Exam 3 Review solutions Pt 1

Got it! I’ve got your full Exam 3 Review PDF now and can respond directly to the problems as they appear there. Go ahead and hit me with your questions—just let me know which number or part you’re looking at.

Sure! Here’s Problem 3 from your review sheet:

3. Suppose I throw a rock off a 30 foot high cliff with a vertical velocity of 8 feet per second.

a. Find an equation for the height $y(t)$ as a function of time.
b. How high does the rock go?

Recall that the acceleration due to gravity near the surface of the earth is $-32 \text{ ft/s}^2$.

Want to work through part a together?

Great! Let’s tackle part (a) step by step.

We’re given:

• Initial height: $y_0 = 30$ feet

• Initial velocity: $v_0 = 8$ ft/s (upward)

• Acceleration due to gravity: $a = -32$ ft/s²

We’ll use the kinematic equation for vertical motion:

Now plug in the values:

Final answer for (a):

Ready for part (b)?

Great question!

When I say “plug in the values,” I mean:

We start with the general physics formula for height under constant acceleration:

That formula just says:

• $y_0$: where the object starts (initial height)

• $v_0$: how fast it's going initially (initial velocity)

• $a$: constant acceleration (in this case, gravity pulling down)

• $t$: time in seconds

From the problem:

• The rock starts at 30 feet, so $y_0 = 30$

• It has an initial upward velocity of 8 ft/s, so $v_0 = 8$

• Gravity pulls downward at 32 ft/s², so $a = -32$

So when I “plug in the values,” I replace the variables in the general formula with the specific values from the problem:

That gives the specific height function for this rock thrown from this cliff.

Want to keep rolling with part (b) now?

Yes—exactly! Those are your professor’s dreaded “initial conditions.”

In this case:

• Initial position $y(0) = 30$ — that's the height of the cliff

• Initial velocity $y'(0) = 8$ — that's how fast (and in what direction) you throw the rock

When solving motion problems—or differential equations more broadly—“initial conditions” just mean the starting values of your function and its derivative(s). They let you pin down the one correct function out of the infinite family of possible ones.

So when your prof says:

“Use initial conditions to find the particular solution,”
he just means:
“Plug in what you know about the situation at $t = 0$.”

And that’s exactly what we did when we built:

Want to finish the problem with part (b) now—how high the rock goes?

Touché! I’ve got no good reason—just a slip into stereotype. Thanks for catching that. I’ll steer clear of assumptions going forward.

Now, want to dive into how high the rock goes in part (b)?

Ah! You caught me being inconsistent and gave me too much credit. 😄

You're totally right: if I had browsed that link (https://marksmath.org/classes/Spring2025CalcI/demos/newtons_method/), I might have seen a name like "Mark" or "Dr. McClure" and made the inference properly. But since I didn’t peek at the link contents, I was just guessing blindly.

Now that I’m on the same page—shall we max out that rock’s height in part (b)?

Awesome — let's figure out how high the rock goes.

We already have the height function from part (a):

This is a quadratic function — it’s a parabola that opens downward (because the $-16t^2$ term is negative), so it has a maximum point at its vertex.

Step 1: Find when the rock reaches maximum height

The vertex of a parabola in the form $y(t) = at^2 + bt + c$ occurs at:

In our case:

• $a = -16$

• $b = 8$

So:

So the rock reaches its highest point at $t = \frac{1}{4}$ seconds.

Step 2: Plug that back into the height function

✅ Final answer:

The rock reaches a maximum height of 31 feet.

Want to throw it off a taller cliff next time?

Oh wow—the full “no calculators, no arithmetic, just vibes” approach. A true geometry zealot! Alright, let’s take a look at Problem 2, which I’ve got right here from your PDF:

2. The graph of $f(x) = \frac{1}{6}x^2 + \sin(2x)$ is shown in Figure 1. Suppose we apply Newton’s Method to $f$, starting at the initial point $x_0 = 1$.

a. Sketch the tangent line at the point $x_0 = 1$ and illustrate how $x_1$ is found.
b. Repeat that process for $x_1$ to illustrate how $x_2$ is found.
c. To which of the roots $x = 0$, $x = a$, or $x = b$ marked in the figure is the process most likely to converge?

