|<ings
Apr24-10, 05:10 PM
This is just to see if I remember? Please confirm, correct any errors, and answer the questions (q's in bold)
1. The problem statement, all variables and given/known data
What geometric transformations will "transform"
f(t) = et \stackrel{transformations}{\rightarrow} \frac{mg}{b} * (1 - e-bt/m)?
2. Relevant equations
f(t) = \frac{mg}{b} * (1 - e-bt/m)
= \frac{mg}{b} - \frac{mg}{b})e-bt/m
3. The attempt at a solution
(1) vertical scaling? or horizontal?
f((-b/m)t) -> e-bt/m
let the new f function be equal to g, another function
i.e. g(t) = e-bt/m
(2) reflection across the x-axis, right?
-1 * g(t) = -e-bt/m
let the new g function be equal to h, another function
i.e. h(t) = -e-bt/m
(3) i think this one is the vertical scaling.
\frac{mg}{b} * h(t) = -\frac{mg}{b}e-bt/m
let the new h function be equal to i, another function
i.e. i(t) = -\frac{mg}{b}e-bt/m
(4) vertical translation
i(t) + \frac{mg}{b} = \frac{mg}{b} - \frac{mg}{b}e-bt/m
= \frac{mg}{b} * (1 - e-bt/m)
= f(t)
also, is the reflection across the y-axis done by multiplying by -1 inside of the done like this?: f(x) \stackrel{reflection across y-axis}{\rightarrow} f(-x). TNX
1. The problem statement, all variables and given/known data
What geometric transformations will "transform"
f(t) = et \stackrel{transformations}{\rightarrow} \frac{mg}{b} * (1 - e-bt/m)?
2. Relevant equations
f(t) = \frac{mg}{b} * (1 - e-bt/m)
= \frac{mg}{b} - \frac{mg}{b})e-bt/m
3. The attempt at a solution
(1) vertical scaling? or horizontal?
f((-b/m)t) -> e-bt/m
let the new f function be equal to g, another function
i.e. g(t) = e-bt/m
(2) reflection across the x-axis, right?
-1 * g(t) = -e-bt/m
let the new g function be equal to h, another function
i.e. h(t) = -e-bt/m
(3) i think this one is the vertical scaling.
\frac{mg}{b} * h(t) = -\frac{mg}{b}e-bt/m
let the new h function be equal to i, another function
i.e. i(t) = -\frac{mg}{b}e-bt/m
(4) vertical translation
i(t) + \frac{mg}{b} = \frac{mg}{b} - \frac{mg}{b}e-bt/m
= \frac{mg}{b} * (1 - e-bt/m)
= f(t)
also, is the reflection across the y-axis done by multiplying by -1 inside of the done like this?: f(x) \stackrel{reflection across y-axis}{\rightarrow} f(-x). TNX