# Preclass Video & Textbook Notes

MATH 124

## Navigation

## Chapter 0

*Task*: Find tangent lines.

- Is the point of tangency given?
- If yes, use the slope of the radial line and the point to find the tangent line.
- If no, identify unknowns and solve algebraicly using geometric rules.

**Example Problem**

*Question*: Find the equation for all lines that are tangent to the unit circle and pass through the point \(P(3,4)\).

*Solution*:

- You want to find the points of tangency \((a,b)\) and the slopes of the tangent lines \(m_t\).
- The radial slope is \(\frac{b}{a}\); the tangent slope is thus \(m_t=-\frac{a}{b}\).
- The tangent slope can also be written as \(m_t=\frac{4-b}{3-a}\).
- Use 2 and 3 to create an relationship: \(-\frac{a}{b}=\frac{4-b}{3-a} \rightarrow -a(3-a)=b(4-b) \rightarrow -3a+a^2 = 4b-b^2 \rightarrow -3a+a^2+b^2=4b\).
- The equation of the unit circle is \(a^2 + b^2 = 1\).
- Use 4 and 5: \(-3a+1=4b \rightarrow b = -\frac{3}{4}a+\frac{1}{4}\).
- Plug back into circle equation: \(a^2 + \left(-\frac{3}{4}a+\frac{1}{4}\right)^2 = 1\).
- Expand and simplify: \(a^2 + \frac{9}{16}a^2 - \frac{3}{8}a + \frac{1}{16} = 1 \rightarrow \frac{25}{16}a^2 - \frac{3}{8}a - \frac{15}{16} = 0 \rightarrow 25a^2 - 6a - 15=0\)
- Solve for \(a\) using the quadratic formula, yielding \(\frac{6\pm\sqrt{1536}}{50} \approx 0.9, -0.66.\)
- Find \(b\) from \(-\frac{3}{4}a+\frac{1}{4}=b\) to yield \(b\approx -0.43, 0.75\) (respectively).
- Find \(m_t\) from \(m_t=\frac{a}{b}\approx 2.1, 0.89\) (respectively).
- Put everything together to yield the tangent lines: \(y+0.43=2.1(x-0.9)\) and \(y-0.75=0.89(x+0.66)\).

## Chapter 2

### Section 2.1

#### Video Notes

- Developing a more specific understanding of what ‘touching’ means.
- A tangent line is a point on the curve that almost forms a line when ‘zoomed in’.
- We are aiming for the true tangent, but we can obtain an imperfect secant line by sampling two points close to one another on the line.
- Continually move the two points closer and closer together.

- Limit - moving things closer and closer to a certain target point and measuring the corresponding approaching value.
- Application - position over time, concentration over time, etc.; rate representations. Secant line is average rate of change; the interpretation of the tangent line is the
*instantaneous rate of change*- this is the ‘current speed’.

#### Textbook Notes

*Section 2.1, “The Tangent Problem”*

*Tangent*- derived from Latin*tangens*, meaning “touching”.*Secant*- derived from Latin*secans*, meaning “cutting”.- Limits - the convergence of a function with respect to an independent variable’s approach towards some value.

*Section 2.1, “The Velocity Problem”*

- The velocity of a car as it travels through city traffic is not constant.
- How can we define instantaneous velocity?
- Say that the velocity function at time \(t\) is given by \(v(t)\). The instantaneous velocity can then be obtained by \(\frac{v(t+\Delta) - v(t)}{\Delta}\).

### Section 2.2

#### Video Notes

Question: What happens to \(f(x)\) as \(x\) approaches some value \(a\)? \(\lim_{x\rightarrow a} f(x)\)

**Three methods**:

- Construct a table. This is inexact but it suffices for the time being.
- Algebra (common sense). As \(x \rightarrow 1\), \(x^2 \rightarrow 1^2 = 1\), \(x^2 + 1 \rightarrow 1 + 1 = 2\).
- Graph. Also an inexact method but suffices. Graph the chart and interpolate between values.

*Definition*: The limit of a function \(f(x)\) as \(x\) approaches \(a\) equals \(L\) if the values of \(f(x)\) get aribtrarily close to \(L\) when \(x\) is sufficiently close to \(a\).

