Question

Learning curves are studied by psychologists interested in the theory of learning. A learning curve is the graph of a function PL(t) that represents the performance level of someone who has trained at a skill for t hours. Thus, the dPL/dt represents the rate at which the PL(t) is taken to be a positive function. If M (a positive constant) is the maximum performance level of which the learner is capable, then which differential equations could be a reasonable model for learning (or more precisely, performance level)?

Answer 1

@ganeshie8 @whpalmer4

Answer 2

@ganeshie8 Sry, i made a mistake but i fixed it

Answer 3

(in the first post i mean)

Answer 4

If M (a positive constant) is the maximum performance level of which the learner is capable

Answer 5

wat does that mean ?

Answer 6

I am not sure. The only other thing it says is "Use your common sense applied to the practical meaning behind each equation to determine which of the following are reasonable."

Answer 7

it means as \(t \to \infty\), \(P_L \to M\)

right ?

Answer 8

the second option doesnt even has a \(M\), so strike the second option off first.

Answer 9

I'd go with first option :

say, \(P_L = y\)

dy/dt = k(M-y)

Integrating gives :

-ln | k(M-y) | = t + C

k(M-y) = e^(-ct)

y = M-e^(-ct-k)

Answer 10

clearly, this function approaches M, as t goes to infinity

Answer 11

I also picked the first option but I just read underneath the problem in my textbook and you can select more than one choice

Answer 12

and also it stays positive in its domain

Answer 13

Oh, then we need to check other two options as well hmm

Answer 14

How would I check?

Answer 15

Maybe the first one is the only correct lol : D

Answer 16

yup !

Answer 17

for third option :
http://www.wolframalpha.com/input/?i=dy%2Fdx+%3D+10*%285-y%29%5E%281%2F2%29

Answer 18

for 4th option :
http://www.wolframalpha.com/input/?i=dy%2Fdx+%3D+10%2F%285-y%29

Answer 19

third option takes negative values also for t >0, so discard

Answer 20

fourth option cannot have infinite domain(t > 0).. so discard it also.

Answer 21

Thnx, you are the best : D Do you have time to help me on another problem?

Answer 22

@ganeshie8

Answer 23

shoot..

Answer 24

1. Using separation of variables technique, solve the following differential equation with initial conditions: 4xsqrt(1-t^2)dx/dt-1=0 and x(0) = -2

a. 2x^2 = arcsint-8
b. 2x^2 = arcsint +8
c. 2x^2 = arccost + 8 - 1/2pi
d. 2x^2 = arccost + 8
e. 2x^2 = -arccos t-8


2.Using separation of variables technique, solve the following differential equation with initial conditions: dy/dx = (yx+5x)/(x^2+1) and y(3) = 5

a. y^2 = Ln(x^2+1)+25 - Ln(10)
b. Ln|y+5| = Ln(x^2 +1)
c. Ln|y+5) = arctan3 + Ln10-arctan3
d. Ln|y+5|=1/2Ln(x^2+1)+1/2Ln(10)
e. y = Ln(x^2+1)+50-Ln(10)

Answer 25

Well, I guess it's two but if you help me on one I can probably apply it to the other

Answer 26

\(\large 4x\sqrt{1-t^2} \frac{dx}{dt}-1=0\) and \(x(0) = -2\)

Answer 27

like that ?

Answer 28

Yes

Answer 29

\(\large 4x\sqrt{1-t^2} \frac{dx}{dt}-1=0\)

\(\large 4x\sqrt{1-t^2} \frac{dx}{dt} = 1\)


\(\large 4x dx = \frac{1}{\sqrt{1-t^2}}dt\)

integrate both sides

\(\large \int 4x dx = \int\frac{1}{\sqrt{1-t^2}}dt\)

Answer 30

evaluate

Answer 31

2x^2 = arcsinx

Answer 32

nope

Answer 33

you meant this :

\(2x^2 = \arcsin(t) + C \)

right ?

Answer 34

Yes

Answer 35

good, solve \(x\) and you're done.

Answer 36

then we plug in -2 and we get 2x^2 = arcsint - 8 ?

Answer 37

oh yeah, we need to solve \(C\)

Answer 38

Okay, it looks like im lost for #2 after all. direct me once again, will you?

Answer 39

\(2x^2 = \arcsin(t) + C\)

since we're given : \(x(0) = -2 \), that means \((0, -2)\) is a point on solution curve :

\(2(-2)^2 = \arcsin(0) + C \implies C= 8\)

Answer 40

so its a positive eight? :O

Answer 41

so the solution is :

\(2x^2 = \arcsin(t) \color{red}{+}8\)

right ?

Answer 42

yup !

Answer 43

dy/dx = (yx+5x)/(x^2+1)

Answer 44

factor out "x" in the numerator,
and separate the variables

Answer 45

x(y+5)

Answer 46

\(\large \frac{dy}{dx} = \frac{yx+5x}{x^2+1} \)


\(\large \frac{dy}{dx} = \frac{x(y+5)}{x^2+1} \)


\(\large \frac{1}{y+5}dy = \frac{x}{x^2+1}dx \)

Integrate both sides

\(\large \int \frac{1}{y+5}dy = \int\frac{x}{x^2+1}dx \)

Answer 47

evaluate

Answer 48

right hand side, u may try u-sub : u = x^2+1

Answer 49

Ln(y+5) = 1/2Ln(x^2+1)

Answer 50

\(\ln |y+5| = \frac{1}{2}\ln | x^2+1| + C\)

Answer 51

And then we do the y(3) = 5 part, do we do this to the numerator?

Answer 52

I got D!

Answer 53

d. Ln|y+5|=1/2Ln(x^2+1)+1/2Ln(10)

Answer 54

Correct !! good job :)

Answer 55

Awesommeeee. I'd medal you thrice if I could lol.

Answer 56

If you don't mind me asking what grade/age are you? ,-, lol

Answer 57

you're awesome :) im done wid studies... working lol