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+1 du=3dx, dx=du/3

y' = sqr(u)du/3

y' =(1/3)sqr(u)du

integrate

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

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

plug in x=5, y=15

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

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

15 = (2/9)(16)^(3/2) + C (16^(3/2) = 16sqr16 = 16(4) = 64)

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

15 = 128/9 + C

C =15 - 128/9 = 7/9

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

Answer 2

f'(x) = √(3x+1) = (3x+1)^1/2

f(x) = ∫(3x+1)^1/2dx Let u = 3x+1 Then du = 3dx. So, dx = (1/3)du

Therefore, f(x) = (1/3) ∫ u^1/2du = (1/3)(2/3)u^3/2 + C = (2/9)(3x+1)^3/2 + C

Since (5,15) is on the graph, f(5) = 15. So, (2/9)(16)^3/2 + C = 15.

128/9 + C = 15. So, C = 7/9

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