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Find the position of the particle at t = 0 seconds and at t = 2\pi seconds. Then, find the equation for the velocity of the particle and see if there are any times on 0 < t < 2\pi where v = 0. If there are, find the position x at those times as well. Then, find the distances of each separate...

2\cos{\alpha} - (1 - 2k) = 0 rearranges to give \cos{\alpha} = \frac{1 - 2k}{2}. Since we know that k\in\[0,\ 1], it follows that \cos{\alpha}\in\left[-\frac{1}{2},\ \frac{1}{2}\right] \quad \implies \quad \alpha \in \left[\frac{\pi}{3},\ \frac{2\pi}{3}\right]. The original form of this...

Take a and b as unit vectors from the origin (O), representing points A and B on the unit circle. The angle between the vectors is 2\theta. Since the distance AB is the length of the vector a - b and it is less than one, the isosceles triangle AOB must have the radii as its longest sides...

The answer is (D)

Yes, the identity |z|^2 = z\bar{z} is in all the texts, and is easily proved. \begin{align*} \text{Let } z &= x + iy \\ \text{And so } \bar{z} &= x - iy \\ \\ \text{RHS } &= z\bar{z} \\ &= (x + iy)(x - iy) \\ &= (x)^2 - (iy)^2 \\ &= x^2 - i^2y^2 \\ &= x^2 - (-1)y^2 \\ &= x^2 + y^2 \\ &= |z|^2...

The only way that g(f(x)) will have a vertical asymptote is if g(f(x)) = g(a), and so there would need to be some x-value such that f(x) = a. Look at the graph, f(x) is always positive or zero, but a is negative. f(x) = a has no solution. So, g(f(x) won't have a vertical asymptote, and the...

Any line through the origin has equation y = mx for some value of m. This line intersects the curve y = x^2 + 3x + 5 when mx = x^2 + 3x + 5 \implies x^2 + (3 - m)x + 5 = 0. If the line is a tangent, then the quadratic will have only one solution, and so the discriminant is zero. Thus...

It is 6!, not 6! x 2. Seat the host in any seat in 1 way (all seats are equivalent under rotation). The seven seats remaining are now all different, but there is only one of them in which the hostess can be seated. The remaining 6 people can be seated in 6! ways. So, # ways = 1 x 1 x 6! = 6!

Note that a much simpler approach is to sketch f(x) = \frac{-8}{x - 2} - \frac{8}{x^2} as both y = \frac{-8}{x - 2} and y = \frac{-8}{x^2} are easily sketched and you can then add them to get y = f(x). The question tells you that v = 0 at x = 1 and so the sketch of v2 v. x is simply the part...

Look at the velocity equation... it is undefined for x = 0 and x = 2, so the particle is always trapped in x > 2, 0 < x < 2, or x < 0. Given x = 1 initially, the domain of movement is no bigger than 0 < x < 2

Taking to the right and upwards as the directions for i and j: The weight force is W = -50g j The tensions are F = F[(cos 53) i + sin 53) j] and T = T[(-cos 40) i + (sin 40) j], where F and T are the magnitudes of the two tensions, respectively. Since the object is stationary, F + T + W = 0...

No. You would use the displacement equations for the positions at each time. You would need to find the velocity equation from v = \frac{dx}{dt} though, to solve v = 0 to find out when the particle is stationary (which is the only place that direction can change).

For total distance travelled, find the position of the particle: at the start of the motion at the end of the motion everywhere it stops (v = 0) during the time interval you are investigating. Then, draw these positions on a number line. So, if: x = 1 at t = 0 x = 5 at t = 5 v = 0 at t = 2...

I can write 1 + i = \sqrt{2}\left(\frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}\right) = \sqrt{2}e^{\frac{i\pi}{4}} using the Principal Argument of 1 + i, which is \frac{\pi}{4}. I can express the same number as 1 + i = \sqrt(2)e^{i\left(\frac{\pi}{4} + 2\pi\right)} =...

The answers given are correct. The angle between OZ1 and OZ2 is \alpha + -\beta = \alpha - \beta. So, to get to the direction of OZ2 from OZ1 requires a clockwise rotation by \alpha - \beta and hence is achieved by multiplying by e^{-i(\alpha - \beta)} = e^{i(\beta - \alpha)}. The angle...

Principal Argument is the argument \in(-\pi,\ \pi] And, the acute angle you mention is -\beta

From the definition in the question, \beta < 0

To help with the drawing, Z3 lies on a circle with OZ1 as radius and O as the centre (because of the equal moduli). Z2 lies in the fourth quadrant and is inside this circle, so after extending OZ2 in both directions, the two possible locations of Z3 are the intersections of this extended line...

The Fundamental Theorem of Calculus states that \frac{d}{dx} \int_a^x f(t)\ \!dt = f(x) Applied to this problem: \begin{align*} \frac{d}{dt} \int_a^{t^2} xf(x)\ \!dx &= \frac{d}{dt^2} \int_a^{t^2} xf(x)\ \!dx \times \frac{d(t^2)}{dt} \qquad \text{using the Chain Rule} \\ &=...

The values are not exact, it depends on you recognising the \sin{37^\circ} \approx \frac{3}{5}. I posted a solution at https://boredofstudies.org/threads/mechanics.405138/#post-7497094