{\partial y} = \frac{\partial f}{\partial u}\frac{\partial u}{\partial y} + \frac{\partial f}{\partial v}\frac{\partial v}{\partial y}\\
&\frac{\partial u}{\partial x} = 2x \quad \frac{\partial v}{\partial x} = 2y\\
&\frac{\partial u}{\partial y} = -2y \quad \frac{\partial v}{\partial y} = 2x\\
&\frac{\partial f}{\partial x} = 2x\,\frac{\partial f}{\partial u} +2y\, \frac{\partial f}{\partial v} \\
&\frac{\partial f}{\partial y} = -2y\,\frac{\partial f}{\partial u} + 2x\,\frac{\partial f}{\partial v}\\
&y\frac{\partial f}{\partial x} + x \frac{\partial f}{\partial y} = 0
\end{aligned}
\] substituting in for the partials of \(f\) with respect to \(x\) and \(y\) in terms of \(u\) and \(v\)
\[
\begin{aligned}
&y\left(2x\frac{\partial f}{\partial u} + 2y\frac{\partial f}{\partial v}\right) + x \left(-2y\frac{\partial f}{\partial u} + 2x\frac{\partial f}{\partial v}\right) = 0\\
&2xy\frac{\partial f}{\partial u} - 2xy \frac{\partial f}{\partial u} + 2y^2 \frac{\partial f}{\partial v} + 2x^2\frac{\partial f}{\partial v} = 0\\
&2(x^2+y^2)\frac{\partial f}{\partial v} = 0
\end{aligned}
\] Assuming \(x^2+y^2\) is not zero (in other words we are stipulating that \(x\) and \(y\) are not both zero) this means that \(\frac{\partial f}{\partial v} = 0\) so \(f\) is then only a function of \(u\) which is the same as saying \(f(x^2-y^2)\).
Find the total derivative of \(f(x,y) = x^2 + 3xy\) with respect to \(x\) given that \(y = \sin^{-1} x\). (From the lecture notes on Total Differential and Total Derivative)
Show that \[ \left(\frac{\partial S}{\partial V}\right)_T = \left(\frac{\partial P}{\partial T}\right)_V \]
(From the lecture on Thermodynamic Relations)
8.2 Week 3
Find the volume of the region bounded by the paraboloid \(z = x^2 + y^2\) and the plane \(z=2y\).
Find the centre of mass of the solid hemisphere bounded by the surfaces \(x^2 + y^2 + z^2 = a^2\) and the \(xy\)-plane, assuming that it has a uniform density \(\rho\).
Identify the curved wedge bounded by the surfaces \(y^2 = 4ax\), \(x+z = a\) and \(z=0\), and hence calculate its volume \(V\).
A tetrahedron is bounded by the three coordinate surfaces and the plane \(x/a + y/b +z/c = 1\) and has a density of \(\rho(x,y,z) = \rho_o(1+x/a)\). Find the average value of the density.
Find the volume integral of \(x^2y\) over the tetrahedral volume bounded by the planes \(x=0,y=0, z=0\), and \(x+y+z=1\). You may turn this problem in for feedback.
Four non-coplanar points A, B, C, D are positioned such that the line AD is perpendicular to BC and BD is perpendicular to AC. Show that CD is perpendicular to AB.
Find the angle between the vectors \({\bf a} = {\bf i} + 2{\bf j} + 3{\bf k}\) and \({\bf b} = 2{\bf i} + 3{\bf j} + 4{\bf k}\).
8.3 Week 4
Two particles have velocities \(v_1 = \hat \imath + 3 \hat \jmath + 6 \hat k\) and \(v_2 = \hat \imath - 2 \hat k\). Find the velocity \({\mathbf u}\) of the second particle relative to the first.
Find the area of the parallelogram with sides \(\mathbf{a} = \hat{\imath} + 2\hat{\jmath} + 3\hat{k}\) and \(\mathbf{b} = 3\hat{\imath} + 4\hat{\jmath} + 5\hat{k}\).
Show that if \(\mathbf{a} = \mathbf{b} + \lambda \mathbf{c}\) for some scalar \(\lambda\) then \(\mathbf{a}\times\mathbf{c} = \mathbf{b} \times \mathbf{c}\).
Find the gradients of: \[ \begin{aligned} &f(x, y, z) = x^2 + y^3 + z^4 \\ &f(x, y, z) = x^2y^3z^4\\ &f(x, y, z) = e^x \sin (y)\, ln (z) \end{aligned} \]
Let \(\mathbf{C} = \mathbf{A} - \mathbf{B}\). Calculate the scalar product of \(\mathbf{C}\) with itself.
Find the angle between the face diagonals of a cube. Assume that the rear corner is at the origin and that the cube sits in the positive quadrant as is of side length one.
Find the directional derivative of \(\phi=x^{2} y+xz\) at \((1,2,-1)\) in the direction \(\mathbf{A}=2 \hat{i}+2 \hat{\jmath}+\hat{k}\)
Lets consider the temperature in a room. The temperature follows \(T=x^{2}-y^{2}+x y z+273\). In which direction is the temp increasing most rapidly at \((-1,2,3)\)?