Presentations
4.1 - 4.8Pacing Tracker
- The quizzes have included questions for 16 objectives. How many have you passed? What are you plans to master those that you haven't mastered yet?
- We've finished units 1 through 3. Have you started your self-directed learning project for each unit?
- Remember you can submit only one SDL project per week. Plan ahead and don't let yourself get behind.
Brain Gains
- A rover moves in a straight line to a new spot that is 3 m north and 4 m west of its current location. The rover's height drops 2 m along the way.
- What is the average slope of the hill in the direction the rover moved?
- What is the average slope of the hill?
Solution
The displacement of the rover has a magnitude of $\sqrt{\Delta x^2+\Delta y^2} = \sqrt{3^2+(-4)^2} = 5$ meters in the $xy$-plane. This means the average slope is $$m = \frac{\text{rise}}{\text{run}} = \frac{-2}{5}.$$ Note that the slope is unitless, as the meters cancel.
The second question does not make sense. Without a direction specified, this question is too ambiguous.
- Let $z= x^3+4xy$. We can think of this as an elevation function that tells us the height $z$ of a hill at a point $(x,y)$. Compute $dz$ in terms of $x$, $y$, $dx$, and $dy$.
Solution
We have $$dz = 3x^2 dx + (4x)(dy)+(4dx)(y).$$ If needed, start with implicit differentiation and then strip off the $dt$'s.
- Write your answer above in the form $dz = (?_1) dx+(?_2) dy$.
Solution
By collecting the terms with a $dx$ in them, and then the terms with a $dy$ in them, we obtain $$dz = (3x^2+4y)dx+(4x)dy.$$
- Write your answer above as the dot product of two vectors, namely $dz= (?????)\cdot (dx,dy)$.
Solution
From inspection, we obtain $$dz = (3x^2+4y,4x)\cdot (dx,dy).$$ You can check this is correct by performing the dot product above and comparing it to the previous problem. This problem is essentially pattern recognition, and trying to find two vectors that give the desired dot product.
- What is the gradient of $z$ for the function above, written $\vec \nabla z(x,y)$? We will often just write $\vec\nabla z$, when the input variables $x$ and $y$ are clear from context.
Solution
The gradient of $z$ is the vector $$\vec \nabla z = (3x^2+4y,4x).$$ Just strip the vector of differentials $(dx,dy)$ off of $dz$. This means we always have $$dz = \vec \nabla z\cdot (dx,dy).$$
In first semester calculus, we had $dy = f' dx$. The gradient $\vec \nabla z$ is the generalization of the derivative to higher dimensions. Note that the gradient here has to parts, which we call partial derivatives.
- What is the partial derivative of $z$ with respect to $x$, written $\ds\frac{\partial z}{\partial x}$ or $z_x$ or $D_x z$?
Solution
The partial derivative of $z$ with respect to $x$ is the quantity $$\frac{\partial z}{\partial x} = 3x^2+4y.$$ This is the slope $\frac{dz}{dx}$ if you let $dy=0$ in $dz = (3x^2+4y)dx+(4x)dy.$ It represents the slope of the function if you change $x$ but hold $y$ constant (hence $dy=0$). We use a stylized "d" (written $\partial$ in $\frac{\partial z}{\partial x}$) to help us remember that just $x$ is changing. The subscript notation can be used (so $z_x$) when no ambiguity arises from other subscripts.
- What is the partial derivative of $z$ with respect to $y$, written $\ds\frac{\partial z}{\partial y}$ or $z_y$ or $D_y z$?
Solution
The partial derivative of $z$ with respect to $y$ is the quantity $$\frac{\partial z}{\partial x} = 4x.$$ This is the slope $\frac{dz}{dx}$ if you let $dy=0$ in $dz = (3x^2+4y)dx+(4x)dy.$ It represents the slope of the function if you change $y$ but hold $x$ constant (hence $dx=0$). We use a stylized "d" (written $\partial$ in $\frac{\partial z}{\partial y}$) to help us remember that just $y$ is changing. The subscript notation can be used (so $z_y$) when no ambiguity arises from other subscripts.
Group problems
- A rover is located on a hill whose elevation is given by $z=f(x,y) = 3x^2+2xy+4y^2$.
- Compute the differential $dz$ in terms of $x$, $y$, $dx$, and $dy$, and write it in the form $dz=(?_1)dx+(?_2)dy.$
- Write your answer above as the dot product of two vectors, i.e., $dz = (??, ??)\cdot(dx, dy).$
- State the gradient of $f$, so $\vec \nabla f$.
- State $\dfrac{\partial f}{\partial x}$ and $f_y$.
- State the differential at the point $P=(1,1)$ (the spot where the rover currently resides) [Check: $dz = 8dx+10dy$].
- What is the slope of the hill at $P=(x,y)=(1,1)$ in the direction $(dx,dy)=(1,0)$?
- What is the slope of the hill at $P=(1,1)$ in the direction $(0,1)$?
- What is the slope of the hill at $P=(1,1)$ in the direction $(3,4)$? [Check: $\frac{64}{5}$. The rise is $dz = 64$, with a run of $5$.]
- The sides of a rectangle $x=3$ ft and $y=2$ ft, with tolerances $dx = .1$ ft and $dy = 0.05$ ft. Use differentials to estimate the tolerance on the area $A=xy$ that results from the given tolerances on $x$ and $y$. So compute $dA$ and then plug in $x$, $y$, $dx$, $dy$, as given. [Check: $dA = 0.35.$]
- Let $g(x,y) =x^2y$.
- Give $g_x$ and $\dfrac{\partial g}{\partial y}$. Then state $\vec \nabla g$.
- Find the directional derivative (slope) of $g$ at $P=(3,1)$ in the direction $(-3,2)$.
- Find the directional derivative of $g$ at $P=(3,1)$ in the direction $(2,-5)$.
- The sides of a box are supposed to be $x=3$ ft by $y=2$ ft by $z=1$ ft, with tolerances $dx = .1$ ft by $dy = 0.05$ ft by $dz=0.02$ ft. Use differentials to estimate the tolerance on surface area $A=2xy+2yz+2xz$ that results from the given tolerances.
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