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# Chapter 3: The Derivative ## 3.1 Tangent Lines and Rates of Change ### The Tangent Problem **Example 1** Find an equation of the tangent line to the parabola $y = x^2$ at the point $P(1, 1)$. **Solution** We will compute the slope of the secant line through $P(1, 1)$ and $Q(x, x^2)$ $m_{PQ} =...

# Chapter 3: The Derivative ## 3.1 Tangent Lines and Rates of Change ### The Tangent Problem **Example 1** Find an equation of the tangent line to the parabola $y = x^2$ at the point $P(1, 1)$. **Solution** We will compute the slope of the secant line through $P(1, 1)$ and $Q(x, x^2)$ $m_{PQ} = \frac{x^2 - 1}{x - 1}$ **Method 1**: Choose values of $x$ close to 1 | x | 0.5 | 0.9 | 0.99 | 0.999 | 1.5 | 1.1 | 1.01 | 1.001 | | :------ | :---- | :---- | :---- | :---- | :---- | :---- | :---- | :---- | | $m_{PQ}$ | 1.5 | 1.9 | 1.99 | 1.999 | 2.5 | 2.1 | 2.01 | 2.001 | The slope appears to be approaching 2. **Method 2**: Simplify the expression for $m_{PQ}$ $m_{PQ} = \frac{x^2 - 1}{x - 1} = \frac{(x - 1)(x + 1)}{x - 1} = x + 1$ Taking the limit as $x$ approaches 1, we see that: $\lim_{x \to 1} m_{PQ} = \lim_{x \to 1} (x + 1) = 1 + 1 = 2$ Thus, the slope of the tangent line is $m = 2$. The equation of the tangent line is: $y - 1 = 2(x - 1)$ $y = 2x - 1$ **Definition** The tangent line to the curve $y = f(x)$ at the point $P(a, f(a))$ is the line through $P$ with slope $m = \lim_{x \to a} \frac{f(x) - f(a)}{x - a}$ provided that this limit exists. If we let $h = x - a$, then $x = a + h$, so the slope of the tangent line can also be written as: $m = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h}$ **Example 2** Find an equation of the tangent line to the hyperbola $y = \frac{3}{x}$ at the point $(3, 1)$. **Solution** $m = \lim_{x \to 3} \frac{f(x) - f(3)}{x - 3} = \lim_{x \to 3} \frac{\frac{3}{x} - 1}{x - 3}$ $= \lim_{x \to 3} \frac{\frac{3 - x}{x}}{x - 3} = \lim_{x \to 3} \frac{-(x - 3)}{x(x - 3)}$ $= \lim_{x \to 3} (-\frac{1}{x}) = -\frac{1}{3}$ Thus, an equation of the tangent line is $y - 1 = -\frac{1}{3}(x - 3)$. $y = -\frac{1}{3}x + 2$ ### The Velocity Problem **Example 3** Suppose that a ball is dropped from the top of the CN Tower, 450 m above the ground. Find the velocity of the ball after 5 seconds. **Solution** From physics, we know that the distance fallen after $t$ seconds is $s(t) = 4.9t^2$. We want to find the velocity after 5 seconds, so we consider the average velocity over the interval $5 \le t \le 5 + h$. Average velocity = $\frac{\text{change in position}}{\text{time elapsed}} = \frac{s(5 + h) - s(5)}{h}$ $= \frac{4.9(5 + h)^2 - 4.9(5)^2}{h} = \frac{4.9(25 + 10h + h^2) - 4.9(25)}{h}$ $= \frac{4.9(10h + h^2)}{h} = 4.9(10 + h)$ Now, we take the limit as $h$ approaches 0 to find the instantaneous velocity: $v(5) = \lim_{h \to 0} 4.9(10 + h) = 4.9(10) = 49 m/s$ **Definition** The instantaneous velocity of an object with position function $s(t)$ at time $t = a$ is $v(a) = \lim_{h \to 0} \frac{s(a + h) - s(a)}{h}$