20241015

• I-Learn, Class Pictures, Learning Targets, Text Book Practice
• Prep Tasks: Unit 1 - Motion, Unit 2 - Derivatives, Unit 3 - Integration, Unit 4 - Vector Calculus

This is Day 6 in Unit 2.

Prep

We're in Unit 2 - Derivatives. Your homework assignment each day is to spend 1-2 hours working on the next 4 tasks from the current unit's prep.

Brain Gains (Rapid Recall, Jivin' Generation)

• Let $f(x,y) = x^2y+7y$. We will focus on what happens at $P=(1,2)$. Note that $f(1,2) = 16$.
  • Compute the differential $df(1,2)$ and gradient $\vec \nabla f(1,2)$.
  • Let $(x,y,z)$ be a point on the tangent plane to $f$ at $(1,2)$. The differentials $dx = x-1$, $dy = y-2$, and $dz = z-16$ represent changes in $x$, $y$, and $z$ on the tangent plane. Give an equation of the tangent plane to $f$ at $(1,2,16)$.
  • Give an equation of the tangent line to the level curve of $f$ at $(1,2)$ (what does $dz$ equal on a level curve).
  • Compute the directional derivative of $f$ at $(1,2)$ in the direction $(3,4)$. [Recall $D_{\vec u}f(P)=\vec\nabla f(P)\cdot \frac{\vec u}{\|\vec u\|}$.]
• Let $f(x,y) = x^2\sin(y)+y^3$.
  • Compute $f_x$ and then compute $\dfrac{\partial}{\partial x}(f_x)$ and $\dfrac{\partial}{\partial y}\left(\dfrac{\partial f}{\partial x}\right)$
  • Compute $f_y$ and then compute $\dfrac{\partial}{\partial x}(f_y)$ and $\dfrac{\partial^2f}{\partial y^2}$
• Find a number $c$ so that the vectors $(1+c,2)$ and $(4,6)$ lie on the same line (are parallel or antiparallel).

Group Problems

1. Let $f(x,y) = 9-x^2-y^2$ with $\vec r(t) = (2\cos t,3\sin t)$.
   1. State $f(\vec r(t))$ and then compute $\frac{df}{dt}$. (Substitute, then differentiate.)
   2. Now instead directly compute the differential $df$ in terms of $x,y,dx,dy$.
   3. Compute the differentials $dx$ and $dy$ in terms of $t$ and $dt$.
   4. Without using the result of part a., use substitution and your work from parts b. and c. to obtain $df$ in terms of $t$ and $dt$. Then state $df/dt$ (which should match what you got in part a.).
2. Let $f(x,y) = 9-x^2-y^2$.
   1. Give an equation of the tangent plane to $f(x,y) = 9-x^2-y^2$ at the point $(2,-3)$. [Find the differential, and then substitute $dx = x-2$, $dy = y-?$, $dz = ?$.]
   2. Give an equation of the tangent line to the level curve of $f(x,y) = 9-x^2-y^2$ at the point $(2,-3)$. [Your answer will be very similar to the previous. What changes?]
3. Let $g(x,y)=x\cos(xy)$.
   1. Compute $g_x$ and $g_y$, and then state the first derivative $Dg(x,y)$.
   2. Compute the second partials $\dfrac{\partial}{\partial x}\left(\dfrac{\partial f}{\partial x}\right)$, $\dfrac{\partial}{\partial y}\left(\dfrac{\partial f}{\partial x}\right)$, $\dfrac{\partial}{\partial x}\left(\dfrac{\partial f}{\partial y}\right)$, and $\dfrac{\partial}{\partial y}\left(\dfrac{\partial f}{\partial y}\right)$.
   3. State the second derivative $D^2g(x,y)$ (it should be a 2 by 2 matrix).
4. Find the directional derivative of $f(x,y)=xy^2$ at $P=(4,-1)$ in the direction $(-3,4)$. [Check: $D_{(-3,4)}f(4,-1) = \vec\nabla f(4,-1)\cdot \frac{(-3,4)}{5}=(-8,16)\cdot \frac{(-3,4)}{5}=88/5$.]
5. Give an equation of the tangent plane to $xy+z^2=7$ at the point $P=(-3,-2,1)$. [Check: $(-2)(x-(-3))+(-3)(y-(-2))+2(1)(z-1)=0$. ]
6. Give an equation of the tangent plane to $z=f(x,y)=xy^2$ at the point $P=(4,-1,f(4,-1))$. [Check: $z-4 = (-1)^2(x-4)+2(4)(-1)(y-(-1))$.]
7. Consider the function $f(x,y,z) = 3xy+z^2$. We'll be analyzing the surface at the point $P=(1,-3,2)$.
   • If $dx=0.1$, $dy=0.2$ and $dz=0.3$, then what is $df$ at $P$.
   • Find the directional derivative of $f$ at $P$ in the direction $(1,-2,2)$.
   • Give an equation of the tangent plane to the level surface of $f$ that passes through $P$.
   • Give an equation of the tangent plane to the level surface of $f$ that passes through $(a,b,c)$.

Today

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