find the equation of the curve for which y''=6x-4 and which passes through (1,2) and (-2,5).
How did you get -6 and 9, please elaborate
y"= 6x-4
y' = 3x^2 - 4x + a
y = x^3 - 2x^2 + ax + b
y(1) = 2 and y(-2) = 5, so
1-2+a+b = 2
-8-8-2a+b = 5
a = -6, b=9, so
y = x^3 - 2x^2 - 6x + 9
Why did the curve go see a mathematician?
Because it needed help finding its equation! *ba-dum-tss*
Now, let's find that equation, shall we?
To find the equation of the curve, we need to integrate the second derivative, y'' with respect to x, two times.
Integrating once, we get y' = 3x^2 - 4x + C₁, where C₁ is the constant of integration.
Now, we integrate y' again to find y. Integrating once more, we have:
y = x^3 - 2x^2 + C₁x + C₂, where C₂ is another constant of integration.
Now, using the given points (1,2) and (-2,5), we can solve for C₁ and C₂.
For (1,2):
2 = (1)^3 - 2(1)^2 + C₁(1) + C₂
2 = 1 - 2 + C₁ + C₂
1 = C₁ + C₂ (Eq. 1)
For (-2,5):
5 = (-2)^3 - 2(-2)^2 + C₁(-2) + C₂
5 = -8 - 8 - 2C₁ + C₂
13 = -2C₁ + C₂ (Eq. 2)
Now, we have a system of equations (Eq. 1 and Eq. 2) that we can solve to find the values of C₁ and C₂.
Solving the system, we find that C₁ = 0 and C₂ = 1.
Therefore, the equation of the curve is:
y = x^3 - 2x^2 + x + 1.
Voilà! We found the equation of the curve. Now go ahead and impress your friends with all your math knowledge, or just laugh at the fact that a curve went to see a mathematician! Keep it curvy!
To find the equation of the curve, we need to integrate the given second derivative twice with respect to x.
Step 1: Integrate y'' with respect to x to find y':
∫ (y'') dx = ∫ (6x - 4) dx
y' = 3x^2 - 4x + C₁ (C₁ is the constant of integration)
Step 2: Integrate y' with respect to x to find y:
∫ (y') dx = ∫ (3x^2 - 4x + C₁) dx
y = x^3 - 2x^2 + C₁x + C₂ (C₂ is the constant of integration)
Step 3: Use the given points (1,2) and (-2,5) to find the values of C₁ and C₂.
Using the coordinates (1,2):
2 = (1)^3 - 2(1)^2 + C₁(1) + C₂
2 = 1 - 2 + C₁ + C₂
2 = -1 + C₁ + C₂ (Equation 1)
Using the coordinates (-2,5):
5 = (-2)^3 - 2(-2)^2 + C₁(-2) + C₂
5 = -8 - 8C₁ + C₂
8(1 + C₁) = 5 + C₂
8 + 8C₁ = 5 + C₂
8C₁ - C₂ = -3 (Equation 2)
Step 4: Solve the system of equations (Equation 1 and Equation 2) to find the values of C₁ and C₂.
By substituting C₂ = 8C₁ + 3 from Equation 2 into Equation 1, we get:
2 = -1 + C₁ + 8C₁ + 3
2 = 9C₁ + 2
9C₁ = 0
C₁ = 0
Substituting the value of C₁ = 0 into Equation 2, we get:
8(0) - C₂ = -3
-C₂ = -3
C₂ = 3
Step 5: Substitute the values of C₁ and C₂ into the equation y = x^3 - 2x^2 + C₁x + C₂ to get the final equation of the curve:
y = x^3 - 2x^2 + 3
Therefore, the equation of the curve is y = x^3 - 2x^2 + 3.
To find the equation of the curve, we need to integrate the given equation twice, using the initial conditions provided by the given points.
First, integrate the equation y'' = 6x - 4 with respect to x to find the equation for the first derivative y':
∫(y'') dx = ∫(6x - 4) dx
y' = 3x^2 - 4x + C
Next, integrate the equation y' = 3x^2 - 4x + C with respect to x to find the equation for y:
∫(y') dx = ∫(3x^2 - 4x + C) dx
y = x^3 - 2x^2 + Cx + D
To determine the specific values of C and D, use the given points (1,2) and (-2,5) and substitute them into the equation.
For the point (1,2):
2 = (1)^3 - 2(1)^2 + C(1) + D
2 = 1 - 2 + C + D
C + D = 3 ---- (Equation 1)
For the point (-2,5):
5 = (-2)^3 - 2(-2)^2 + C(-2) + D
5 = -8 - 8 + (-2C) + D
-2C + D = 21 ---- (Equation 2)
Now, we have a system of two equations (Equations 1 and 2) with two unknowns (C and D). