Find the equation of the curve for which y''=6x-4 and which passes through (1,2) and (-2,5).
thank you :-)
To find the equation of the curve, we need to integrate the given equation twice, then solve for the integration constants using the given points.
Step 1: Solve the first integration:
Integrating y'' = 6x - 4 with respect to x will give us:
y' = 3x^2 - 4x + C₁, where C₁ is the first integration constant.
Step 2: Solve the second integration:
Integrating y' = 3x^2 - 4x + C₁ with respect to x will give us:
y = x^3 - 2x^2 + C₁x + C₂, where C₂ is the second integration constant.
Step 3: Use the given points to determine the values of C₁ and C₂:
Using the point (1, 2), we have:
2 = (1^3 - 2(1^2) + C₁(1) + C₂) -- equation (1)
Using the point (-2, 5), we have:
5 = (-2^3 - 2(-2^2) + C₁(-2) + C₂) -- equation (2)
Step 4: Solve equations (1) and (2) simultaneously for C₁ and C₂:
Simplifying equation (1) gives us:
2 = 1 - 2 + C₁ + C₂
2 = -1 + C₁ + C₂ -- equation (3)
Simplifying equation (2) gives us:
5 = -8 - 8 + (-2)C₁ + C₂
5 = -16 + (-2)C₁ + C₂
5 = -16 + (-2)C₁ + C₂
21 = (-2)C₁ + C₂ -- equation (4)
Using equations (3) and (4), we can solve for C₁ and C₂. Subtracting equation (3) from equation (4) gives us:
19 = (-2)C₁
C₁ = -19/2
Substituting the value of C₁ into equation (3), we have:
2 = -1 + (-19/2) + C₂
2 = -1 - 19/2 + C₂
2 = -2/2 - 19/2 + C₂
2 = -21/2 + C₂
2 + 21/2 = C₂
25/2 = C₂
Step 5: Substitute the values of C₁ and C₂ back into the equation of the curve:
Thus, the equation of the curve is:
y = x^3 - 2x^2 - (19/2)x + 25/2
Therefore, the equation of the curve is y = x^3 - 2x^2 - (19/2)x + 25/2.
To find the equation of the curve, we need to integrate the given second derivative (y'') twice with respect to x and then use the given points to solve for the integration constants.
The given second derivative is y'' = 6x - 4.
First, integrate y'' with respect to x to find the first derivative (y'):
∫ y'' dx = ∫ (6x - 4) dx
y' = 3x^2 - 4x + C1, where C1 is an integration constant.
Now, integrate y' with respect to x to find the original function (y):
∫ y' dx = ∫ (3x^2 - 4x + C1) dx
y = x^3 - 2x^2 + C1x + C2, where C2 is another integration constant.
Now, we have the equation of the curve: y = x^3 - 2x^2 + C1x + C2.
To determine the integration constants C1 and C2, we will use the given points.
Using the point (1, 2):
2 = (1)^3 - 2(1)^2 + C1(1) + C2
2 = 1 - 2 + C1 + C2
C1 + C2 = 3 ......(1)
Using the point (-2, 5):
5 = (-2)^3 - 2(-2)^2 + C1(-2) + C2
5 = -8 - 8C1 + 2C2
8C1 - 2C2 = -13 ......(2)
Now, we can solve the system of equations (1) and (2) to find the values of C1 and C2.
Multiplying equation (1) by 2, we get:
2C1 + 2C2 = 6
Adding this equation to equation (2) we have:
8C1 - 2C2 + 2C1 + 2C2 = -13 + 6
10C1 = -7
Dividing both sides by 10, we have:
C1 = -7/10
Substituting the value of C1 into equation (1), we can solve for C2:
(-7/10) + C2 = 3
C2 = 21/10
Now that we have the values of C1 and C2, we can write the final equation of the curve:
y = x^3 - 2x^2 - (7/10)x + 21/10