- I-Learn, Class Pictures, Learning Targets, Text Book Practice
- Prep Tasks: Unit 1 - Motion, Unit 2 - Derivatives, Unit 3 - Integration, Unit 4 - Vector Calculus
This is Day 8 in Unit 2.
Prep
We're in Unit 2 - Derivatives. Your homework assignment each day is to spend 1-2 hours working on the next 4 tasks from the current unit's prep.
Brain Gains (Rapid Recall, Jivin' Generation)
- Consider the function $z=\sin(x)+e^y$, where $x=3t$ and $y=t^2$. Compute $\frac{dz}{dt}$.
Solution
There are two ways to do this.
- Substitution gives $z=\sin(3t)+e^{t^2}$. Differentiation (using the chain rule) then gives $$\frac{dz}{dt}=\cos(3t)\frac{d}{dt}(3t)+e^{t^2}\,\frac{d}{dt}(t^2)=\cos(3t)3+e^{t^2}\,2t.$$
- Differentials give $dz = \cos(x)dx+e^ydy$, with $dx = 3dt$ and $dy=2tdt$. Substitution then gives $$dz = \cos(3t)3dt+e^{t^2}\,2tdt\quad\text{or}\quad\frac{dz}{dt}=\cos(3t)3+e^{t^2}\,2t.$$
In both cases, we obtained the same solution of $$\frac{dz}{dt}=\underbrace{\cos(3t)}_{f_x}\underbrace{3}_{\frac{dx}{dt}}+\underbrace{e^{t^2}}_{f_y}\underbrace{2t}_{\frac{dy}{dt}}.$$ Writing the solution above symbolically gives us the chain rule $$\frac{dz}{dt} = f_x\frac{dx}{dt}+f_y\frac{dy}{dt} = \frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt}.$$
- Suppose $dz = e^{x^2}dx+\cos(2y)dy$, $x=3t$, and $y=t^2$. Compute $\frac{dz}{dt}$.
Solution
This time we don't know what the function $z$ equals, so we cannot first substitute and then differentiate. We do know however that $f_x = e^{x^2}$ and $f_y = \cos(2y)$. We can compute differentials and then substitute.
- Note that $dx = 3dt$ and $dy = 2tdt$. Substitution then gives $$dz = e^{(3t)^2}3\,dt+\cos(2(t^2))2t\,dt \quad\text{and so}\quad \frac{dz}{dt} = e^{(3t)^2}3+\cos(2(t^2))2t.$$
Again, the the solution above symbolically gives us the same chain rule $$\frac{dz}{dt} = f_x\frac{dx}{dt}+f_y\frac{dy}{dt} = \frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt}.$$
- For the function $f(x,y)= 2x^2+3xy+4y^2-5x+2y$, find all critical points of $f$. In other words, find when the first derivative equals zero.
Solution
The first derivative is $Df(x,y) = \begin{bmatrix} 4 x + 3 y -5 & 3 x + 8 y +2\end{bmatrix}$. We need to solve the system of equations $$4 x + 3 y -5=0 \quad 3 x + 8 y +2=0.$$ The solution can be found using substitution, elimination, or Mathematica, yielding the only critical point as $(x,y)=(2,-1)$.
f[x_, y_] := 2 x^2 + 3 x y + 4 y^2 - 5 x + 2 y
Df = D[f[x, y], {{x, y}}]
Solve[Df == 0, {x, y}]
- For the function $f(x,y)= 2x^2+3xy+4y^2-5x+2y$, compute the second derivative $D^2f(x,y)$ at each critical point above.
Solution
The second derivative is $D^2f(x,y) = \begin{bmatrix}4&3//3&8\end{bmatrix}$. At this point $(2,-1)$, we have $D^2f(2,-1) = \begin{bmatrix}4&3//3&8\end{bmatrix}$, which didn't change anything because the second derivative is constant for this function.
We can check our result with Mathematica.
f[x_, y_] := 2 x^2 + 3 x y + 4 y^2 - 5 x + 2 y
D2f = D[f[x, y], {{x, y}}, {{x, y}}]
D2f /. {x -> 2, y -> -1}
%//MatrixForm
- Determine the eigenvalues of $D^2f(2,-1)$, and use your result to classify the critical point as a local max, local min, or saddle point.