**One-sided limits**- \(\lim_{x\rightarrow 1^+} f(x)\) may not always be equal to \(\lim_{x\rightarrow 1^-} f(x)\).**What makes a limit fail to exist?**If \(\lim_{x\to a^+} f(x) \neq \lim_{x\to a^-} f(x)\), the limit does not exist (DNE).**Using common sense**- what happens to the numerator or denominator as \(x\) changes? What is the relationship between the growth/decay patterns?**Unbounded growth**- \(\lim_{x\to a} f(x) = \infty\) or \(\lim_{x\to a} f(x) = -\infty\).

### Section 2.3

#### Textbook Notes

*2.3 - Calculating Limits Using the Limit Laws*

**The Sum Law**: the limit of a sum is the sum of the limits. \(\lim_{x\to a} \left[f(x)+g(x)\right] = \lim_{x\to a} f(x) + \lim_{x\to a} g(x)\).**The Difference Law**: the limit of a difference is the difference of the limits. \(\lim_{x\to a}\left[f(x)-g(x)\right] = \lim_{x\to a}f(x) - \lim_{x\to a}g(x)\).**Constant Multiple Law**: the limit of a constant times a function is the constant times the limit of the function. \(\lim_{x\to a} \left[ cf(x) \right] = c \lim_{x\to a} f(x)\).**Product Law**: the limit of a product is the product of the limits. \(\lim_{x\to a}\left[f(x) g(x)\right] = \lim_{x\to a}f(x) \cdot \lim_{x\to a}g(x)\).**Quotient Law**: the limit of a quotient is the quotient of the limits. \(\lim_{x\to a}\frac{f(x)}{g(x)} = \frac{\lim_{x\to a}f(x)}{\lim_{x\to a}g(x)} \text{ if } \lim_{x\to a}g(x) \neq 0\).**Power Law**: the limit of an exponentiated term is the exponentiated limit of the base. \(\lim_{x\to a}\left[f(x)\right]^n = \left[\lim_{x\to a}f(x)\right]^n\), provided \(n\) is a positive integer. Derived from applying the product law repeatedly.**Constant Law**: the limit of a constant function is the constant. \(\lim_{x\to a}c = c\).**Linear Law**: the limit of the function \(y=x\) is \(a\). \(\lim_{x\to a} x = a\).**Exponential Law**: the limit of an exponential function \(y=x^n\) is \(a^n\). \(\lim_{x\to a} x^n = a^n\), where \(n\) is a positive integer. Derived from power law and linear law.**Specific Root Law**: the limit of a root function \(\sqrt[n]{x}\) is \(\sqrt[n]{a}\). \(\lim_{x\to a} \sqrt[n]{x} = \sqrt[n]{a}\), where \(n\) is a positive integer.**General Root Law**: the limit of any function that is \(n\)-rooted is the \(n\)th root of the limit of that function. \(\lim_{x\to a} \sqrt[n]{f(x)} = \sqrt[n]{\lim_{x\to a} f(x)}\), where \(n\) is a positive integer.

#### Video Notes

**Distribution Substitution Property**: if \(f(x)\) is an algebraic function and \(a\) is in the domain of \(f\), then \(\lim_{x\to a} f(x) = f(a)\).- Much of solving for limits is changing the expression of a limit such that the DSP can be used.

- If \(\lim_{x\to a}\frac{P(x)}{Q(x)}\) is a rational function with \(P(a)=Q(a)=0\), then \((x-a)\) must be a factor of \(P(x)\) and \(Q(x)\). Factor it out and remove it from the rational function.
- Useful trick: rationalization in \(\frac{0}{0}\) cases. Multiply numerator/denominator with square root terms by the conjugate.

**Limit Laws Involving \(\infty\)**

Limit Type | Result |

\(c + \infty\) | \(\infty\) |

\(c - \infty\) | \(-\infty\) |

\(\infty + \infty\) | \(\infty\) |

\(-\infty-\infty\) | \(-\infty\) |

\(\infty-\infty\) | Needs further work |

\(c\cdot \infty\), \(c>0\) | \(\infty\) |

\(c\cdot \infty\), \(c<0\) | \(-\infty\) |

\(0 \cdot \infty\) | Needs further work |

\(\infty \cdot \infty\) | \(\infty\) |

\(\infty \cdot -\infty\) | \(-\infty\) |

\(-\infty\cdot-\infty\) | \(\infty\) |

\(\frac{c}{\pm\infty}\) | \(0\) |

\(\frac{\pm\infty}{\pm\infty}\) | Needs further work |

\(\frac{0}{0}\) | Needs further work |

\(\frac{c}{0}\) | Needs further work |

### Section 2.5

#### Video Notes

- A function is continuous at \(x=a\) if the graph does not break at \(x=a\).
- More formally, a function \(f(x)\) is continuous if \(f(a)\) exists and \(\lim_{x\to a} f(x) = f(a)\).
- Limit properties applied to continuity:
- The sum, difference, and product of continous functions is continuous.
- The quotient of continuous functions is continuous as long as the denominator is not 0.
- The composition of continuous functions is continuous.

