Assignment 7
Problems
Function Sketching
Consider the function \(f\colon \mathbb{R}^2\to\mathbb{R}\) given by the following formula: \[f(x,y)=x^3+y^3-3xy+14\] In this problem, we work to understand \(f\) without any technology: to showcase the power that calculus has brought us.
- Find the critical points of \(f\), and classify these as max, mins or saddles via the second derivative test.
- Use this classification to sketch the level curves of \(f\) like we did in class. Start by drawing the level curves near each max, min and saddle. Then use that level curves cannot cross (away from a saddle) and there are no other max/mins to fill in the rest of the picture.
- Use this sketch of the level curves of \(f\) and the information about which points are maxima and minima to sketch the gradient of \(f\).
Optimization
You are once again working on the engineering team developing new rocket engines for the Artemis mission, where on a previous homework you submitted recommendations to your team members on how to modify the nozzle width and fuel temperature to increase efficiency. Now, you are tasked with taking this efficient design and trying to maximize the thrust during the initial burn to leave earth’s atmosphere, which is where most fuel gets used up.
Having already set the nozzle diameter and fuel temperature, you look to study the effect of two other variables: the overall radius of the rocket and the slope of the nose-cone (both of which greatly affect air resistance) on the net thrust. Your colleagues run a computational simulation of the atmospheric physics, and determine that the net force varies as a function of the radius \(r\) and slope \(s\) as
\[F(r,s)=8+4rs-4r^4-4s^4\]
Find the critical points of \(F\): these are the configurations that you and your engineering team should investigate more closely.
After figuring this out, what radius and slope should you recommend to the manufacturing team? (Remember radius should be a positive number, as its a length).
Constrained Optimization
On the last problem, you helped an engineering team for the NASA Artemis moon mission optimize the width and slope of their rocket’s nose cone for aerodynamic efficiency.
However, after a brief celebration you receive news back from the manufacturing team: your solution simply will not work! While it certainly is the most efficient with regard to air friction, the nose cone is but one piece of a very large and complicated rocket, and there’s other constraints that need to be taken into account.
Indeed, there is a fixed cap on the total weight of the rocket overall, and a fixed percentage of that can be used on materials in the manufacture of the nose cone. You are allotted exactly 800kg of material from which to design the nose cone, and asked to produce a design weighing exactly this much (so that other teams know ahead of time the nose cone’s final weight, and can plan accordingly).
The original nosecone design violates this, and indeed the team at NASA suggests that you throw out the entire model entirely and re-design the nosecone using lighter-weight specifications. Your new model has a different dependency of thrust on radius and efficiency:
\[F(r,s)=4-r^2-s^2+3rs\]
With this new model, you go back to your modeling software and ask the computer to estimate the weight of a nose cone with radius \(r\) and slope \(s\): it tells you that the weight can be modeled by
\[W(r,s)=100(s^2+r^2)\] (where we require that \(r> 0\) and \(s\geq 0\) as they are a distance and a slope)
Given this thrust equation and weight constraint, what should you bring back to the manufacturing team as the best \((r,s)\) parameters for aerodynamic flight?
Solutions
Function Sketching
To sketch the level sets and gradient of this function, we first must find the critical points and classify them. Taking the graident,
\[\nabla f = \langle 3x^2-3y,3y^2-3x\rangle\]
So the system of equations we need to solve is (after dividing by the 3):
\[\left\{x^2=y,\,y^2=x\right\}\]
Substituting the first equation into the second we see
\[y^2=(x^2)^2=x^4=x\]
Thus either \(x=0\) (and hence \(y=0\)) or \(x^3=1\). But this only has the solution \(x=1\) (and hence \(y=1\)), so there are two critical points
\[\{(0,0),(1,1)\}\]
To classify them, we need to apply the second derivative test:
\[f_{xx}=6x\hspace{1cm}f_{xy}=-3\hspace{1cm}f_{yy}=6y\]
Thus the Hessian matrix and its determinant are
\[H=\pmat{6x &-3\\-3&6y}\hspace{1cm}D=\det H=36xy-9\]
Evaluating this at the critical points gives \(D(0,0)=-9\), so \((0,0)\) is a saddle, and \(D(1,1)=36-9=25\), so \((1,1)\) is either a max or min. To figure out we look at the sign of \(f_{xx}\) or \(f_{yy}\). Both of these are positive, so its a minimum.
To sketch the level sets, we can start with the local model of a minimum (concentric rings) at \((1,1)\) and of a saddle (an X shape) at \((0,0)\), then fill in the rest of the plot making sure no other level sets cross or close up into circles.
