20220224

• 4.10 - Trent
• 4.11 - Max
• 4.12 - Megan
• 4.16 -
• 4.17 -
• 4.18 -
• 4.19 -
• 4.20 -

Two options:

1. Substitute, so $f(\vec r(t)) = a(ct+d)+b(et+f)$, and then differentiate.
2. 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.

We have

• $f_x = 8xy +0 $, and
• $\frac{\partial f}{\partial y} = 4x^2(1)+3y^2$.

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 $$ f_{xx}=\ds\frac{\partial}{\partial x}\left(\frac{\partial f}{\partial x}\right), \quad f_{xy}=\ds\frac{\partial}{\partial y}\left(\frac{\partial f}{\partial x}\right), \quad f_{yx}=\ds\frac{\partial}{\partial x}\left(\frac{\partial f}{\partial y}\right), \quad f_{yy}=\ds\frac{\partial}{\partial y}\left(\frac{\partial f}{\partial y}\right). $$

• $(4,-2,-\frac{5}{2})$.

Your graph should have 3 parabolas, namely $y=x^2$, $y=x^2+1$, and $y=x^2-4$.

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)

The vectors are orthogonal. So as long as neither is the zero vector, then the angle is 90 degrees.