Math 215 - Multivariate Calculus - Schedule

Rapid Recall

Consider the function $f(x,y)=y-x^2$. On a single $xy$ grid, graph the level curves with heights $0$, $1$, and $-4$.

Solution

Are the vectors $(4,6)$ and $(6,9)$ parallel? In general, how do we tell if $\vec u$ and $\vec v$ are parallel?

Solution

Yes. Note that $\frac{3}{2}(4,6)=(6,9)$. Sine one vector is just a multiple of the other, they are parallel. In general, to know if $\vec u$ and $\vec v$ are parallel, we just need to know if $\vec v = c\vec u$ for some constant $c$.

Solution

Either substitute (getting $f(\vec r(t)) = a(ct+d)+b(et+f)$ and then differentiate, or compute differentials ($df=adx+bdy$, $dx=cdt$, and $dy=edt$) and then substitute to get $$\dfrac{df}{dt} = ac+be.$$

In your answer to the previous problem, locate and label each of $f_x$, $f_y$, $\frac{dx}{dt}$, and $\frac{dy}{dt}$, giving a general formula for $\frac{df}{dt}$ in terms of the four quantities.

Solution

We have $$ \frac{df}{dt} = \underbrace{a}_{f_x}\underbrace{c}_{\frac{dx}{dt}}+\underbrace{b}_{f_y}\underbrace{e}_{\frac{dy}{dt}} = f_x\frac{dx}{dt}+f_y\frac{dy}{dt}. $$

Construct a surface plot of $f(x,y)=y-x^2$.
Construct a contour plot of $f(x,y)=4-|y|$.
Construct a surface plot of $f(x,y)=4-|y|$.
Consider the elevation function $f(x,y)=xy^2$ and the path $\vec r(t) = (2t-1,t^2)$.
- Compute $f(\vec r(t))$ and then compute $\frac{df}{dt}$.
- Find $df$ in terms of $dx$ and $dy$. Then find $dx$ and $dy$ in terms of $dt$.
- Use substitution to find $df$ in terms of $dt$. Then state $\frac{df}{dt}$.
- In you previous work, label each of $f_x$, $f_y$, $\frac{dx}{dt}$ and $\frac{dy}{dt}$.
Repeat the previous with $f(x,y)=e^x\sin y$ and $\vec r(t) = (t^2,t^3)$.
Repeat the previous with $f(x,y)=ax+by$ and $\vec r(t) = (ct+d, et+f)$.
If you finish early, make up your own function $f(x,y)$ and path $\vec r(t)$ and repeat.