Web Assign Review

Review of the web assign practice for the mid term review

Vectors stuff

We are given:

$x^2+y^2=64$

$z=x^{-}{y}^2$

Find the curve of intersection

standard parameterization of a circle of radius 8

$x=8\cos t$

$y=8\sin t$

once we have the first 2, the third comes for free

$z=64{\cos }^{2}t-64{\sin }^{2}t$

alternatively

$\vec{r}(t)=⟨8\cos t,8\sin t,64\cos 2t⟩$

we use the cosine double angle here

$\cos (2t)={\cos }^{2}t-{\sin }^{2}t$

$z=64\cos (2t)$

another problem example

$x^2+y^2=36$

$z=xy$

vector

$\vec{r}(t)=⟨6\cos t,6\sin t,18\sin (2t)⟩$

Curvature practice

e/trig problem

$\vec{r}(t)=⟨e^t\cos t,e^t\sin t,3t⟩$

find the curvature at point (1,0,0)

we need to find the value of t that corresponds to this point

t=0 is a good example

curvature equations that we need:

$K=\frac{|{\vec{r}}^{\prime }\times {\vec{r}}^{\prime \prime }|}{|{\vec{r}}^{\prime }{|}^3}$

$K=|{\vec{r}}^{\prime \prime }(\partial )|$

$K=|\frac{dT}{d\partial }|$

$K=\frac{|{\vec{T}}^{\prime }|}{|{\vec{r}}^{\prime }|}$

plugging in for ${\vec{r}}^{\prime }$

${\vec{r}}^{\prime }=⟨e^t\cos t-e^t\sin t,e^t\sin t+e^t\cos t,3⟩$

$=⟨e^t(\cos t-\sin t),e^t(\sin t+\cos t),3$

${\vec{r}}^{\prime }=⟨e^t(\cos t-\sin t)+e^t(-\sin t-\cos t),e^t(\sin t+\cos t)+e^t(\cos t-\sin t),0⟩$

$=⟨e^t(\cos t-\sin t-\sin t-\cos t),e^t(\sin t+\cos t+\cos t+\sin t),0$

$=⟨-2e^t\sin t,2e^t\cos tt,0⟩$

$r^{\prime }(0)=⟨1,1,3⟩$

$r^{\prime \prime }(0)=⟨0,2,0⟩$

$|{\vec{r}}^{\prime }(0)|=\sqrt{1+1+9}=\sqrt{11}$

i & j & k\\ 1 & 1 & 3 \\ 0 & 2 & 0 \end{bmatrix} = \langle-6,0,2 \rangle$$ $$|\vec{r}' \times $$ ## Polar Coordinates Problem Find the area in the top right quadrant and outside of the smaller circle $$x^2 + y^2 = 16, \ \ \ x^2 +y^2 = 4x$$ on desmos: > [!Info] Desmos > > <iframe src="https://www.desmos.com/calculator/aj9wwbqagm" width=600 height="400" ></iframe> we can get r from the point of intersection $$r = 4\cos \theta$$ settin up the double integral: $$\int \int_{\mathcal{D}} x dA = \int_{0}^{\frac{\pi}{2}} \int _{4\cos \theta}^4 r^2 \cos \theta \ dr \ d\theta$$ $$I_{1}: \ \ \ \left. \frac{r^3}{3}\cos \theta \right|_{4\cos \theta}^4$$ $$\frac{64}{3}\cos \theta - \frac{64}{3}\cos^4 \theta$$ $$\int_{0}^{\frac{\pi}{2}} \frac{64}{3} \cos \theta - \frac{64}{3} \cos^4 \theta \ d\theta$$ $$\frac{64}{3} - \frac{64}{3}\int_{0} ^ \frac{\pi}{2} \cos^4 \theta \ d\theta$$ $$\frac{64}{3} - \frac{64}{3} \int \left( \frac{1+\cos 2 \theta}{2} \right)^2 \ d\theta$$ $$\frac{64}{3} - \frac{16}{3} \int_{0}^ \frac{\pi}{2} (1+\cos 2 \theta)^2 \ d\theta$$ ## Tangent Planes / Normal Lines Problem $$x + y + z = 3e^{xyz}$$ Find the tangent plane and normal line at the point (0,0,3) write as a function $$f(x,y,z) = x+y+z-3e^{xyz}$$ do chain rule and make into a vector $$\nabla f = \left. \langle 1 - 3yze^{xyz}, 1-3xze^{xyz}, 1-3xye^{xyz} \rangle \right|_{(0,0,3)}$$ $$\nabla f = \langle 1,1,1 \rangle$$ $$\langle 1,1,1 \rangle \cdot \langle x-0, y-0, z-3 \rangle = 0$$ $$x+y+z-3=0$$ ### Equation of the tangent plane of this $$\nabla f \cdot \nabla r = 0$$ **this is usefull on test** $$\frac{x-0}{1}=\frac{y-0}{1} = \frac{z-0}{1}$$ ## random practice i forgot $$r_{1} = 1.2$$ $$h_{1}=-2.8$$ $$v_{1} = ? \ when \ r=135, h =186$$ $$V = \frac{\pi r^2 h}{3}$$ chain rule problem ```mermaid flowchart TD V --> r V --> h r --> t1 h --> t2 ``` $$V_{t} = V_{r}r_{t}+V_{h}h_{t}$$