Try FOH (Faculty Office Hours) - https://www.youtube.com/watch?v=yQq1-_ujXrM
Brain Gains
- If we know $r=-4$ and $\theta = \pi$, find $x$ and $y$.
Solution
The solution is $(x,y)=(4,0)$.
You can do this problem visually.
- Start on the $x$-axis and rotate 180 degrees till you are facing west. Then walk backwards (east) 4 units landing you at $(4,0)$.
- Go west 4 units, and then rotate the segment from (0,0) to (-4,0) 180 degrees, fixing the origin, to land at $(4,0)$.
You can also just compute directly
- $x=-4\cos\pi = 4$.
- $y=-4\sin\pi = 0$.
- For $z=3xy^2+2x$, find the derivative $\frac{dz}{dt}$, provided both $x$ and $y$ are functions of $t$.
Solution
We obtain $\frac{dz}{dt} = 3x(2y)\frac{dy}{dt}+3\frac{dx}{dt}y^2+2\frac{dx}{dt}$.
Note, this means $$\begin{align*} dz &= 3x(2y)dy+3(dx)y^2+2dx\\ &= (3y^2+2)dx + (6xy)dy\\ &= \langle 3y^2+2,6xy\rangle \cdot \langle dx,dy\rangle .\end{align*}$$ We call $\langle 3y^2+2,6xy\rangle$ the gradient of $z$ and write $\vec\nabla z = \langle 3y^2+2,6xy\rangle$.
- Find the differential $dy$ of the function $y = x^3+2x$ in terms of $x$ and $dx$.
Solution
The derivative is $\frac{dy}{dx} = 3x^2+2$. This gives the differential as $$dy = (3x^2+2)dx.$$
Let's compare the differentials $dy = (3x^2+2)dx$ and $dz = \langle 3y^2+2,6xy\rangle \cdot \langle dx,dy\rangle$. In both cases we have a function of the form $\text{output} = f(\text{input})$. The differential gives $$\begin{align*} dy &= (3x^2+2)dx\\ dz &= \langle 3y^2+2,6xy\rangle \cdot \langle dx,dy\rangle\\ d(\text{output}) &= \text{thing}\cdot d(\text{input}) .\end{align*}$$ The "thing" above is the derivative, and turns out to be a matrix in all dimensions.
Little changes in the output can be approximated by multiplying the derivative by little changes in the input. The derivative is precisely the object that helps us understand how changing input variables results in a change in the output.
- In polar coordinates, we have $x=r\cos\theta$ and $y=r\sin\theta$. Use this to rewrite the equation $y=x^2$ using polar coordinates (so obtain a polar equation of the parabola).
Solution
Since $y=x^2$, we have $(r\sin\theta) = (r\cos\theta)^2$. It's customary to solve for $r$, which gives $\ds r = \frac{\sin\theta}{\cos^2\theta}$.
- Give a Cartesian equation of the polar curve $\ds r = \frac{8}{2\cos\theta+5\sin\theta}$.
Solution
One way to tackle this is to rewrite the above equation in the form $$ 2r\cos\theta+5r\sin\theta = 8.$$ We then substitute $x=r\cos\theta$ and $y=r\sin\theta$ to obtain $$ 2(x)+5(y) = 8.$$ It's a line.
Group Discussion
After each problem, or each part, remember to let someone else take a turn being scribe.
- Plot the polar points with $(r,\theta)$ given by $(2,0)$, $(4,\pi/2)$, $(-4,\pi/2)$, $(2,\pi/6)$, $(-2,\pi/6)$.
- Give a polar equation of the curve $2x+3y=4$. (So substitute $x=r\cos\theta$ and $y=r\sin\theta$, and then solve for $r$.)
- For the equation $z=x^2y+3y^2$, explain why $\ds\frac{dz}{dt} = 2xy\frac{dx}{dt}+x^2\frac{dy}{dt}+6y\frac{dy}{dt}$.
- Give a Cartesian equation of the polar curve $r=\tan\theta\sec\theta$. (Use $x=r\cos\theta$ and $y=r\sin\theta$ to work backwards. Start by rewriting the trig functions in terms of sines and cosines.)
- Compute the differential $dA$ for the area function $A=xy$. (Find $dA/dt$ first, if needed, and then multiply by $dt$.)
- We know $x=r\cos\theta$ and $y=r\sin\theta$. Compute $dx$ in terms of $r, \theta,dr, d\theta$. (If you need to, assume that everything depends on $t$, compute derivatives, then multiply by $dt$.)
- Plot the curve $r=3-2\sin\theta$.
|
Sun |
Mon |
Tue |
Wed |
Thu |
Fri |
Sat |