- Compute the determinant of $\begin{bmatrix}0&2&-1\\4&5&-2\\1&3&1\end{bmatrix}$.
- Find the area of a triangle with vertices $(0,0),(a,b), (c,d)$. (Use determinants.)
- Solve the system of equations $x+2y-3z=2$, $y-2z=1$. You might want to start by letting $z=t$, and then solving for $y$ and then $x$ in terms of $t$.
- Suppose we know $x=at+b$, $y=ct+d$, and $z=et+f$. Write this in the form $\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}?\\?\\?\end{bmatrix}t+\begin{bmatrix}?\\?\\?\end{bmatrix}$.
- Let $P(x)=ax^2+bx+c$. Let $f(x)=\ln(x+1)$. Solve the system of equations $P(0)=f(0) $, $P'(0)=f'(0) $, and $P' '(0)=f''(0) $ for the coefficients $a,b,c$.
- The volume of a box is $V=lwh$. Compute $dV$ in terms of $l, w,h, dl, dw, dh$. Then write your answer as the matrix product $$ dV = \begin{bmatrix}?&?&?\end{bmatrix}\begin{bmatrix}dl\\dw\\dh\end{bmatrix}.$$
- Compute $\int \arcsin x dx$.
Some Solutions:
- Software will give you this answer quickly. Here's a WolframAlpha solution.
- The area of a parallelogram is the determinant of a matrix whose columns are vector that form the edges of the parallelogram. Halving that give the area of the triangle, so the solution is $\frac{1}{2}\det \begin{bmatrix}a&c\\b&d\end{bmatrix}=\frac{ad-bc}{2}$.
- Let $z=t$ which means the second equation becomes $y-2t=1$ or $y=1+2t$. Replacing both $y$ and $z$ with what they equal in terms of $t$ into the first equation gives us $x+2(1+2t)-3t=2$, which means $x=2+3t-2-4t=-t$. The solution is $(x,y,z) = (-t, 1+2t, t)$.
- The solutions is simply $\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}a\\c\\e\end{bmatrix}t+\begin{bmatrix}b\\d\\f\end{bmatrix}$.
- The first two derivatives are $P'(x)=2ax+b$, $P''(x)=2a$, $f'(x)=(1+x)^{-1}$, and $f' '(x)=-(1+x)^{-2}$. Evaluating each at zero, the three equation given become $c=0$, $b=1$, $2a=-1$, or $a=-1/2$. So $P(x) = -\frac{1}{2}x^2+(1)x+0$.
- The differential is $dV = whdl+lhdw+lwdh.$ In matrix form we have $$ dV = \begin{bmatrix}wh&lh&lw\end{bmatrix}\begin{bmatrix}dl\\dw\\dh\end{bmatrix}.$$
- The integral is $x\arcsin(x)+\sqrt{1-x^2}$
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