\[ x^2 + y^2 = 1. \]
The point \(P=(0,1)\) is on this circle. Reason geometrically to
determine the slope of the line tangent to \(x^2 + y^2 = 1\) at \(P\). The slope is \(\answer {0}\).
Two young mathematicians discuss the standard form of a line.
Check out this dialogue between two calculus students (based on a true story):
- Devyn
- Riley, I think we’ve been too explicit with each other. We should try to be more implicit.
- Riley
- I. Um. Don’t really…
- Devyn
- I mean when plotting things!
- Riley
- Okay, but I still have no idea what you are talking about.
- Devyn
- Remember when we first learned the equation of a line, and the
“standard form” was \[ ax+by = c \]or something, which is totally useless for graphing. Also a circle is\[ x^2 + y^2 = r^2 \]or something, and here \(y\) isn’t even a function of \(x\).
- Riley
- Ah, I’m starting to remember. We can write the same thing in two ways.
For example, if you write \[ y = mx + b, \]then \(y\) is explicity a function of \(x\) but if you write\[ ax + by = c, \]then \(y\) is implicitly a function of \(x\).
- Devyn
- What I’m trying to say is that we need to learn how to work with these “implicit” functions.
Consider the unit circle
Consider the unit circle
2025-01-06 19:51:10 \[ x^2 + y^2 = 1. \]
The point \[ P=\left (\frac {\sqrt {2}}{2},\frac {\sqrt {2}}{2}\right ) \]
is on this circle. Reason geometrically to
determine the slope of the line tangent to \(x^2 + y^2 = 1\) at \(P\). The slope is \(\answer {-1}\).