# Tutor profile: Michael H.

## Questions

### Subject: Geometry

Show that if a line d is not contained in plane z, then the intersection of d and z is the empty set or it is formed of a single point.

Let's make the assumption that the intersection of d and z yields the set {A,B} if and only if d belongs to the set of z. It will contradict the hypothesis if and only if the intersection of d and z = {A} or if it equals the empty set. This means that if the line exists in the plane z then there are an infinite number of points, so if the line doesn't exist in the plane, it will either intersect it exactly once, or not intersect it at all.

### Subject: Calculus

Find the maximum volume of a rectangular prism that's inscribed within a sphere.

This problem is a classic case of when to use Lagrange multipliers because of the maximization of a volume. The volume of the rectangular prism is given by V(x,y,x)=xyz, all that's left is to determine a constraint. To maximize the volume, the diagonal of the rectangular prism should be the same length as the diameter of the sphere. The diameter of the rectangular prism is given by x^2+y^2+z^2=d^2, substituting d for 2r and rearranging to let the function equal o we get x^2+y^2+z^2-4r^2=0. First we need to find the gradient of V(x,y,z) which is equal to <yz,xz,xy> and the gradient of our constraint is <2x,2y,2z>. Setting up the Lagrange multiplier means that yz=(lambda)2x, xz=(lambda)(2y), and xy=(lambda)(2z). This means that (yz/2x)=(xz/2y)=(xy/2z). Solving this equation yields that x=y=z, so substituting this into the constraint equation yields x^2+x^2+x^2-4r^2=0. Solving for x yields x=(2r/sqrt(3)). Since x=y=z all three terms will equal this, plugging these values back into the volume equation yields (8r^3/3sqrt(3)).

### Subject: Physics

Suppose a rocket is sitting still in space far away from any bodies of mass. The rocket then begins to travel at a constant velocity in the positive x direction of 2400 m/s. If the initial weight of the rocket with fuel is 4800 kg, and it's weight without fuel is 2400 kg, find the velocity of the rocket once all of the fuel is used up.

To begin the problem it's important to understand that Newton's second law doesn't directly apply to the problem. A rocket is propelled forward by the combustion of its fuel source, which means that the mass of the rocket is going to vary with time. Let v represent the velocity of the rocket traveling in the positive x direction and v(ex) be the velocity of the exhaust. In this system the velocity of the fuel burned is going to be v-v(ex). The total momentum of this part is going to be (-dm)(v-v(ex)), remember that we are losing the mass of the fuel and that it's just a small amount of fuel at an infinitesimal small time. The momentum of the rocket itself is given by (m+dm)(v+dv) where dv is the small increase in velocity and (m+dm) is the mass of the rocket with the small amount of fuel unburned. Since momentum is conserved in this system mv=(m+dm)(v+dv)+(-dm)(v-v(ex)). Following some algebraic simplification the equation can be written as mdv=-dmv(ex)-dmdv. Both dm and dv are two infinitesimal small quantities, so within the context of this problem it'll be alright to ignore them. This now yields the equation mdv= -dmv(ex). Dividing each side by the mass of the rocket yields dv=(-dmv(ex)/m). We now have an expression for the instantaneous velocity of our rocket in relation to its changing mass, the only thing left to do is add up these infinite dv's, which is the definition of integration. Integrate dv with limits of the intial veloctiy to the final velocity and dm with limits of the initial mass to the final mass. Upon integration our equation becomes v-v(initial)=v(ex)ln(m(initial)/m). Since the initial velocity is 0, the final equation will be v=v(ex)ln(m(initial)/m). Now plugging in our values we get v=(2400 m/s)ln(4800 kg/2400 kg)=1663 m/s. This means that the final velocity of the rocket once the fuel is used up is 1663 m/s.

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