Tangent lines

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Tangent lines are fundamental for understanding calculus.

Several of the sections in this portion of the course deal with finding derivatives and equations of tangent lines. A few points to remember:

The slope of a tangent line to a curve is always given by $\dd [y]{x}$ .

When we are dealing with parametric representations or polar representations of the curve, we must figure out how to compute $\dd [y]{x}$ using the information we are given!

If we have a point $(x_0, y_0)$ on a curve, the equation of the tangent line (if it exists) at that point in Cartesian coordinates is: $y - y_0 = m_\text {tan}(x - x_0)$ where $m_\text {tan}$ is the slope of the tangent line at that point.

We will find various different ways to find this slope depending on how we describe the curve!

The purpose of this assignment is to review some of the procedures and concepts related to tangent lines from a first course in calculus that will be necessary to answer questions from the upcoming sections.

I understand I do not understand

The slope of the tangent line is given by $m_\text {tan} = \lim _{x\to a} \frac {f(x) - f(a)}{x-a}.$ Given two points, $(a, f(a))$ and $(x,f(x))$ where $x\ne a$ , the slope of the secant line that joins these points is given by: $m_\text {sec} = \frac {f(x) - f(a)}{x-a}$ The limit (assuming it exists) of this quotient as $x$ approaches $\answer {a}$ is the slope of the tangent line.

Given a function $f$ ,

$f'(x)$ is the slope of the tangent line. $f'(x)$ gives a formula for the slope of the tangent line for any value of $x$ in the domain of $f$ . $f'(x)$ gives a formula for the tangent line for any value of $x$ in the domain of $f$ . $f'(x)$ is the instantaneous rate of change.

Let’s see if we can explain why the slope of the tangent line is the instantaneous rate of change:

The average rate of change on an interval $[a,x]$ is given by the of the . The limit of the average rate of change as $\answer {x}$ goes to $\answer {a}$ gives the averageinstantaneous rate of change.

Let $f(3) = 1$ , $f'(3)=-4$ , and $g(x) = f(x)+x^2$ . Find the tangent line to $y= g(x)$ at $x=3$ . $y= \answer {2(x-3)+10}$

Let $f(3) = 1$ , $f'(3)=-4$ , and $g(x) = x\cdot f(x)$ . Find the tangent line to $y= g(x)$ at $x=3$ . $y= \answer {-11(x-3)+3}$

Remember the product rule.

Let $f(3) = 1$ , $f'(3)=-4$ , and $g(x) = f(x)^2$ . Find the tangent line to $y= g(x)$ at $x=3$ . $y= \answer {-8(x-3) + 1}$

Remember the chain rule.