- Consider the curve $r=3-3\cos \theta$. Yesterday you drew the curve in the $xy$ plane and got a heart shaped object.
- Compute $dx$ and $dy$ for the curve above (they will be in terms of $\theta$ and $d\theta$). Since $x=r\cos\theta$, for this curve you have $x=(3-3\cos\theta)\cos\theta$. You can use this quickly compute $dx$. A similar computation will get you $dy$.
- Give a vector equation of the tangent line to the curve above at $\theta=\pi/2$.
- Compute the slope $dy/dx$ at $\theta=\pi/2$, and then give a Cartesian equation of the tangent line to the curve at $\theta=\pi/2$.
- Set up an integral formula to compute the arc length of this curve. Don't actually compute the integral, rather just set it up.
- Now Consider the change of coordinates $x=2u-v$, $y=u+2v$.
- Draw the curve $u^2+v^2=4$ in the $uv$ plane. Give the area inside the curve in the $uv$ plane.
- Draw the curve above in the $xy$ plane. (So make a table listing $u,v,x,y$ values. You know a few points for the curve in the $uv$ plane, so pick those and then find the corresponding points in the $xy$ plane.)
- Compute $dx$ and $dy$, and then write them in the matrix form $$ \begin{bmatrix} dx\\dy \end{bmatrix}= \begin{bmatrix} ?&?\\?&? \end{bmatrix} \begin{bmatrix} du\\dv \end{bmatrix}.$$
- Compute the determinant of the matrix above, and then make a guess for the area inside the curve in the $xy$ plane.
- A parametrization of the curve in the $uv$ plane is $u=2\cos t, v=2\sin t$. Compute $du$ and $dv$ in terms of $t$ and $dt$, and then $dx$ and $dy$ in terms of $t$ and $dt$.
- Given an equation of the tangent line to the curve in the $xy$ plane at $t=\pi/2$ (so $(u,v) = (0,2)$, or $(x,y) = (0-2,0+4)$).
- Set up an integral to compute the arc length of the curve in the $xy$ plane.
- If you finish early, then repeat the problem above with another change of coordinates of the form $x=au+bv$, $y=cu+dv$ for different values of $a,b,c,d$.
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