Chapter 0 wrap-up

Readiness review

Closing thought

Every tool here is something you will soon do in three dimensions instead of two. A vector is a length and a direction, added and scaled one component at a time. A parametric curve is a point moving as $t$ changes, and its length comes from adding up tiny straight pieces. Polar coordinates trade left-right and up-down for a distance and an angle. On the integral side, substitution and integration by parts undo the chain rule and the product rule. An integral with a moving upper limit is itself a function. And symmetry can settle some integrals before you compute anything.

So you can already do the flat version of what's coming. Take $\mathbf{v} = \langle 6, 8 \rangle$ . Its length is $\sqrt{36 + 64} = 10$ , so $\langle 0.6, 0.8 \rangle$ is the unit vector pointing the same way. Chapter 1 adds a third coordinate. Points become triples $(x, y, z)$ , and vectors gain a third component along with two new ways to multiply them.

Chapter review

One practice set per topic this chapter taught — same fixed, ungraded ladders as the practice panels inside each section.

Practice: Readiness check: vectors, curves, and polar coordinates (6 problems · doesn’t affect your progress)

What is the magnitude of $\langle -5, 12 \rangle$ ?

Correct! 🎉 Not quite.

Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.

Why

$\sqrt{(-5)^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13$ . Squaring removes the minus sign.

Correct answer

Also accepted:

Why

Which vector is the unit vector pointing in the same direction as $\langle 3, -4 \rangle$ ?

Correct! 🎉 Not quite.

Why

$|\langle 3, -4 \rangle| = 5$ , so divide each component by $5$ . Dividing by $25$ uses the squared length by mistake. Flipping both signs gives the opposite direction. $\langle 1, -1 \rangle$ has length $\sqrt{2}$ , and it points in a different direction.

$\langle \tfrac{3}{5}, -\tfrac{4}{5} \rangle$

The curve $x = \cos t$ , $y = \sin t$ runs for $0 \le t \le \pi$ . What does it trace?

Correct! 🎉 Not quite.

Why

At $t = 0$ the point is $(1, 0)$ , at $t = \frac{\pi}{2}$ it is $(0, 1)$ , and at $t = \pi$ it is $(-1, 0)$ . $y = \sin t \ge 0$ throughout, so this is the upper half, traced counterclockwise. A full circle would need $t$ to run to $2\pi$ .

Make a small table: $t = 0$ , $\frac{\pi}{2}$ , $\pi$ .

The upper half of the unit circle, counterclockwise from $(1, 0)$ to $(-1, 0)$

Which polar pair names the same point as $\left(2, \frac{\pi}{6}\right)$ ?

Correct! 🎉 Not quite.

Why

$\frac{7\pi}{6} = \frac{\pi}{6} + \pi$ points the opposite way, and $r = -2$ walks backwards along it. That lands on the original point. $\left(-2, \frac{\pi}{6}\right)$ and $\left(2, \frac{7\pi}{6}\right)$ each flip only once, so they reach the opposite point. $-\frac{\pi}{6}$ is the reflection across the $x$ -axis.

A negative $r$ means walking along the opposite ray, the one pointing $\pi$ further around.

$\left(-2, \frac{7\pi}{6}\right)$

Convert the rectangular point $(-2, -2)$ to polar form with $r > 0$ and $0 \le \theta < 2\pi$ .

Correct! 🎉 Not quite.

Why

$r = \sqrt{4 + 4} = 2\sqrt{2}$ . Adding the sizes of the coordinates to get $4$ does not give a distance. $\tan \theta = 1$ , and the point is in the third quadrant, so $\theta = \frac{5\pi}{4}$ . $\frac{\pi}{4}$ is the first-quadrant answer that a calculator's arctangent gives. $\frac{3\pi}{4}$ is in the second quadrant.

$\tan \theta = \frac{y}{x} = 1$ has two solutions in $[0, 2\pi)$ . Which quadrant is $(-2, -2)$ in?

$\left(2\sqrt{2}, \frac{5\pi}{4}\right)$

Which integral gives the length of the curve $y = x^2$ from $x = 0$ to $x = 1$ ?

Correct! 🎉 Not quite.

Why

$f'(x) = 2x$ , so $[f'(x)]^2 = 4x^2$ and the length is $\int_0^1 \sqrt{1 + 4x^2}\,dx$ . Squaring $f$ instead of $f'$ gives $x^4$ , dropping the square root leaves no length formula at all, and $1 + 2x$ forgets to square the derivative.

