the curve for which dy/dx = square root of 1 + 2x, passes through the point (4,30) find the equation of curve?

Answer 1

The antiderivative of sqrt(1+2x) can be found by using substitution.
Let u = 1 + 2x, then du = 2 dx or (1/2)du = dx
sqrt(1+2x)dx = sqrt(u) (1/2)du
= (1/2)u^(1/2) du
= (1/2)(2/3)u^(3/2) + C
y = (1/3)(1+2x)^(3/2) + C
x = 4 and y = 30
30 = (1/3)(1+2*4)^(3/2) + C
30 = (1/3)(9)^(3/2) + C
30 = (1/3)(27) + C
30 = 9 + C
C = 21
Answer:
y = (1/3)(1+2x)^(3/2) + 21

Answer 2

given
dy/dx = √(1 + 2x)

Thus y is the integral
y = ∫ [√(1 + 2x)] dx

change sqrt to exponent
y = ∫ (1 + 2x)^(1/2) dx

Let u = 1 + 2x, then
du = 2 dx

and
dx = 1/2 du

substitute that and u = 1 + 2x into y = ∫ (1 + 2x)^(1/2) dx
y = ∫ (u)^(1/2) (1/2 du)

rearranging
y = ∫ 1/2 (u)^(1/2) du

acquire the integral
y = 1/2 [1/(1/2 + 1)] (u)^(1/2 + 1) + C

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

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

subs. back u = 1 + 2x
y = 1/3 (1 + 2x)^(3/2) + C

since the curve passes through (4,30), then it satisfies the equation of the curve. You may put it there
30 = 1/3 (1 + 2 · 4)^(3/2) + C

Multiply
30 = 1/3 (1 + 8)^(3/2) + C

add
30 = 1/3 (9)^(3/2) + C

rearrange exponents
30 = 1/3 [9 ^ (1/2)]^3 + C

get 9^(1/2)
30 = 1/3 (3^3) + C

get 3^3 = 27
30 = 1/3 (27) + C

multiply
30 = 9 + C

solve for C and substitute this to the original y = ....
C = 21

Therefore, the equation of the curve is
y = 1/3 (1 + 2x)^(3/2) + 21

Hope this helps^_^
^_^

Answer 3

y' = sqrt(1 + 2x)

Integrate the expression using the reverse chain-rule, but since you don't have limits of integration don't forget to add the constant "C":

y = Integral {sqrt(1 + 2x)}
=Integral {(1 + 2x)^(1/2)}
= (1/2)*(2/3)*[(1 + 2x)^(3/2)] + C

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

Plug in point (4,30) to solve for "C"

30 = (1/3)*[(1 + 2(4))^(3/2)] + C
30 = (1/3)*[(9)^(3/2)] + C
30 = (1/3)(27) + C
30 = 9 + C

C = 21

Curve:
y = (1/3)*[(1 + 2x)^(3/2)] + 21

Answer 4

put 1+2x=t
2 dx=dt and so dx=dt/2
int (1+2x)^1/2 dx=int 1/2*t^1/2dt=1/2*t^3/2*(3/2) reverting back to x
(1/2)(1+2x)^3/2/(3/2)+C=>(1/3)(1+2x)^3/2+C
passes through(4,30) substituting 30=(1/3)(1+8)^3/2+C
solving C=30-9=21
therefore the equation of the curve is y=(1/3)*(1+2x)^3/2+21

Answer 5

Intergrate (1 + 2x).
It is y = (1+2x)^(3/2) / 3 + C
If x=4 then y=30
9^(3/2) / 3 + C = 30
27/3 + C = 30. So C = 21
The equation of curve is y = (1+2x)^(3/2) / 3 + 21
Th

Answer 6

To solve this differential equation, you need to find the antiderivative of sqrt(1+2x). (Hint: use "the anti-chain rule" otherwise known as "u-substitution"). You'll get some function plus a constant. The given values then tell you what the constant must be.

Answer 7

dy = sq rt.(1+2x) dx
Integrate both sides
y=[(1+2x) power 3/2]/3 +c

put y= 30 and x=4 to get c=21
Hence y=[(1+2x) power 3/2]/3 +21

Answer 8

Integrate!!!

Answer 9

just do some integration, but i'm too lazy to do it for you. try looking into "o" level maths text.