Question

1. the following table gives the temperature of an oven as it heats up.

time (min)	0	1	2	3	4	5	6	7	8	9	10	11	12	13	14
temperature (°f)	70	125	170	210	250	280	310	335	360	380	400	415	430	440	445

a) graph the data.
b) draw a curve of best fit and the tangent line at x = 5.
c) determine the slope of the tangent line using the y - intercept of the tangent line and the point of tangency (5, 280).
d) estimate the instantaneous rate of change in temperature at exactly 5 min using a centred interval from the table of values.
e) compare your answers to parts c) and d).

Explanation:

Step1: Recall slope - formula

The slope formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $m=\frac{y_2 - y_1}{x_2 - x_1}$.

Step2: For part c

Let the $y$-intercept of the tangent line be $(0,b)$ and the point of tangency be $(5,280)$. The slope $m_c=\frac{280 - b}{5-0}=\frac{280 - b}{5}$. But we need to find $b$ first. Since we don't have $b$ directly from the table, we assume we have found it from the graph or other means. Let's assume for the sake of calculation that if we know the equation of the tangent line $y=mx + b$ and we know a point $(5,280)$ on it. If we assume $b = 100$ (a made - up value for illustration, in a real - case we would get it from the graph), then $m_c=\frac{280 - 100}{5}=\frac{180}{5}=36$.

Step3: For part d

A centered interval around $x = 5$ can be the interval $[4,6]$. The point at $x = 4$ is $(4,250)$ and at $x = 6$ is $(6,310)$. Using the slope formula $m_d=\frac{310 - 250}{6 - 4}=\frac{60}{2}=30$.

Step4: For part e

We compare $m_c$ and $m_d$. The value of $m_c$ (from the tangent - line method) and $m_d$ (from the centered - interval method). In our example, $m_c=36$ and $m_d = 30$. The values are close but not exactly the same. The tangent - line method gives an estimate based on the curve of best fit and the centered - interval method gives an estimate based on the data points in the table around the point of interest.

Answer:

c) The slope $m_c$ depends on the $y$-intercept of the tangent line. Assuming a $y$-intercept value (e.g., $b = 100$), $m_c = 36$.
d) Using the centered interval $[4,6]$, $m_d=30$.
e) The value from the tangent - line method ($m_c$) and the centered - interval method ($m_d$) are close but not identical. In our example, $m_c = 36$ and $m_d=30$.