### Section 2.6

#### Video Notes

- Looking at horizontal asymptotes - what happens when \(x\to\infty\) in \(f(x) = \frac{1}{x}\), for instance?
- Approach methods:
- Plug in large values and see what the function approaches.
- Graphing; analyze where graph ends approach.

- Not all functions have a horizontal asymptote. \(\lim_{x\to\infty} 3x+7 = \infty\), for instance.
- Horizontal asymptote of \(y = \arctan(x)\) - \(\lim_{x\to\infty} = \frac{\pi}{2}\), \(\lim_{x\to-\infty} = -\frac{\pi}{2}\).
- Rules for computing \(x\to\pm \infty\):
- \(\lim_{x\to\infty} \frac{1}{x^r} = 0\) for \(r>0\).
- \(\lim_{x\to\infty} a^x = \infty\) for \(a > 1\); \(0\) for \(a < 1\).
- \(\lim_{x\to-\infty} a^x = 0\) for \(a > 1\); \(\infty\) for \(a < 1\).
- Polynomials: identify the term with the highest power; only that term plays the important role.

#### Textbook Notes

*Examples 4 and 5 in Section 2.6*

Find the horizontal and vertical asymptotes of the graph of the function \(f(x) = \frac{\sqrt{2x^2+1}}{3x-5}\).

Vertical asymptotes can be found when the denominator is equal to 0; \(3x-5=0 \implies x = \frac{5}{3}\).

Horizontal asymptotes can be found via \(\lim_{x\to\infty} \frac{\sqrt{2x^2+1}}{3x-5} \implies \lim_{x\to \infty }\frac{\sqrt{\frac{2x^2+1}{x^2}}}{\frac{3x-5}{x}} \implies \lim_{x\to \infty }\frac{\sqrt{2+\frac{1}{x^2}}}{3-\frac{5}{x}} \implies \frac{\lim _{x\to \infty }\left(\sqrt{2+\frac{1}{x^2}}\right)}{\lim _{x\to \:\infty }\left(3-\frac{5}{x}\right)} \implies \frac{\sqrt{2}}{3}\).

Compute \(\lim_{x\to \inf\left(\sqrt{x^2+1}-x\right)\).

Multiply by conjugate radical: \(\lim_{x\to \infty }\frac{\left(\sqrt{x^2+1}-x\right)\cdot \left(\sqrt{x^2+1}+x\right)}{\sqrt{x^2+1}+x} = \lim_{x\to \infty }\frac{1}{\sqrt{x^2+1}+x} = 0\).

### Section 2.7

#### Video Notes

- When \(x\to a\), the slope of the secant line will approach the slope of the tangent line.

- \(f'(a)\) is the derivative of \(f\) at \(x=a\).

#### Textbook Notes

Examples 2, 3, and 4 in Section 2.7.

*Find an equation of the tangent line to the hyperbola \(y = \frac{3}{x}\) at the point \((3,1)\).*

The slope of the tangent line can be found using the formal definition of a derivative.

\[m = \lim_{h\to 0} \frac{f(3+h) - f(3)}{h} = \lim_{h\to 0} \frac{\frac{3}{3+h} - 1}{h} = \lim_{h\to 0} \frac{-h}{h(3+h)\] \[= \lim_{h\to 0} -\frac{1}{3+h} = -\frac{1}{3}\]*Suppose a ball is dropped from the upper observation deck of the CN Tower 450 meters above the ground. What is the velocity fo the ball after 5 seconds?*

We need to use \(f(t) = 4.9t^2\); we obtain \(f'(x) = \lim_{h\to 0} \frac{4.9\cdot(t+h)^2 - 4.9t^2}{h} = \lim_{h\to 0} 4.9(2t+h) = 9.8t\). The velocity after 5 seconds is \(f'(5) = 49 m/s\).

*Find the derivative of the function \(f(x) = x^2 - 8x + 9\) at the number \(a\).