Arc length is $\int_a^b \sqrt{1 + [f'(x)]^2}\,dx$ . Find $f'(x)$ first, then square it.

$\int_0^1 \sqrt{1 + 4x^2}\,dx$

Practice: Readiness check: integration techniques and symmetry (4 problems · doesn’t affect your progress)

To evaluate $\int 2x \cos(x^2)\,dx$ by substitution, what should $u$ be?

Correct! 🎉 Not quite.

Why

Choose the inner piece whose derivative sits elsewhere in the integral. With $u = x^2$ , $du = 2x\,dx$ , and the integral becomes $\int \cos u\,du = \sin(x^2) + C$ .

$u = x^2$

Evaluate $\int_0^1 4x^3 (x^4 + 1)\,dx$ .

Correct! 🎉 Not quite.

Why

With $u = x^4 + 1$ , $du = 4x^3\,dx$ , and the integral becomes $\int_1^2 u\,du = \frac{u^2}{2}\Big|_1^2 = 2 - \frac{1}{2} = 1.5$ .

Correct answer

Also accepted:

Why

Let $u = x^4 + 1$ . Change the limits too: $x = 0$ gives $u = 1$ , and $x = 1$ gives $u = 2$ .

1.5

Use integration by parts to find $\int x e^x\,dx$ .

Correct! 🎉 Not quite.

Why

With $u = x$ , $dv = e^x\,dx$ : $du = dx$ , $v = e^x$ . So $\int x e^x\,dx = x e^x - \int e^x\,dx = (x - 1)e^x + C$ . Check: the derivative of $(x - 1)e^x$ is $e^x + (x - 1)e^x = x e^x$ .

Take $u = x$ (it gets simpler when differentiated) and $dv = e^x\,dx$ .

$(x - 1)e^x + C$

Evaluate $\int_{-2}^{2} (x^3 - 5x + 4)\,dx$ .

Correct! 🎉 Not quite.

Why

$x^3 - 5x$ is odd, so it contributes $0$ over the symmetric interval. The constant $4$ is even, so it contributes $2\int_0^2 4\,dx = 16$ .

Correct answer

Also accepted:

Why

Split the integrand into an odd part and an even part. The odd part integrates to $0$ over $[-2, 2]$ .

Practice: Readiness check: accumulation functions and the chain rule (4 problems · doesn’t affect your progress)

Let $A(x) = \int_2^x \sqrt{1 + t^3}\,dt$ . What is $A'(x)$ ?

Correct! 🎉 Not quite.

Why

FTC Part 1 says the derivative of an accumulation function is the integrand, evaluated at the upper limit. The lower limit $2$ doesn't matter. Differentiating the integrand itself is a different operation, and $A$ is a function of $x$ , not a fixed number.

$\sqrt{1 + x^3}$

Let $A(x) = \int_1^x 2t\,dt$ . What is $A(3)$ ?

Correct! 🎉 Not quite.

Why

$A(3) = \int_1^3 2t\,dt = t^2 \Big|_1^3 = 9 - 1 = 8$ . The running total starts at $x = 1$ , so $A(1) = 0$ .

Correct answer

Also accepted:

Why

Let $y = (3x + 1)^4$ . What is $\frac{dy}{dx}$ at $x = 0$ ?

Correct! 🎉 Not quite.

Why

$\frac{dy}{dx} = 4(3x + 1)^3 \cdot 3 = 12(3x + 1)^3$ . At $x = 0$ that is $12$ . Forgetting the inner factor $3$ gives $4$ .

Correct answer

Also accepted:

Why

Outer derivative times inner derivative: $4(3x + 1)^3 \cdot 3$ .

Let $G(x) = \int_0^{x^2} \cos t\,dt$ . What is $G'(x)$ ?

Correct! 🎉 Not quite.

Why

$A'(u) = \cos u$ by FTC Part 1, and the chain rule multiplies by the derivative of the inner function $x^2$ . So $G'(x) = \cos(x^2) \cdot 2x$ . Stopping at $\cos(x^2)$ drops that factor.

$G(x) = A(x^2)$ , where $A(u) = \int_0^u \cos t\,dt$ . Use the chain rule.

$2x \cos(x^2)$

Take the chapter quiz On to Chapter 1 →

← Symmetry and even/odd shortcuts Vectors and the geometry of space →