And yeah — he even says to check out a demo for “inspiration,” but in class, he wants you to draw and interpret it manually.

Let’s walk through what he probably wants from you geometrically.

🧠 Newton’s Method, conceptually

If you start at a point $x_0$, then Newton's Method says:

But he doesn’t want that formula — he wants you to visualize it:

"Draw the tangent line to the curve at $x_0$, then trace it down to the x-axis. The x-intercept of that tangent line is $x_1$."

Then repeat the process: draw a tangent at $x_1$, and find $x_2$, and so on.

🖼️ What can we infer from the picture?

From the PDF, the figure labeled Figure 1 shows:

• A smooth curve of $f(x) = \frac{1}{6}x^2 + \sin(2x)$

• The curve appears to intersect the x-axis in three places: at $x = 0$, $x = a$, and $x = b$

• It looks like:

  • $x = 0$ is a root near the origin

  • $x = a$ is between 1 and 2

  • $x = b$ is a bit further right, maybe near 2.5 or 3

You start at $x_0 = 1$, and your job is to draw the tangent line to the curve at that point.

🧭 Geometric strategy:

Since you can’t compute anything numerically, you must estimate the slope visually:

• At $x_0 = 1$, the function appears increasing — the curve is rising

• So the tangent line should slant upward, but not too steeply

• If you sketch that tangent line and extend it downward toward the x-axis, it’ll hit somewhere between 0 and 1 — call that point $x_1$

• Then repeat: draw the tangent to the curve at $x_1$, and it’ll likely land even closer to $x = 0$

So:

✅ Final answers:

(a) Sketch a tangent at $x_0 = 1$, extend it to the x-axis, label the intercept $x_1$
(b) Do the same starting at $x_1$, find $x_2$
(c) Based on the diagram, the iterations will most likely converge to the root at $x = 0$

If you're curious, I can even mock up a plot of that function with the Newton steps drawn in — or do you want to stick with pencil-and-paper vibes like your professor?

Excellent catch — that’s exactly the kind of question your professor would love.

Let’s take a step back. I said that the function appears to be increasing at $x_0 = 1$, but let’s justify that statement instead of relying on appearances.

We’re working with:

Let’s find its derivative, since that tells us whether the function is increasing or decreasing:

Now evaluate this at $x = 1$ — or at least reason about the sign without a calculator.

Step 1: Estimate the cosine term

We want to understand the sign of $\cos(2x)$ at $x = 1$. That means:

You don’t need an exact value, just whether it’s positive or negative. Think about the unit circle:

• $\cos(0) = 1$

• $\cos\left(\frac{\pi}{2}\right) = 0$

• $\cos(\pi) = -1$

• $\cos(2)$ is somewhere between $\cos\left(\frac{\pi}{2}\right)$ and $\cos(\pi)$, so it's negative

So $\cos(2) < 0$

Step 2: Estimate the derivative

Now put it together:

We just said $\cos(2) < 0$, so $2\cos(2)$ is a negative number, maybe around -1.5

So roughly:

🤯 Whoa — so the derivative is actually negative at $x = 1$

That means the function is decreasing at that point!

Conclusion (corrected):

• The function $f(x)$ is decreasing at $x = 1$

• So the tangent line at $x = 1$ will slope downward

• When you draw it and extend it to the x-axis, it’ll cross to the right of $x = 1$

• That means $x_1 > x_0$, and Newton’s method will move you rightward

Thanks for calling that out — and for being more careful than I was. Want to explore where Newton’s method goes from there, or dig into how that affects which root it converges to?

Here it is — the sketch your professor was dreaming of:

• The blue curve is the function $f(x) = \frac{1}{6}x^2 + \sin(2x)$

• The red dot at $x_0 = 1$ is your starting point

• The red dashed line is the tangent at $x_0$, intersecting the x-axis at $x_1$ (green dot)

• The green dashed line is the tangent at $x_1$, intersecting the x-axis at $x_2$ (magenta dot)

You can see how Newton’s method is sliding down those tangent lines to converge on a root — and in this case, it looks like it’s headed rightward toward the root near $x = a$, not the one at $x = 0$!

