Find the equation of the curve whose slope at any point (x,y) is √3x+1 and passes through (5,15).

Answer 1

y'= sqr(3x) + 1

let u=3x, du= 3dx, dx = du/3

y' = sqr(u) + 1

dy/dx = sqr(u) + 1

dy = (sqr(u) + 1)dx

dy = (sqr(u) +1)du/3

integrate

y = (1/3)(u^(3/2) +u + C

y = (1/3)(3x)^(3/2) + 3x + C

plug in (5,15) x=5, y=15

15 = (1/3)(3(5))^(3/2) +3(5) + C

simplify and solve for the constant term

45 = 15^(3/2) +15 + C

30 = 15(sqr(15)) + 15 + C

divide by 15

2 = sqr15 +1 + K

K = 1-sqr15

y = (1/3)(3x)^(3/2) +3x + 1-sqr15

UNLESS you really meant

y' = sqr(3x+1)

then

let u= 3x+1, du= 3dx, dx=du/3

y'= sqr(u)

dy = sqr(u)du/3

dy = (1/3)sqr(u)du

integrate

y = (1/3)(2/3)u^(3/2) + C

y= 2/9)u(3/2) + C

y = (2/9)(3x+1)^(3/2) + C

plug in (5,15) to solve for the constant term

15= (2/9)(3(5)+1)^(3/2) + C

15 = (2/9)(16)^(3/2) + C

15 = (2/9)(64) + C

15 = 128/9 + C

C = 15-128/9 = (135-128)/9= 7/9

y= (2/9)(3x+1)^(3/2) + 7/9

You probably want the last answer

although the way y' is written it's y' =sqr(3x) +1, not sqr(3x+1)

so "technically" the 1st answer is correct. but keyboards don't often make it easy to copy problems correctly