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I've got the ordered pairs $(1491947996, 15.7)$ and $(1491948898, 12.9)$. The X values in each set indicate the seconds from Unix Epoch and the Y values in each set indicate the temperature in Celsius.

I'm trying to determine the X-coordinate where $Y = 15$ (degrees Celsius), which will tell me the precise time in seconds when a temperature dropped below a safe-zone of 15°C.

Naturally, $y = mx + b$, and $x = \frac{y-b}{m}$. I've got the slope, which is $0.003$, using $\frac{y2 - y1}{x2 - x1}$, or $\frac{12.9 - 15.7}{ 1491948898 - 1491947996}$

I guess what I don't know is how to define $y$ or $b$ in this case. I thought $y$ would certainly be $15$, but I guess I'm lost. It's been a long time since my last algebra course.

Dando18
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Jon Mitten
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4 Answers4

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What you want is called the "point-slope form" of a line. This is, for a given slope $m$ and a given point on the line $(x_0,~ y_0)$:

$$y-y_0 = m(x-x_0)$$

law-of-fives
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Algebraically, which means assuming your data points are exact, once you have $m$ you can just plug one of your points into $y=mx+b$ and find $b$. Your slope has the wrong sign and is not exact. Alpha gives about $-0.003104$. Once you have $b$, you are correct and should substitute $y=15$ and solve for $x$.

Alternately, once you have $m$ you can use $y-y_1=m(x-x_1)$. Since your first point is close you can say $-0.7=-0.003104(x-1491947996)$ and find $x=1491948222$

As you say, the huge $x$ numbers are scary. If you believe they were accurately measured times, you don't have to worry about it. It would be better to subtract $1491947996$ or some convenient number from both $x$ values. That just represents changing the zero of time to a convenient time instead of using the Unix zero. The scary part comes because you worry that you need such relative accuracy for the difference of the $x$ values to be significant, but what you really measured was the difference in the $x$ values, then you added a big number on to them.

Ross Millikan
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You probably will find it easiest to use the two point form, since that's your starting point. Once you have the equation you can substitute 15 for y and get the value for X, in seconds since Jan 1, 1970. You really don't care about b, the temperature at the beginning of 1970.

stretch
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![image

Offset the measurements by $$ x_{0} = 1491947995 $$ The transformation is $\tilde{x} = x - x_{0}.$ (As noted by @Ross Millikan.)

Now the data looks like this: $$ p = \left\{ 1, 15.7 \right\}, \qquad q = \left\{ 903, 12.9 \right\} $$

The equation for a line in slope-intercept form: $$ y(x) = mx + b $$

The slope is computed via $$ m = \frac{p_{y} - q_{y}} {p_{x} - q_{x}} $$

The intercept can be computed from either $p$ or $q$: $$ m = p_{y} - m p_{x} $$

To find the solution point, solve for $x$: $$ 15 = m x + b $$

Slope $m$

$$ m = \frac{p_{y} - q_{y}} {p_{x} - q_{x}} \approx -0.00310421286031042 $$

Intercept $b$

$$ m = p_{y} - m p_{x} \approx 15.70310421286031 $$

Solution

$$ \color{red}{s} = \left\{ \frac{15 - b}{m}, 15 \right\} = \color{red}{\left\{ 226.5 , 15 \right\}} $$

dantopa
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