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question:Let a, b and c be positive integers such that gcd(a,b)=1 and a and b do not divide c. Prove that if ab − a − b < c < ab then the equation ax+by=c has exactly one positive solution x,yinmathbb{ Z}.
answer:Existence: let (x,y)inmathbb Ntimesmathbb Z be such that ax+by=c and x minimal. Then, x-b<0 hence by=c-axge c-a(b-1)>-b hence yge0. Uniqueness: if there were two positive solutions, there would exist two solutions of the form (x,y)inmathbb N^2 and (x-b,y+a)inmathbb N^2. But then c=ax+byge ab+b0 would contradict the assumption c<ab.
question:Determine the concavity and inflection points for the function {eq}f(x) = x^{3} - frac{3}{2}x^{2};quad ~[-1,2] {/eq}.
answer:To determine the concavity and inflection points, we need to find the second derivative of the function: {eq}f''(x)= frac{d}{dx}left(3x^2-3xright)=6x-3{/eq} Setting the second derivative equal to zero, we find the potential inflection point: {eq}6x-3=0 Rightarrow x=frac{1}{2}{/eq} Now, we can determine the concavity of the function by testing the sign of the second derivative in the intervals {eq}[-1,frac{1}{2}]{/eq} and {eq}[frac{1}{2},2]{/eq}: {eq}f''(-1) = -9 < 0 Rightarrow{/eq} Concave down on {eq}[-1,frac{1}{2}]{/eq} {eq}f''(1) = 3 > 0 Rightarrow{/eq} Concave up on {eq}[frac{1}{2},2]{/eq} Therefore, the function has an inflection point at {eq}x=frac{1}{2}{/eq}, where it changes concavity from concave down to concave up.
question:Why are the dynamics of L_x and L_y neglected in the analysis of a rigid body's rotation about the z-axis in introductory textbooks, like Halliday, Resnick, and Walker, even though they are non-zero and can change due to forces applied in arbitrary directions?
answer:In school-level textbooks, the dynamics of rigid body rotation are often simplified for ease of understanding. When a rigid body rotates about a fixed z-axis, it is common to choose principal axes, which are axes where the body's moments of inertia (I_x, I_y, I_z) are maximized or minimized. In this coordinate system, the angular momentum vector vec{L} can be expressed as the sum of the moments of inertia multiplied by their corresponding angular velocities: vec{L} = I_xomega_xhat{i} + I_yomega_yhat{j} + I_zomega_zhat{k}. For a rotation purely about the z-axis, omega_x and omega_y are zero, resulting in vec{L} being aligned with the hat{k} direction. However, when forces are applied in arbitrary directions, torques in the x- and y-axes (tau_x and tau_y) can arise, causing changes in L_x and L_y. Despite these components being non-zero, introductory textbooks often focus on L_z and its relationship with the torque tau_z along the axis of rotation (frac{dL_z}{dt} = Idot{omega}) to simplify the complex nature of rigid body dynamics.
question:What is the resultant force acting on a body when a force of 15 N acts towards the north and another force of 10 N acts towards the south?
answer:To find the resultant force, we consider the forces in the direction of the north and south. Since the forces are in opposite directions, we can simply subtract the smaller force (south) from the larger force (north). {eq}begin{align*} F_{Resultant} &= F_{North} - F_{South} F_{Resultant} &= 15 N - 10 N F_{Resultant} &= 5 N end{align*} {/eq} The magnitude of the resultant force is 5 N, and it acts in the direction of the net force, which is towards the north.