Solution
The eigenvalues are the solutions to the equation $(4-\lambda)(8-\lambda)-(3)^2=0$. The left hand side can be simplified and factored to yield $\lambda^2-12\lambda+23=0$. The quadratic formula yields $\lambda = \frac{12\pm\sqrt{144-4(1)(23)} }{2}$. We could simplify this to $\lambda = 6\pm\sqrt{13}$, but it's actually not needed to answer the question. Note that $\sqrt{144-4(1)(23)}<12$, which means both eigenvalues will be positive (we won't subtract a large enough number from 12 to make one eigenvalue negative). This means the vector field of the function, near the critical point, has all vectors pointing outwards. The critical point corresponds to a local minimum.
Here we use Mathematica to (1) compute the eigenvalues, (2) show a contour plot with gradient plot, and (3) show a plot of the surface.
f[x_, y_] := 2 x^2 + 3 x y + 4 y^2 - 5 x + 2 y
Df = D[f[x, y], {{x, y}}]
Solve[Df == 0, {x, y}]
D2f = D[f[x, y], {{x, y}}, {{x, y}}];
D2f // MatrixForm
D2f /. {x -> 2, y -> -1} // Eigenvalues
xB = {x, 0, 4};
yB = {y, -3, 1};
Show[ContourPlot[f[x, y], xB, yB], VectorPlot[Df, xB, yB]]
Plot3D[f[x, y], xB, yB, MeshFunctions -> {#3 &}]
Group Problems
- For the function $f(x,y)=x^2+4xy+3y^2-10x-18y$, verify that the first derivative $Df(x,y)$ and second derivative $D^2f(x,y)$ are $$Df(x,y) = \begin{bmatrix}2x+4y-10&4x+6y-18\end{bmatrix}\quad\text{and}\quad
D^2f(x,y) = \begin{bmatrix}\begin{matrix}2\\4\end{matrix}&\begin{matrix}4\\6\end{matrix}\end{bmatrix}.
$$
- Solve $Df(x,y)=\begin{bmatrix}0&0\end{bmatrix}$, to find the critical points of this function. [Check: $(x,y)=(3,1)$.]
- Find the eigenvalues of $D^2f(3,1)$. [Check: $\lambda = 4\pm\sqrt{20} = 4\pm 2\sqrt{5}$.]
- Does the function $f$ have a local max, local min, or saddle at $(3,1)$? Explain.
- Copy/paste the Mathematica code from the last question of today's Brain Gains, and then modify that code to work for this function.
- We will find the points on the curve $g(x,y)=xy^2=16$ that minimize the function $f(x,y)=x^2+y^2$.
- Compute $\vec \nabla f$ and $\vec \nabla g$.
- To find the points where $\vec \nabla f$ and $\vec \nabla g$ are either parallel or anti-parallel, we need to solve $\vec \nabla f=\lambda\vec \nabla g$ together with $g(x,y) = 16$. Write the three equations that result from needing to solve this system (you should get $2x=\lambda y^2$, $2y = \lambda 2xy$, and $xy^2=16$.)
- Solve the system above for $x$ and $y$ (a value for $\lambda$ will appear, but we don't need it). [Check: $x=2$ and $y=\pm \sqrt{8}=\pm 2\sqrt{2}$, $\lambda = $ something. ] We don't need the value for $\lambda$, so one suggestion is to start by eliminating $\lambda$ (via substitution). Alternately, try using Mathematica's Solve[] command to get the solution.
- Consider the function $f(x,y,z) = 3xy+z^2$. We'll be analyzing the level surface that passes through the point $P=(1,-3,2)$.
- Compute the differential $df$, and then evaluate the differential at $P$.
- For a level surface, the output remains constant (so $df=0$). If we let $(x,y,z)$ be a point on the surface really close to $P$, then we have $dx=x-1$, $dy=y-(-3)$ and $dz = z-?$. Plug this information into the differential at $P$ to obtain an equation of the tangent plane.
- Give an equation of the tangent plane to the level surface of $f$ that passes through $(1,2,-3)$.
- Give an equation of the tangent plane to the level surface of $f$ that passes through $(a,b,c)$.
- What relationship exists between the gradient of $f$ at $P$ and the tangent plane through $P$?
- Suppose a plane passes through the point $(a,b,c)$ and has normal vector $(A,B,C)$. Give an equation of that plane.
- Give an equation of the tangent plane to $xy+z^2=7$ at the point $P=(-3,-2,1)$.
- Give an equation of the tangent plane to $z=f(x,y)=xy^2$ at the point $P=(4,-1,f(4,-1))$.
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