For section 003, the average score on midterm 3 was 66.64 with a 65 median. I know that doesn’t sound amazing, but I’m impressed with how well the class did, considering how brutal the test was. Good job! Tests will be passed back on Friday. The ordered scores are as follows:

21, 50, 55 (2), 56, 58, 61 (2), 62, 63, 65 (6), 66, 67, 69, 72, 74 (2), 76, 77, 79 (3), 80 (2), 81 (2)

Again, the scores may not look pretty to you, but I think you guys did a great job.

This post will change slightly over time. Solutions will become available on Monday, April 9 to the reviews I have written.

Practice Integrals (pdf) and Practice Integrals Solutions (pdf)

The following comes in two parts, corresponding to chapter 16 and 17 respectively. I’ll be posting solutions on Monday morning.

midterm 3 review part 1 (pdf) | midterm 3 review part 2 (pdf)

I gave up on the scanner. (It’s 11 years old.) Here are some images from my phone of my solutions:

**Example:** Parameterize the line through $P=(1,0)$ and $Q=(3,2)$ so that $P$ and $Q$ correspond to $t=1$ and $t=3$, respectively.

**Solution 1:** This is more-or-less the approach that we took in class. We look at the vector formed by going from point $P$ to point $Q$, which is $\vec{PQ}=2\vec{i}+2\vec{j}$. We create a line parallel to this emanating from $P$ and we get something that looks like

\[\vec{r}(t)=P+t\vec{PQ}=\langle 1,0\rangle+t\langle 2,2\rangle\]

The only problem with this is that it doesn’t satisfy the condition that $P$ occurs when $t=1$ and $Q$ occurs when $t=3$. We have component functions of the form

\[x=1+2t\quad\text{and}\quad y=0+2t=2t.\]

We can perform our standard algebraic manipulations here, multiplying $t$ by a scalar and adding some scalar to $t$. We know that increasing $t$ by 1 means that both $x$ and $t$ should increase by 1, and so we replace $t$ by $\frac{1}{2}t$ in the above component functions. (Notice that this just makes us traverse along the path from $P$ to $Q$ more slowly.) We now have component functions of the form

\[x=1+t\quad\text{and}\quad y=t\]

If we replace $t$ now by $t-1$, we get something that actually works. (What we’re doing here is just picking the line up and moving it along itself.) We now have

\[x=1+(t-1)=t\quad\text{and}\quad y=t-1.\]

**Solution 2: **Another way of doing this is to start with the displacement vector $\vec{OP}$ (from the origin to $P$), which we find to be $1\vec{i}+0\vec{j}$, and again consider the displacement vector $\vec{PQ}$ as being added on. We want $t=1$ to be the point along the line where we are at $P$, and so we know that we’ll have something along the lines of

\[\vec{r}(t)=\vec{OP}+\frac{t-1}{a}\vec{PQ}\]

since $t=1$ causes this to simply output the vector corresponding to $P$. Now, we need to know how to scale the right side of the equation (i.e., to traverse the line from $P$ to $Q$ at the appropriate pace). If $a=1$, then plugging in $t=3$ gives $\vec{OP}+2\vec{PQ}$, which overshoots $Q$. We find that $a=2$ works. So, we have

\[\vec{r}(t)=\vec{OP}+\frac{t-1}{2}\vec{PQ}=\langle 1,0\rangle+\frac{t-1}{2}\langle 2,2\rangle=t\vec{i}+(t-1)\vec{j},\]

which is exactly what we expected.

Please come to office hours (or at the very least send someone from your group) if you have questions on this assignment.