Presentations
4.4 - 4.11Pacing 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
- Consider the function $f(x,y)=x^2y+5y$. Compute $\vec \nabla f(x,y)$.
Solution
- $\vec \nabla f(x,y)= (2xy, x^2+5)$.
You can do this several ways.
- One option is to first compute the differential $dz=2xydx+x^2dy+5dy$ and factor to get $dz=2xydx+ (x^2+5)dy$. Then extract the partials to get the gradient above. This option is sometimes much more time consuming that the next option.
- Another option is to compute the partials directly, without ever computing the differential.
- If $y$ is constant, then the derivative of $x^2y+5y$ with respect to $x$ is $f_x=2xy+0$. Note that $5y$ is a constant which is why the derivative was zero.
- If $x$ is constant, then the derivative of $x^2y+5y$ with respect to $y$ is $f_y=x^2(1)+5(1)$.
- For $f(x,y)=x^2y+5y$, compute $D_{ (-2,3) }f(1,1)$, the derivative of $f$ in the direction of $(-2,3)$. This is the same as the slope of the function at $(1,1)$ in the direction $(-2,3)$, which you can find using $\frac{dz}{\sqrt{dx^2+dy^2}}$.
Solution
Using directional derivative notation, we have
- $D_{ (-2,3) }f(1,1) = \vec \nabla f(1,1)\cdot \dfrac{ (-2,3) }{|(-2,3)|} = (2,6)\cdot \dfrac{ (-2,3) }{\sqrt{4+9}} = \dfrac{14}{\sqrt{13}}$.
Using differential notation, we have
- $\ds\frac{dz}{\sqrt{dx^2+dy^2}} = \frac{2xydx+ (x^2+5)dy}{\sqrt{dx^2+dy^2}}.$
Plugging in $(x,y)=(1,1)$ and $(dx,dy)=(-2,3)$ gives
- $\ds\frac{dz}{\sqrt{dx^2+dy^2}} = \frac{ (2)(-2)+ (6)(3) }{\sqrt{(-2)^2+(3)^2}} = \frac{14}{\sqrt{13}}.$
- For $z=f(x,y)=x^2y+5y$, give a Cartesian equation of the contour (level curve) that passes through the point $(1,1)$.
Solution
- We know $f(1,1) = 6$, so an equation is $6=x^2y+5y$. While not needed, you can solve for $y$ to obtain $y = \frac{6}{x^2+5}.$
- Draw the vertical cross section of the surface $z=x^2y+5y$ that occurs from letting $y=0$, then $y=1$, then $y=2$.
Solution
We just need to plot the three parabolas
- $z=0$,
- $z=x^2+5$, and
- $z=2x^2+10$.
The links below point to WolframAlpha.
- Draw the vertical cross section of the surface $z=x^2y+5y$ that occurs from letting $x=0$, then $x=1$, then $x=2$.
Solution
We draw the three lines
- $z=5y$,
- $z=6y$, and
- $z=9y$.
The links below point to WolframAlpha.
Group problems
- Let $g(x,y) =xy^3$.
- Compute $dg$.
- State $g_x$ and $\dfrac{\partial g}{\partial y}$. Then state $\vec \nabla g$.
- Find the directional derivative (slope) of $g$ at $P=(3,1)$ in the direction $(-3,2)$.
- Find the directional derivative of $g$ at $P=(3,1)$ in the direction $(2,-5)$.
- Consider the function $z=f(x,y)=x^2+y^2-4$.
- Construct a contour plot of $f$. So let $z=0$ and draw the resulting curve in the $xy$ plane. Then let $z=5$ and draw the resulting curve in the $xy$ plane. Then pick other values for $z$ and draw the resulting curve in the $xy$ plane. If you get a bunch of concentric circles, you're doing this right. On each circle you draw, write the height of that circle.
- Construct a 3D surface plot of the function.
- Consider the function $z=4-y^2$.
- Construct a 2D contour plot.
- Construct a 3D surface plot.
- Consider the function $f(x,y)=4-|x|$.
- Construct a 2D contour plot.
- Construct a 3D surface plot.
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