Rapid Recall
- The elevation is $z=y+x^2$, along a path $y-2x=5$. Find the $(x,y)$ location of any maxes or mins.
Solution
With $f(x,y) = y+x^2$ and $g(x,y)=y-2x=5$, we have $\vec \nabla f = (2x,1)$ and $\vec \nabla g = (-2,1)$. The system we must solve is $$2x=\lambda(-2), 1=\lambda 1, y-2x=5.$$ The middle equation gives $\lambda =1$, which when plugged into the first equation gives $x=-1$. In the last equation, this give $y=3$.
- The elevation is $z=y+x^2$, along the path $\vec r(t) =(t,2t+5)$. Find the $(x,y)$ location of any maxes or mins.
Solution
Since $\vec r(t) =(t,2t+5)$, we know $x=t$ and $y=2t+5$. Substitution gives $z=(2t+5)+(t)^2$. We compute $\frac{dz}{dt} = 2+2t$. This derivative equals zero when $t=-1$, so we know $(x,y) = \vec r(-1) = ((-1),2(-1)+5) = (-1,3)$. Note that this is almost identical to the last problem.
- For the function $f(x,y)=x^2+xy+y^2-2y$, determine the location of any maxes, mins, or saddles, and classify each location appropriately using eigenvalues.
Solution
The gradient is $\vec \nabla f(x,y) = (2x+y,x+2y-2)$. The gradient equals zero when $2x+y=0$ and $x+2y-2=0$. The first equation tells us $y=-2x$. Plugging this into the second equation gives $x+2(-2x)=2$, or $x=-2/3$. Back substitution tells us $y=4/3$. The only critical point is hence $(x,y) = (-2/3,4/3)$.
The second derivative is $D^2f(x,y) = \begin{bmatrix}2&1\\1&2\end{bmatrix}$. The matrix does not change at the critical point, so this is also $D^2f(-2/3,4/3)$. The eigenvalues are the solution to the equation $(2-\lambda)(2-\lambda)-1=0$. Note that $$(2-\lambda)(2-\lambda)-1 = \lambda^2-4\lambda+3 = (\lambda-3)(\lambda-1).$$ The eigenvalues are $\lambda = 3$ or $\lambda = 1$. Since both are positive, the gradient points away from the critical point. This means the critical point corresponds to a minimum.
Group problems
- A box lies inside the rectangle $ [-2,6]\times [1,5] $ (so $-2\leq x\leq 6$ and $1\leq y \leq 5$ ).
- Compute the integral formula $\ds\frac{\int_{-2}^{6}\int_1^5 x dydx}{\int_{-2}^{6}\int_1^5 1 dydx}.$
- Compute the integral formula $\ds\frac{\int_{-2}^{6}\int_1^5 y dydx}{\int_{-2}^{6}\int_1^5 1 dydx}.$
- What physical quantities do the two integrals above compute?
- Compute the integral formula $\ds\frac{\int_1^5 \int_{-2}^{6}y dxdy}{\int_1^5 \int_{-2}^{6}1 dxdy}.$
- Draw the region described by the bounds of each integral.
- $\ds\int_{0}^{2}\int_{2x}^{4}dydx$
- $\ds\int_{0}^{4}\int_{0}^{y/2}dxdy$
- $\ds\int_{0}^{3\pi/2}\int_{0}^{2+2\cos\theta}rdrd\theta$
- $\ds\int_{-3}^{3}\int_{0}^{9-x^2}\int_{0}^{5}dzdydx$
- $\ds\int_{0}^{1}\int_{0}^{1-z}\int_{0}^{\sqrt{1-x^2}}dydxdz$
- Set up an integral formula to compute each of the following:
- The mass of a disc that lies inside the circle $x^2+y^2=9$ and has density function given by $\delta = x+10$
- The $x$-coordinate of the center of mass (so $\bar x$) of the disc above.
- The $z$-coordinate of the center-of-mass (so $\bar z$) of the solid object in the first octant (all variables positive) that lies under the plane $2x+3y+6z=6$.
- The $y$-coordinate of the center-of-mass (so $\bar y$) of the same object.
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