Search results

Yeah, so it means sketch it for x < 0.

Probably the negative reals. What was the context?

We basically need to draw a decent diagram. The "longest diagonal" is the diagonal joining two opposite corners of the box. Let the "height" be 4, width 3, and length 12, and draw it so the 3 x 12 face is on the bottom. Take say the diagonal from the "top back left" vertex of the box to the...

Re: MATH2601 Linear Algebra/Group Theory Questions $\noindent Here's some hints. If $B$ is a diagonalisable matrix such that $B^{2} = A$, then write $B = PDP^{-1}$ (so $P$ is an invertible matrix whose columns are eigenvectors of $B$, with corresponding eigenvalues in the diagonal entries of...

Re: MATH2601 Linear Algebra/Group Theory Questions No, the value for the trace and determinant of a real or complex matrix does not uniquely specify the matrix. $\noindent For $2\times 2$, let $A = \begin{bmatrix}a & b \\ c & d\end{bmatrix}$ and suppose the trace is given to be $\alpha$ and...

Re: Several Variable Calculus $\noindent Some further hints. On the sphere $\mathcal{S}$, we have $\mathbf{x}\cdot \mathbf{x} = \left\|\mathbf{x}\right\|^{2} = a^{2}$ and $\left\|\mathbf{x}\right\| = a$, so $\mathbf{F} = \phi(a)\mathbf{x}$ at any point $\mathbf{x}$ on the sphere. Also...

Re: Several Variable Calculus $\noindent The outward unit normal at any point $\mathbf{x}$ on the sphere is $\frac{\mathbf{x}}{a}$.$

For the first one, multiply top and bottom of the LHS by tan(x). This makes the numerator become tan(x) + 1, and the denominator is tan(x)(1 + tan(x)), so the fraction becomes 1/(tan(x)) = cot(x). For the second one, multiply top and bottom of the LHS by 1 – sin(α), and remember the difference...

Re: Several Variable Calculus $\noindent In this one, the surface is one where the points on it have one of the variables (in this case $z$) given explicitly as a function of the other two. In other words, the surface is of the form $z = g(x,y)$, where $x,y$ vary in some region...

Re: Several Variable Calculus $\noindent Let the surface be $\mathcal{S}$. We know that the area can be expressed using a surface integral as$ $$\mathrm{Area}(\mathcal{S}) = \iint _{\mathcal{S}}1 \, \mathrm{d}S.$$ $\noindent We just need to figure out how to parametrise $\mathcal{S}$. Here's...

Re: Why is english compulsory if if has no relevance to the real world (read more bel That wouldn't explain why English in particular is compulsory, only that some subject should be compulsory.

Are you sure it starts at rest? If the rocket starts at rest on the surface of the earth, it's not going to move... (Unless you assume there's no ground or something.)

Re: Several Variable Calculus It might be easier to use cylindrical coordinates. The way rho varies is different depending on what range phi is in. When phi is such that your point in the region lies within the "ice cream cone" part of the region, then rho will vary from 0 to a. For phi...

Re: Several Variable Calculus Inspection would be the best way to determine the limits probably (you can draw a diagram to help if you decide to use inspection).

$\noindent The general result is, for constants $a > 0$ and $m \neq 0$,$ $$\int a^{mx + b}\, \mathrm{d}x = \frac{1}{m\ln a}a^{mx + b} + C.$$ $\noindent To show this, it suffices to show that $\frac{ \mathrm{d} }{ \mathrm{d} x}\left(a^{mx+b}\right) = m( \ln a)\left(a^{mx+b}\right)$. Because of...

$\noindent Since $x^{2} \leq x^{2} + 4$ for all $x \in \mathbb{R}$, we have $\frac{x^{2}}{x^{2} + 4} \leq 1$ for all $x \in \mathbb{R}$, which implies $y \leq 1$ for all $y \in T$. Also, given any $0 < \varepsilon < 1$, take $x$ to be any real number with $x > \sqrt{\frac{4}{\varepsilon} - 4}$...

$\noindent The set $T$ is bounded above (note that $y \leq 1$ for all $y\in T$, for example). Also, $T$ is bounded below (note that $y \geq 0$ for all $y \in T$). Since $0 \in T$, as $0 = \frac{0^{2}}{0^{2} + 4}$ and $0 \in \mathbb{R}$, we have that $0$ is in fact the minimum of $T$, which...

You can also do (i) by differentiating the geometric series 1/(1-x) = 1 + x + x^2 + x^3 + ..., for |x| < 1, differentiating the RHS term by term (recalling that a power series can be differentiated term-by-term within its interval of convergence). To do the inductive step in (ii)...

Re: maths questions help Note that in general, y' = 0 and y" = 0 at some point is not sufficient for that point to be an inflection (for example, consider y = x^4 at x = 0). In this example, the third derivative is non-zero at x = -1, which implies it is an inflection. Or you can note that...

Depends on your definition of "harder" or "better".