- 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 9 in Unit 2.
Brain Gains (Rapid Recall, Jivin' Generation)
- Consider the function $$x^3 - y^3 - 12 x + 3 y - 4.$$
- Compute the directional derivative of the function at the point $(1,2)$ in the direction $(3,4)$.
- Give an equation of the tangent plane to the function at the point $(1,2, f(1,2))$.
- Give an equation of the tangent line to the level curve that passes through the function at the point $(1,2)$.
- Locate all critical points of the function.
- One critical point is $(2,1)$. Use the eigenvalues of $D^2f(2,1)$ to classify this critical point as a maximum, minimum, or saddle point. We'll use software to examine the other 3 points.
Solution
The gradient is $\vec \nabla f(x,y) = (-12 + 3 x^2, 3 - 3 y^2)$, which gives $\vec \nabla f(1,2) = (-9,-9)$. We then have $$D_{ (3,4) }f(1,2) = \vec \nabla f(1,2)\cdot \frac{ (3,4) }{ |(3,4)| } = \frac{-63}{5}.$$ We can check with Mathematica.
f[x_, y_] := x^3 - y^3 - 12 x + 3 y - 4
Df = D[f[x, y], {{x, y}}]
Df /. {x -> 1, y -> 2}
% . {3, 4}/Norm[{3, 4}]
The function at $(1,2)$ gives $f(1,2) = -17$. An equation of the tangent plane to $f$ at $(1,2)$ (using $dz = f_x dx+f_y dy$) is $$z-(-17) = -9 (x-1) -9 (y-2).$$ The corresponding tangent line to the level curve passing through $(1,2)$ is found by replacing $dz$ with 0, giving $$0 = -9 (x-1) -9 (y-2).$$
The critical points are found by setting the derivative equal to zero. This requires we solve $-12 + 3 x^2=0$ and $3 - 3 y^2=0$. This means $x^2=4$ and $y^2=1$, which yields the four points $(2,1)$, $(-2,1)$, $(2,-1)$, and $(-2,-1)$. The second derivative is $$D^2 f(x,y) = \begin{bmatrix} 6 x & 0 \\ 0 & -6 y \end{bmatrix}.$$ We can quickly see that $$D^2 f(2,1) = \begin{bmatrix} 12 & 0 \\ 0 & -6 \end{bmatrix}.$$ The eigenvalues of this matrix satisfy $(12-\lambda)(-6-\lambda)-0=0$. The left hand side is already factored, so the eigenvalues are 12 and -6. Because they differ in sign, the point $(2,1)$ corresponds to a saddle point.
Here we use Mathematica to locate the critical points and corresponding eigenvalues of the hessian.
f[x_, y_] := x^3 - y^3 - 12 x + 3 y - 4
Df = D[f[x, y], {{x, y}}]
critpoints = Solve[Df == 0, {x, y}]
D2f = D[f[x, y], {{x, y}}, {{x, y}}];
D2f // MatrixForm
Table[D2f /. critpoints[[i]] // MatrixForm, {i, 1, Length[critpoints]}]
Table[D2f /. critpoints[[i]] // Eigenvalues, {i, 1, Length[critpoints]}]
xB = {x, -4, 4};
yB = {y, -4, 4};
Show[ContourPlot[f[x, y], xB, yB], VectorPlot[Df, xB, yB]]
Plot3D[f[x, y], xB, yB, MeshFunctions -> {#3 &}]
- Consider the function $w(x,y)$, where $x$ and $y$ both depend on $r$ and $h$. Compute $\frac{\partial w}{\partial r}$.
- Let $f(x,y)=x^2+y$, and $g(x,y)=4x+3y$. Find the locations of all local maxima and minima of $f$ on the curve $g(x,y) = 12$. In other words, solve the system $\vec \nabla f = \lambda \vec \nabla g$ and $g(x,y)=12$.
Solution
We have $\vec \nabla f = (2x,1)$ and $\vec \nabla g = (4,3)$. The equation $\vec \nabla f = \lambda \vec \nabla g$ gives us $2x=\lambda\cdot 4$ and $1 = \lambda 3$. The second equation tells us $\lambda =1/3$, and the first equation tells us $x=\lambda\cdot 2=2/3$. Substitution into $4x+3y=12$ tells us $y=(12-8/3)/3$.
Group Problems
- Consider the function $f(x,y)= x^3+3xy-y^3$. This function has two critical points, namely $(0,0)$ and $(1,-1)$.
- Compute the gradient $\vec \nabla f(x,y)$.
- Compute both $\vec \nabla f(0,0)$ and $\vec \nabla f(1,-1)$. Your work should show that both $(0,0)$ and $(1,-1)$ are critical points. (What value should you obtain, and do you obtain it?)
- Compute the second derivative $D^2f(x,y)$. Then compute both $D^2f(0,0)$ and $D^2f(1,-1)$, the second derivative at these critical points.
- Classify each critical point as a maximum, minimum, or saddle point, by computing the eigenvalues of the second derivative at that point.
- A rover travels along the curve $4x+y=3$. The elevation near the rover is given by $z=y-x^2$. Use Lagrange multipliers to locate the $(x,y)$ coordinates where the rover reaches maximum height. [Check: $(x,y)= (-2,11)$.]
- Suppose $h$ is a function of $p$ and $q$, where $p$ and $q$ are both functions of $x$, $y$, and $z$. Compute the partial derivatives of $h$ with respect to $x$, $y$, and $z$.
- 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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