Question

A curve passes through the point (0, 5) and has the property that the slope of the curve at every point P is twice the y-coordinate of P. What is the equation of the curve? (Use x as the independent variable.)

Answer 1

f'(x) = 2x i believe

Answer 2

Amistre, after that can you look at my other question here please, http://openstudy.com/#/updates/4eaf2766e4b0cb917620b8a8

Answer 3

and it wants the initial condition of f(0)=5

Answer 4

It's not 2x :(

Answer 5

It just wants the equation of the curve. Does that mean mx+b form? or does that mean the function?

Answer 6

the ycoord of P is simply f(x) with any luck ... so i could be reading it a little off

Answer 7

a line is generally not a curve, but in some text they consider it a "curve"

Answer 8

f'(x) = 2 f(x) by chance?

Answer 9

Well the answer format is

y(x) = [ box ]

Answer 10

y(x) = x^2 + C ; when x=0, y = 5 such that C = 5

y(x) = x^2 + 5; and the slope is y'(x) = 2x

the slope at x=6 is 12 while the ycoord is 36 so yeah, that aint it

Answer 11

im thinking its y= e^2x but i gotta dbl chk my hunch

Answer 12

Not e^2x
:(

Answer 13

y = e^2x + C

y' = 2 e^2x fits

when x=0, y=5 means the C = 5

\[y = e^{2x} + 5\]
is what makes sense to me

Answer 14

^^not that either..

Hmm, this one is really tricky.

Answer 15

it is tricky, but that makes it better than algebra :)

"slope of the curve at every point P is twice the y-coordinate of P"

the derivative at every x is 2f(x) is how i see it

f'(x) = 2 f(x)

f'(x)/f(x) = 2

ln(f(x)) = 2x+C

f(x) = e^(2x+C)

this might be getting close

Answer 16

when x=0; e^C = 5

C = ln(5)

f(x) = e^(2x) * e^(ln(5))

f(x) = e^(2x) * 5

f(x) = 5e^(2x)

that might be best

Answer 17

5e^2x was right.

Thanks, I think I see how you did it. I'll have to come back to it later and study what you did.