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As everyone has said, just do Advanced. Otherwise you have really poor aligning for a course of similar difficulty.

With the circle thing, I believe it would affect it because the circle's shape means that it as it enters a magnetic field, the area entering is not constant per unit length entered (because cross-sectional lines of a circle aren't constant in length). And ε = –N.dΦ/dt. Since the flux is Φ(t) =...

It's a combination of resisted motion and the Faraday-Lenz Law I believe. I see the sample answers didn't explain the curved nature of the graph, although that is probably due to the fact that this actually would need calculus to explain, and that's not assumed knowledge for HSC Physics. For...

I'm guessing he used calculus then, 4U mechanics is mostly calculus-based (apart from some circular motion stuff, which doesn't seem relevant for that Q.).

Was that the one Drsoccerball used calculus on, or was that a different Q.?

It was used in the 2011 HSC 4U paper question in the end about polynomials, referring the maximum of the moduli of the coefficients or something if I recall correctly. (And M is often used in other contexts to refer to something big or the max. of things etc.)

Epsilon is also traditionally used to represent small things, which is basically why it's used in the limits, since we think of the "tolerance" as being made as small as we like. (There is this maths joke "Let epsilon be smaller than zero.")

(Those question marks in my above post were meant to be a delta. Test: δ. When I try to edit that post, the stuff I wrote disappears.)

The limit exists because when they say "the solution" to the inequation, it means that |f(x) – 1| < ε iff x is in the union of those intervals. For the limit as x → +∞, the left interval is irrelevant (although that one actually shows that the limit as x → -∞ is also 1). The right one is...

I think he typo'ed or got confused with his notation or something; the vector MA should be half the vector BA, which he labelled as b. This problem can be done more easily if we define the origin to be at one of the vertices (say A), and then use the convention of calling the position vector...

Correct, not surjective.

F = ma is not correct when the mass is changing with time. The correct equation is: F = m(dv/dt) + v(dm/dt) (as you can see, this reduces to F = ma when the mass is constant, which is when dm/dt = 0). I've posted more about this in the HSC Physics Marathons.

Pretty much all of the hyperbolic trig. ones are the same as their circular trig. counterparts in terms of derivatives. The only one that changes is that sech(x) gets a negative sign when differentiated (and of course that cosh(x) doesn't get a negative when differentiated; this is essentially...

For the csch it works the same, because when we multiply top and bottom by csch(x) + coth(x) (which is analogous to what we'd do to integrate csc(x) in the circular trig. cases), we see that the derivatives of csch(x) and coth(x) match their circular trig. analogs. I.e. (coth(x))′ = -csch2(x)...

They are related; the part containing the lambda in the non-homogeneous case is precisely the solution to the homogeneous case (not by coincidence). In general, if we have a consistent (i.e. solvable) non-homogeneous system Ax = b, the solution will be the form x = xp + xh, where xp is any...

Convert the linear system into an augmented matrix, then get this to row echelon form using Gaussian Elimination. You can then check whether solutions exist by observing the right-hand column. If solutions exist, you can proceed to find them in parametric form by setting any non-leading column...

I think it needs to be proved. Check your HSC 4U textbook for the proof (it should be in both the Patel and Arnold & Arnold books). Alternatively, you can check it here from this old thread, which does it for the ellipse: . The proof for the hyperbola is basically identical, just replace b^2...

$\noindent The hyperbola is: $\frac{x^2}{4^2}-y^2 = 1$. Now, use the fact that the equation of the chord of contact from the point $\left(x_0,y_0\right)$ to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is $\frac{x_0 x}{a^2}-\frac{y_0y}{b^2}=1$, assuming such a chord exists (which'll happen...

$\noindent This question was answered here $\Big{(}$With $a$ needing to be in $\left[0,1-\frac{1}{n}\right]$$\Big{)}$:$ http://community.boredofstudies.org/238/extracurricular-topics/321437/intermediate-value-theorem-question.html.

If f(0) = 0 or f(1) = 1, we are done (taking c = 0 or c = 1 respectively does the job). Otherwise, apply the intermediate value theorem to g (which is continuous as it is the difference of two continuous functions), noting that g(0) and g(1) are of opposite signs, because g(0) = f(0) and g(1)...