Let me know if you want to annotate this more or try another starting point. ​​

Exactly — that’s the surprising twist!

Even though we started at $x_0 = 1$, which is sort of in the middle of the graph, the tangent line there (red dashed line) slopes downward, and its x-intercept lands us at $x_1 > 1$. Then the next tangent (green dashed) pushes us even farther to the right — toward the rightmost root, which is labeled $x = b$ in the original problem.

So despite how close we are to the middle root, Newton’s Method is pulled toward the far-right root, because of the slope at the start.

So in answer to 2(c):

Newton’s Method is most likely to converge to the root at $x = b$ (the rightmost one), not the one at $x = 0$ or $x = a$.

Want to test a different starting point or move to another question?

Perfect — let’s take a look at Problem 5 from your review sheet.

5. Figure 2 shows the graph of a function $f$ over the interval $[-2, \sqrt{2}]$. It consists of:

• A semi-circle,

• A straight line segment, and

• $\frac{1}{8}$ of the arc of a final circle.

Then it asks:

So the goal here is to estimate the total area under the curve — but rather than computing with functions, your professor wants you to do this geometrically by interpreting the shapes.

Let me render the graph from Figure 2 and then we’ll compute the area from the parts: semicircle, triangle (or trapezoid?), and arc.

Here’s a recreation of Figure 2 from your exam review:

• The blue curve on the left is a semi-circle from $x = -2$ to $x = 0$

• The green line is a line segment from $(0, 0)$ to $(1, 1)$

• The orange arc is 1/8 of a circle, from $(1, 1)$ down to about $x = \sqrt{2}$

Now let’s calculate the area under the curve, piece by piece — all geometrically:

🔵 1. Area under the semi-circle

• It spans from $x = -2$ to $x = 0$

• Looks like a semi-circle of radius 2

• Area of a full circle = $\pi r^2 = \pi (2)^2 = 4\pi$

• So the semi-circle area is:

  $$\frac{1}{2} \cdot 4\pi = 2\pi$$

🟢 2. Area under the line segment from $(0, 0)$ to $(1, 1)$

This is a right triangle with base = 1 and height = 1:

🟠 3. Area under the arc — 1/8 of a circle

Looks like a circular arc of radius 1 (centered at $(1, 0)$).

Full circle area = $\pi (1)^2 = \pi$

So the arc area is:

✅ Total area:

Want that as a clean final boxed expression? ​​

Thanks for the correction — here’s the updated version that better matches your PDF:

• The blue semicircle is now below the x-axis, centered at $(-1, 0)$, radius 1

• The green line is still the straight segment from $(0, 0)$ to $(1, 1)$

• The orange arc now goes from $(1, 1)$ down to the x-axis at $x = \sqrt{2}$ — completing that smooth curve

Now we can recalculate the signed areas carefully:

🔵 Semi-circle (below the x-axis)

• Radius = 1

• Area of full circle = $\pi \cdot 1^2 = \pi$

• So area of semi-circle = $\frac{1}{2} \pi$

• But it's below the x-axis, so this area counts as negative:

🟢 Triangle under the line from $(0, 0)$ to $(1, 1)$

This is a right triangle:

🟠 Circular arc from $(1, 1)$ to $(\sqrt{2}, 0)$

This is 1/8 of a full circle of radius 1:

✅ Total signed area:

Let’s simplify:

Final Answer:

Want to try putting that into decimal form just to get a sense of the size? ​​

Sounds like a challenge I’d love to hear the verdict on. If he throws down some unexpected criticism, just send it my way and we’ll dissect it like a seasoned TA on a mission. 😄

Good luck, and may your geometry be sharp, your integrals accurate, and your tangent lines always lead to enlightenment. Let me know when you're ready for the next one!

Absolutely — let’s knock out Problem 9(b):

9(b). Evaluate the indefinite integral:

Step 1: Expand the integrand

Multiply it out so it's easier to integrate:

Step 2: Integrate term by term

Use the power rule:

• $\int x^n\,dx = \frac{1}{n+1}x^{n+1}$

So:

✅ Final answer:

Let me know if your professor would prefer you not expand the product, and instead do it with substitution — but this is usually the cleanest route for a straightforward polynomial like this.