Given this picture, we can draw the gradient by remembering its orthogonal to level sets, and points in the direction of maximum increase. Since we know that saddles go upwards along one axis and downwards along the other, knowing where the minimum is sorts out the directions of everything else:
Optimization
To optimize the behavior of this rocket, we are searching for maxima in \((r,s)\). That means we again must begin by finding critical points and classifying them. Starting with the gradient
\[\nabla F = \langle 4s-16r^3,4r-16s^3\rangle\]
As the critical points are where the gradient vanishes, this leads to the system of equations
\[\left\{s=4r^3,\, r=4s^3\right\}\]
Similar to the previous problem, we can subsitute one of these into the other to get an equation in a single variable:
\[s=4r^3=4(4s^3)^3= 4^4 s^9\]
Thus, either \(s=0\) (and hence \(r=0\) too) or we can divide by \(s\) and see that
\[1=4^4 s^8\,\,\implies\,\,\frac{1}{4^4}=s^8\]
Taking the fourth root of both sides gives
\[s^2=\frac{1}{4}\,\,\implies\,\, s=\pm\frac{1}{2}\]
Plugging these into the equation \(r=4s^3\) to get \(r\) in each case, we find three critical points:
\[\left\{(0,0),\left(\tfrac{1}{2},\tfrac{1}{2}\right),\left(\tfrac{-1}{2},\tfrac{-1}{2}\right)\right\}\]
Now to classify these we must apply the second derivative test:
\[H=\pmat{f_{xx}& f_{xy}\\ f_{xy}& f_{yy}}=\pmat{-48r^2 & 4\\ 4& -48s^2}\]
\[D=\det H = (-48r^2)(-48 s^2)-(4)(4)=48^2 r^2s^2-4^2\]
Plugging in the critical points, we find that
\[D(0,0)=-16\hspace{1cm}D(\tfrac{1}{2},\tfrac{1}{2})=144\hspace{1cm}D(\tfrac{-1}{2},\tfrac{-1}{2})=144\]
Thus \((0,0)\) is a saddle, and we need to look closer at the sign of \(f_{rr}\) to determine the other two extrema. In each case \(f_rr=-12>0\) so they are each maxima.
BUT: remember that \(r\) is a radius in the problem, and must be a positive number, so we can actually ignore this critical point - it doesn’t correspond to a buildable rocket. This means we should suggest the parameters
\[(r,s)=\left(\frac{1}{2},\frac{1}{2}\right)\]
to NASA, where the maximal value is \[F(1/2,1/2)=\frac{31}{4}\]
Constrained Optimization
To solve the constrained optimization problem, we need to employ the method of Lagrange multipliers. We need to find the gradient of the efficiency function as well as of the constraint.
\[\nabla F = \langle -2r+3s,-2s+3r\rangle\]
The constraint on weight is \(100(s^2+r^2)=800\), so we can simplify the numbers a bit by first dividing by 100 to get \(s^2+r^2=8\). Now, \(\nabla W = \langle 2r,2s\rangle\) and so the system of equations to be solved is
\[\nabla F = \lambda \nabla W\]
\[\pmat{3s-2r\\ 3r-2s}=\pmat{2\lambda r\\ 2\lambda s}\]
One solution to this system is \(r=0\) and \(s=0\) as then both equations just become \(0=0\) which is true. When they are nonzero, we can solve each equation for \(2\lambda\) by dividing by a variable and set them equal:
\[\frac{3s-2r}{r}=2\lambda = \frac{3r-2s}{s}\]
This eliminates \(\lambda\) which is good - we don’t care about its value anyway - and gives us an equation just in terms of \(r\) and \(s\). Multiplying through by the denominators we can simplify a bit
\[(3s-r)s=(3r-s)r\] \[3s^2-rs=3r^2-rs\] \[3r^2=3s^2\] \[r^2=s^2\]
Taking the square root of both sides we find \(r=\pm s\), which we can then use in the constraint. In either of these cases, \[r^2+s^2=r^2+r^2=2r^2=8\,\,\implies\,\, r^2=4\,\,\implies \,\,r=\pm 2\]
This gives us four more critical points: when \(r=2\) we can have \(s=\pm 2\) so we get \((2,2)\) and \((2,-2)\), similarly when \(r=-2\), where we get \((-2,-2)\) and \((-2,2)\).
Thus, the five critical points for this system are
\[\{(0,0),(2,2),(2,-2),(-2,2),(-2,-2)\}\]
But - not all of these are physically relevant to the problem! We need both \(r>0\) and \(s\geq 0\) for the model to make sense, and only one of these satisfies both of these constraints:
\[(r,s)=(2,2)\]
The new efficiency of this model is
\[F(2,2)=4-(2)^2-(2)^2+3(2)(2)=4-4-4+12=8\]