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Lecture

# 3.1 tangent lines.pdf

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University of Toronto St. George

Mathematics

MAT237Y1

Dan Dolderman

Fall

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Calc 1
Mon 9/15/08
3.1 - Tangent Lines, Velocity, and General Rates of Change
So, recall back when we ▯rst started limits, I introduced to you the idea of a tangent line.
Basically, you take a look at the secant lines, and take the limit as the second point approaches
the ▯rst. In other words, the slope of the tangent line is given by
lim f(x) ▯ f(x 0
x!x 0 x ▯ x
0
where (x0;f(x 0) is the point at which we want to know the slope of the tangent line.
Well, sometimes it’s a bit easier to take a limit to zero, than to have x ! x . So, another way of
0
thinking about this same tangent line is by rather than having x !0x , we’re going to have the
di▯erence between x and x go to zero. Let h = x ▯ x . Then the slope of the tangent line is
0 0
f(x 0 h) ▯ f(x 0
lim
h!0 h
Example: ▯nd the equation of the tangent line of f(x) = 3x at x 0 2.
Solution: To ▯nd the equation of the tangent line, we need two things: the slope of the tangent
like, and the point. Well, the point is not too bad, that’s just (2;f(2)) = (2;12). Now we need the
slope:
f(2 + h) ▯ f(2) 3(2 + h) ▯ 3(2)2
lim = lim
h!0 h h!0 h
3(4 + 4h + h ) ▯ 12
= lim
h!0 h
12 + 12h + 3h ▯ 12
= lim
h!0 h
2
12h + 3h
= h!0 h
h(12 + 3h)
= lim
h!0 h
= lim 12 + 3h
h!0
= 12
Therefore, the equation of the tangent line is
y ▯ 12 = 12(x ▯ 2)
I can’t even begin to tell you how many times you are going to see the question, \▯nd the
equation of the tangent line," so you better get used to it!
1 1
Example: Find the equation of the tangent line of f(x) = x at the point (0 x ).
0
Solution: We’re going to keep x 0rbitrary this time and see what happens.
1 1 1 x0 ▯ x0+h
x0+h▯ x0 x0(0 +h) x0(0 +h)
h!0m h = h!0m h
▯h
= lim
h!0 x 0(x 0 h)
▯1
= lim
h!0 x 0x 0 h)
▯1
=
(x0)
So, our equation is
y ▯ 1 = ▯1 (x ▯ x )
x0 (x0)2 0
and if you really want to you can plug in an actual point 0or x . But we just basically found the
equation of the tangent line for all points!
Displacement: This is exactly what you think it is from physics. It’s the change in position over
time, or, in other words, it’s the di▯erence between a particle’s ▯nal and initial position.
So, if the function of displacement is f(t), then the displacement over an interval [t ;t + h] where
0 0
t0is the initial time (usually it’s zero)0is f(t + h0 ▯ f(t ).
Because displacement is the ▯nal position minus the start, you can have negtive displacent! It all
has to do with directions, it’s not saying that one is actually travelling backwards or something.
It’s just saying one is going, say, west.
So, that being said,
Average Velocity: how fast a particle is moving over a time interval [t ;t + h], with position
0 0
function f(t), given by the equation
displacement f(x + h) ▯ f(x )
v = = 0 0
time elapsed h
Instantaneous Velocity: how fast a particle is moving at a particular instant of time. Basically,
it’s the limit of the average velocities as h ! 0. Sound familiar? Hmm...this seems to be just like
the slope of the tangent line to a curve. Only our curve in this case is the position function!
f(x0+ h) ▯ f(x0)
vinst lh!0 h
Examplep Find the average velocity over [1;4] and the instantaneou

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