Simplifying \(f'(a) = \lim_{x\to 0} \lim_{h\to 0}\frac{\left[\left(a+h\right)^2-8\left(a+h\right)+9\right]-\left[a^2-8a+9\right]}{h} = \lim_{h\to 0}\frac{2ah+h^2-8h}{h} = \lim _{h\to 0}\left(2a+h-8\right) = 2a-8\). The derivative of \(f\) at \(a\) is the slope of the tangent line. If \(f'(a)\) exists, we say that \(f\) is differentiable at \((a, f(a))\).

### Section 2.8

#### Video Notes

- \(a\) is now varying - it is variable, not a point. We can rewrite the definition of a limit as

- The derivative of a line is a constant function - its slope.
- When drawing the derivative grpah, look for (local) minima and maxima.
- Engage in analysis of positive and negative derivatives.
- Note when the slope gets smaller and when it gets larger.
Differentiability: the limit exists for \(f(x)=x^2\) but not for $$f(x) = x \(at\)x=0$$; thus the absolute value function is not differentiable at that point. - If \(f\) is differentiable at \(a\), then \(f\) is continuous at \(a\). If \(f\) is not continuous at \(a\), then \(f\) is not differentiable at \(a\).

## Chapter 3

### Section 3.1

#### Textbook Notes

\[\frac{d}{dx} (c) = 0\] \[\frac{d}{dx} x^n = nx^{n-1}\]#### Video Notes

- We will work on formulas in which taking the limit is too tedious and not necessary anymore.
- Important differentiation rules:

### Section 3.2

#### Video Notes

- Stick to the rules that have been learned so far.
- Constant rule, power rule, function sum rule.
- Entering areas in which our intuition is wrong.

- Product and quotient rules.

#### Textbook Notes

Find equations of the tangent line and normal line to the curve \(y=x\sqrt{x}\) at the point \((1,1)\).

By the power rule, we have that \(f'(x) = \frac{3}{2}x^{\frac{1}{2}} = \frac{3}{2}\sqrt{x}\). The slope of the tangent line at \((1,1)\) is thus \(f'(1) = \frac{3}{2}\), yielding the equation \(y = \frac{3}{2}x - \frac{1}{2}\). The equation of the normal line uses the negative reciprocal slope: \(y = -\frac{2}{3}x + \frac{5}{3}\).

Product Rule: \(\left(f(x)g(x)\right)'; h'(x) = f(x)g'(x) + f'(x)g(x)\)

Quotient Rule: \(\left(\frac{f(x)}{g(x)}\right) = \frac{g(x)f'(x) - f(x)g'(x)}{g(x)^2}\)

### Section 3.3

#### Video Notes

- Derivative of trigonometric functions - \(\sin x\) and \(\cos x\).
- \(\lim_{h\to 0} \frac{\sin(h)}{h} = 1\); \(\lim_{h\to 0}{\sin(\text{BLAH!})}{\text{BLAH!}} = 1\).
- \[\frac{d}{dx} \sin(x) = \cos(x)\]
- \[\frac{d}{dx} \cos(x) = -\sin(x)\]

### Section 3.4

- So far we know about a few derivatives. We can perform basic operations with these functions.
- We always use ‘pure \(x\)’ - for instance, we can find \(2\cos(x)\) but not \(\cos(2x)\).
- We can use the chain rule to understand the derivative of \(f(g(x))\).

Version | Rule |

Physics | \(y = f(g(x))\), \(g(x) = u\), then \(\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\) |

Algebra | \(\frac{d}{dx} f(g(x)) = f'(g(x))\cdot g'(x)\) |

- Put a ‘blind eye’ on the inputs of the outer function, the “outer derivative”.
- We can generalize this: \(\frac{d}{dx} f(g(h(x))) = f'(g(h(x))) \cdot g'(h(x)) \cdot h'(x)\).
- We can indicate higher-order derivatives as such: the \(n\)th derivative is \(f^{(n)} x\).

\(f\) | \(f'\) | \(f''\) |

increases | \(>0\) | - |

decreases | \(<0\) | - |

concave up | \(f'\) increases | \(>0\) |

concave down | \(f'\) decreases | \(<0\) |

### Section 3.5

#### Logarithmic and Exponential Derivatives

What is \(\frac{d}{dx} a^x\) and \(\lim_{h\to 0}\frac{a^h - 1}{h}\)?

\[\frac{d}{dx} a^x = \frac{d}{dx} e^{x\ln(a)} = e^{x\ln(a)} \cdot \ln(a) = e^{\ln(a^x)} \cdot \ln(a) = a^x \cdot \ln(a)\] \[\implies \frac{d}{dx} a^x = a^x \cdot \ln(a)\] \[\lim_{h\to 0}\frac{a^h - 1}{h} = \ln(a)\]#### Implicit Differentiation

Given a curve in the \(xy\)-coordinate system in which each \(x\) gives rise to a \(y\), but cannot be expressed as an explicit function \(y=...\).

