Presentations
11-18Pacing Tracker
- The quizzes have included questions for 16 objectives. How many have you passed? What are you plans to master those that you haven't mastered yet?
- We've finished units 1 through 3. Have you started your self-directed learning project for each unit?
- Remember you can submit only one SDL project per week. Plan ahead and don't let yourself get behind.
Brain Gains
- Let $f(x,y)=ax+by$ and $\vec r(t) = (ct+d, et+f)$. Compute $\frac{df}{dt}$.
Solution
Two options:
- Substitute, so $f(\vec r(t)) = a(ct+d)+b(et+f)$, and then differentiate.
- Compute differentials, so $df=adx+bdy$, $dx=cdt$, and $dy=edt$, and then substitute.
Either way, we end up with $$\dfrac{df}{dt} = ac+be.$$ Symbolically, we have $$\frac{df}{dt} = \underbrace{a}_{f_x}\underbrace{c}_{\frac{dx}{dt}}+\underbrace{b}_{f_y}\underbrace{e}_{\frac{dy}{dt}} = f_x\frac{dx}{dt}+f_y\frac{dy}{dt}. $$ We call this the chain rule.
- For the function $f(x,y) = 4x^2y+y^3$, compute $f_x$ and $\ds\frac{\partial f}{\partial y}$.
Solution
We have
- $f_x = 8xy +0 $, and
- $\frac{\partial f}{\partial y} = 4x^2(1)+3y^2$.
- For the function $f(x,y) = 4x^2y+y^3$, compute $\ds\frac{\partial}{\partial x}\left(\frac{\partial f}{\partial y}\right)$.
Solution
Since we know $\frac{\partial f}{\partial y} = 4x^2(1)+3y^2$, then we have $$\ds\frac{\partial}{\partial x}\left(\frac{\partial f}{\partial y}\right) =\ds\frac{\partial}{\partial x}\left(4x^2(1)+3y^2\right) =8x. $$ We call the above the second partial derivative of f, first with respect to $y$, and then with respect to $x$. You'll see it appear in the homework very soon. For this function $f(x,y)$, there are four total second partial derivatives of $f$, namely $$ f_{xx}=\ds\frac{\partial}{\partial x}\left(\frac{\partial f}{\partial x}\right), \quad f_{xy}=\ds\frac{\partial}{\partial y}\left(\frac{\partial f}{\partial x}\right), \quad f_{yx}=\ds\frac{\partial}{\partial x}\left(\frac{\partial f}{\partial y}\right), \quad f_{yy}=\ds\frac{\partial}{\partial y}\left(\frac{\partial f}{\partial y}\right). $$
- State $a$ and $b$ so that $\vec u=(-8,4,5)$ and $\vec v=(4,a,b)$ lie along the same line.
Solution
- $(4,-2,-\frac{5}{2})$.
- Consider the function $f(x,y)=y-x^2$. On a single $xy$-grid, graph the level curves with heights $0$, $1$, and $-4$. In other words, construct a contour plot that includes the contours 0, 1, -4.
Solution
Your graph should have 3 parabolas, namely $y=x^2$, $y=x^2+1$, and $y=x^2-4$.
- Using the same function $f(x,y)=y-x^2$, construct a 3D surface plot.
Solution
There are several ways to construct this surface plot. When $x$ is held constant, the graph should consist of lines. When $y$ is held constant, the graph consists of parabolas opening downwards (negative $z$). When $z$ is held constant, the graphs are parabolas opening towards larger $y$ values. We'll graph this together in class.
var('x','y','z') #Define your variables
f(x,y) = y-x^2 #State the function
xbounds = (x,-3,3)
ybounds = (y,-3,3)
#This gives you a generic contour plot (a plot consisting of several level curves)
pcontour=contour_plot(f,(x,-2,2), (y,-2,2),colorbar=True)
show(pcontour)
#This section allows you to plot as many level curves as you want
level_curve_values=[0,1,-4]
p=implicit_plot(f(x,y)==level_curve_values[0],xbounds,ybounds)
for zlevel in level_curve_values:
p+=implicit_plot(f(x,y)==zlevel,xbounds,ybounds)
show(p)
#Uncomment this section if you want a 3D image of the surface.
p3D=plot3d(f,xbounds,ybounds, mesh=True)
for zlevel in level_curve_values:
p3D+=implicit_plot3d(f(x,y)==zlevel, xbounds,ybounds,(z,zlevel-0.1,zlevel+0.1),thickness=4, color = "red")
show(p3D)
#This section draws the gradient of your function.
gradf = f.diff()(x,y)
gradplot=plot_vector_field(gradf,xbounds,ybounds)
show(gradplot)
#This section combines your level curve plot and your gradient plot.
show(p+gradplot)
- If we know $\vec u\cdot \vec v=0$, then what is the angle between $\vec u$ and $\vec v$?
Solution
The vectors are orthogonal. So as long as neither is the zero vector, then the angle is 90 degrees.
Group problems
Here's a link to the Mathematica Notebook ContourSurfaceGradient.nb.
- Consider the function $f(x,y)=4-|x|$.
- Construct a 2D contour plot. Label your contours with their corresponding height.
- Construct a 3D surface plot.
- Construct both the above with software.
- Consider the elevation function $f(x,y)=e^x\sin y$ and the path $\vec r(t) = (t^2,t^3)$.
- Compute $f(\vec r(t))$ and then compute $\frac{df}{dt}$.
- Find $df$ in terms of $x$, $y$, $dx$, and $dy$. Then find $dx$ and $dy$ in terms of $t$ and $dt$.
- Use substitution to state $df$ in terms of $dt$ and divide by $dt$ to obtain $\frac{df}{dt}$. [You should have the same answer as the first part.]
- In your solution for $\frac{df}{dt}$, label each of $f_x$, $f_y$, $\frac{dx}{dt}$ and $\frac{dy}{dt}$ to verify that $$\frac{df}{dt} = f_x\frac{dx}{dt}+f_y\frac{dy}{dt}.$$
- Consider the function $z=4-y^2$.
- Construct a 2D contour plot by hand. So pick several values for $z$ and plot the resulting curves. If you end up with lots of horizontal lines in the $xy$-plane, you're doing this correctly. Write the height on each horizontal line you draw.
- Construct a 3D surface plot by hand.
- Construct both the above with software.
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