velvet kettle
I would be able to find the P'(2) and Q'(7) if F(x) and G(x) were given in their function form rather than graph, but I don't know how to translate the graph into function form. Could someone walk me through how to translate the graph into it's function
What does the step up for the rate problem look like awell ?
idle slateBOT
static valve
use the product rule (and quotient rule)
and sub in the values of F' G' F and G at x=2 and x=7
using the graphs
velvet kettle
Thanks for the fast reply! I'm assuming I can get the value of F and G by just find the y-value when x = 2 or 7, but how do I find F' or G' ?
static valve
the points are well chosen
look at F'(2) is obv 0
its flat
G is a straight line ar x=2
so you can calculate the slope easily
same applies for x=7
static valve
its good then
static valve
dont be fooled by physics
they just want you to differentiate R with respect to time
using quotient rule
you can differentiate both sides of the equation
then solve for dR/dt
and sub in all the values
velvet kettle
I have no idea
velvet kettle
I have no idea
what website or book is this from?
Its an activity that my teacher gave us
Dont know where he got it
idle slateBOT
I would be able to find the P'(2) and Q'(7) if F(x) and G(x) were given in their...
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ebon smelt
\begin{align*}
\frac1R&=\frac1{R_1}+\frac1{R_2}\
R&=\frac1{\frac1{80}+\frac1{100}}=\frac1{1/80+1/100}=\frac{400}9\
-\frac1{R^2}\dv{R}{t}&=-\frac1{R_1^2}\dv{R_1}{t}-\frac1{R_2^2}\dv{R_2}{t}\
\dv{R}{t}&=R^2\qty(\frac1{R_1^2}\dv{R_1}{t}+\frac1{R_2^2}\dv{R_2}{t})\
&=\qty(\frac{400}9)^2\qty(\frac1{80^2}\cdot0.3+\frac1{100^2}\cdot0.2)\
&=\boxed{\frac{655}{162}\frac{\Omega}{\text s}}\end{align*}
glad solsticeBOT
mtt07734
ebon smelt
.solved