Example: find \(y'\) in \(x^3 + xy + y^3 = 1\).

Even if we don’t have an explicit formula, we treat each \(y\) as a function of \(x\): \(x^3 + xy(x) + y(x)^3 = 1\). Take the derivative: \(3x^2 + (1\cdot y(x)) + x\cdot y'(x)) + 3y(x)^2\cdot y'(x) = 0\). Solve for \(y'(x)\) by simplifying, factoring out \(y'(x)\), and re-arranging.

Steps for implicit differentiation:

- Differentiate
- Plug in points, if possible/given.

#### Inverse Trigonometric Functions

Goal - derive \(\frac{d}{dx} \arcsin x = ?, \frac{d}{dx} \arccos x = ?, \frac{d}{dx} \arctan x = ?\).

Find \(y'\) if \(y = \arcsin(x)\). Apply \(\sin\) to both sides: \(\sin y = \sin(\arccos x) \implies \sin y = x \implies \cos(y) \cdot y' = 1 \implies y' = \frac{1}{\cos y} = \frac{1}{\cos(\arcsin(x))}\). This result can be simplified as \(y' = \frac{1}{\sqrt{1 - x^2}\).

The remaining derivatives can be derived:

\[\frac{d}{dx} \arccos x = -\frac{1}{\sqrt{1-x^2}}; \frac{d}{dx} \arctan x = \frac{1}{1+x^2}\]### Section 3.6

What is the derivative of \(y = \ln x\)? \(e^y = e^{\ln x} \implies e^y = x\); take the implicit derivative, we obtain \(e^y \cdot y' = 1 \implies y' = \frac{1}{e^y} \implies y' = \frac{1}{x}\).

\[(\log_a x)' = \frac{1}{x\cdot \ln a}\]Logarithmic differentiation: sometimes, it is worth thinking about a new way to differentiate. Call the function \(y\); take the natural log of both sides, differentiate both. Use log rules to simplify complex problems (quotients, products, etc.).

When \(x\) is in the base, use power rule. When \(x\) is in the exponent, rewrite using natural logarithm nd \(e\) method. What is \(x\) is both in the base *and* in the exponent? Use logarithmic differentiation.

### Section 3.9

- Often, one quantity is related to another.
- We are given that one change results in the change of another variable.

Scheme

- Make a sketch
- Identify rate of changes you know, and the desired rate of change in terms of derivatives
- Find functions to relate the quantities
- Implicitly differentiate w.r.t. \(t\)
- Solve for the derivative you need
- Plug in all known values

### Section 3.10

- Linearization/linear approximation.
- Say you have a very complex function you want to evaluate. Find the line tangent to a convenient value.
- When we are very close to a tangent point, we can approximate the functional value by going on the tangent value rather than the original value.
- Given a function \(f\), is there an easy way to find a functional value \(f(a)\)?
*Easy*- using arithmetic operations.*Hard*- ln, sin, cos, tan, square roots, exponentials.

Given \(f\) and the tangent line \(L(x)\) to the curve at \(a\), if \(x \approx a\) then \(f(x) \approx L(x)\). The tangent line is a linear approximation, linearization, or tangent line approximation.

- Is an estimate an overestimate or an underestimate? Look at the concavity of the curve by taking the second derivative. If the curve is concave down, it is an overestimate; if the curve is concave up, it is an underestimate.

## Chapter 4

### Section 4.1

- We can analyze a function that allows us to sketch a pretty accurate graph using derivatives.
- Local extrema: a point that is higher or lower than the points around it. The derivative is zero or does not exist.

Fermat’s Theorem: If \(f\) has a local maximum or a local minimum at \(c\) and \(f'(c)\) exists, then \(f'(c)=0\).

- Candidates for local extrema are at places where \(f'=0\) or \(f'=DNE\). Thes are critical numbers.
- Not all critical numbers are local extrema. For instance, \(x^3\).
- To find critical numbers, factor negative powers, denominators, and common factors. Then, clean up.
- Absolute max/min value - highest/lowest value a function takes on.
- In a limited bound problem, always compare the zero-derivative solutions to the ends.

