Rapid Recall
- For the function $f(x,y)=x^2y^3$, compute $f_x$. Then compute the partial of $f_x$ with respect to $y$, which we write symbolically as $f_{xy}$, or $(f_x)_y$, or $\frac{\partial}{\partial y}\left(\frac{\partial f}{\partial x}\right)$ or $\frac{\partial^2 f}{\partial y\partial x}$.
Solution
We have $f_x = (2x)y^3$. We then compute the partial of this with respect to $y$ to obtain $f_{xy}=(2x)(3y^2)$.
- The hyperbola $x^2-y^2=5$ passes through the point $(3,2)$. Differentials tell us $2xdx-2ydy = 0$. Give an equation of the tangent line to this curve at $(3,2)$.
Solution
Let $(x,y)$ be a point on the tangent line. Then change in $x$ from $(3,2)$ to $(x,y)$ is $dx = x-3$, with $dy=y-2$. Plugging $x=3$, $y=2$, $dx=x-3$ and $dy=y-2$ into the differential gives $$2(\underbrace{3}_{x})\underbrace{ (x-3) }_{dx}-2(\underbrace{2}_{y})\underbrace{ (y-2) }_{dy} = 0.$$
- A parallelogram has edges given by the vectors $(2-\lambda, 1)$ and $(3, 4-\lambda)$. Find $\lambda$ so that the area of the parallelogram is zero.
Solution
Remember the area formula $A=|ad-bc|$. Employing this formula gives $$A=(2-\lambda)(4-\lambda)-3 = (\lambda^2-6\lambda+8)-3 = \lambda^2-6\lambda+5 = (\lambda-5)(\lambda-1).$$ Since we want $A=0$, then we have either $\lambda=5$ or $\lambda=1$.
- Find an $x$ value so that the two vectors $\vec u = 2{\bf i}+(3+x){\bf j}$ and $\vec v= 1{\bf i}+(2-x){\bf j}$ are parallel.
Solution
We need $(2,3+x) = \lambda (1,2-x)$ (one vector needs to be a multiple of the other). This gives two equations to solve, namely $$2=\lambda 1, \quad \text{and}\quad (3+x)=\lambda(2+x).$$ The first equation tells us $\lambda=2$. The second equation then becomes $3+x = 2(2-x)$, or $3+x=4-2x$. Rearranging gives $3x=1$, or $$x=1/3.$$
Group Problems
- We will find the points on the curve $g(x,y)=xy^2=16$ that minimize the function $f(x,y)=x^2+y^2$.
- Compute $\vec \nabla f$ and $\vec \nabla g$.
- To find the points where $\vec \nabla f$ and $\vec \nabla g$ are either parallel or anti-parallel, we need to solve $\vec \nabla f=\lambda\vec \nabla g$ together with $g(x,y) = 16$. Write the three equations that results from this.
[Check: Did you get $2x=\lambda y^2$, $2y = \lambda 2xy$, and $xy^2=16$.
- Solve the system above (you should get $x=2$ and $y=\pm \sqrt{8}=\pm 2\sqrt{2}$).
- In the $xy$-plane, draw the curve $xy^2=16$, and then also add to your plot several level curves of $f$.
- As a group, discuss what the equation $\vec \nabla f=\lambda\vec \nabla g$ means in the context of this problem.
- Consider the function $f(x,y,z) = 4x^2+4y^2+z^2$. We'll be analyzing the surface at the point $P=(1/2,0,\sqrt{3})$.
- Compute $f(1/2,0,\sqrt{3})$. Then draw the level surface that passes through this point. So draw the ellipsoid $4=4x^2+4y^2+z^2$. [Hint: Divide both sides by 4 first. ]
- Compute the gradient $\vec\nabla f(x,y,z)$, and then give $\vec\nabla f(P)$.
- Compute the differential $df$, and then the differential at $P$.
- For a level surface, the output remains constant (so $df=0$). If we let $(x,y,z)$ be a point on the surface really close to $P$, then we have $dx=x-1$, $dy=y-0$ and $dz = z-?$. Plug this information into the differential to obtain an equation of the tangent plane to the surface.
- Give an equation of the tangent plane to the level surface of $f$ that passes through $(1,2,-3)$.
- Give an equation of the tangent plane to the level surface of $f$ that passes through $(a,b,c)$.
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