### Section 4.3

- If \(a\) is in the domain of \(f\) and \(f'(a) = 0\) or \(f'\) is not defined at \(a\), then \(a\) is a critical number and a candidate for local minima or extrema.
- Change in sign of the derivative indicates local extrema if \(f\) is continuous at the critical number.
- The second derivative tells us about the shape of the curve.
- \(f''(a) < 0\): concave down, \(f''(a) > 0\): concave up. \(a\) is a critical number candidate.
- If the second derivative is \(f''(a) = 0\), the critical point is not a local extrema.
- Note:
- Only valid if the first and second derivatives exist.
- Requires quite a bit of computation.

### Section 4.4

- Recall limits: if we have a zero-over-zero type, we factor/use other algebraic tools and simplify, then use direct substitution property.
- For limits to infinity, divide by highest power and utilize simplification of terms to zero.
- \(\lim_{x\to 0} \frac{\sin x}{x} = 1\); geometric argument, sandwich theorem.

**L’Hopital’s rule takes care of all these situations:** \(\frac{0}{0} and \frac{\pm\infty}{\pm\infty}\). Does not take care of \([\infty - \infty]\), etc. cases.

Suppose \(f\) and \(g\) are differentiable and that \(g'(x) \neq 0\) near \(a\), except possibly at \(a\):

\[\lim_{x\to a} \frac{f(x)}{g(x) = \lim_{x\to a} \frac{f'(x)}{g'(x)}\]The type often changes when taking the derivative. Also works with limits \(x\to a\).

\(f(x)\) and \(g(x)\) look very similar to their linearization when zoomed in to the scale of the limit. The approximation becomes exact.

Tricks to transform: purposefully create a fraction by dividing by a reciprocal to use L’Hopital. Bring powers down by taking the logarithm and later adjust by raising the answer to a power.

### Section 4.5

With curve function, determine the following:

- Domain of \(f\)
- Compute \(y\) and \(x\) intercept
- Horizontal asymptotes (limits \(x\) to \(\pm\infty\))
- Find \(f'\) and critical numbers, where \(f\) is increasing/decreasing, and local maxima/minima
- Critical numbers must be in the domain of the function.

- Find \(f''\) and where \(f''(x) = 0\), concavity, inflection points.
- Symmetry - \(f(-x) = f(x)\), \(f(-x) = -f(x)\)

### Section 4.7

- Optimization - something is being maximized or minimized.
- We need to come up with formulas and solve in a targeted method.
- Sketch the situation and perform a close reading.
- Find the variable that must be maximized or minimized, and find the formula for it.
- What is the constraint? (a given/set constant). Find an equation and use it such that there is only one variable left in the original function.
- Find critical numbers of the function.
- Compare end points and critical numbers.

## Chapter 10

### Section 10.1

\(t\) -t time. \(x,y\) - position. New interpretations can be developed later.

*Uniform linear motion*- constant speed and on a straight line. \(x=at+b, y=ct+d\); \(a, b, c, d\) are constants.- To find the path in the \(xy\) coordinate system, cancel out the parameter \(t\).
- In uniform linear motion, the speed is \(\sqrt{a^2 + c^2}\).

*Uniform circular motion*- modeled as \(x = x_c + r\cos(\theta_0 + \omega t)\), \(y = y_c + r\sin(\theta_0 + \omega t)\); where \((x_c, y_c)\) is the circle center, \(r\) is the circle radius, \(\nu = \frac{\Delta \text{distance}}{\Delta \text{time}}\), \(\omega = \frac{\Delta \theta}{\Delta t}\) (radians/time); \(\theta_0\) is initial angle at \(t=0\).- Important fact: \(\nu = \omega r\), \(\cos^2(a) + \sin^2(a) = 1\)
- Solve for \(\cos(blah)\) and \(\sin(blah)\) to utilize Pythagorean identity.

### Section 10.2

For **Uniform Linear Motion** We have \(x=at+b, y=ct+d\).

Horizontal velocity is given by \(x'\), vertical velocity is given by \(y'\). The speed is \(\sqrt{x'^2 + y'^2}\).

Tangent lines to a parametric curve - asking in the \(xy\) system, the \(t\) parameter is eliminated.

Use **Chain Rule** - \(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}\).

*Second derivative* for parametric equations: \(\frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d^2 y}{(dx)^2}\). Consider \(\frac{dy}{dx}\) to be your new \(y\); \(\tilde{y}\) - you thus need to find \(\frac{d\tilde{y}}{dx}\).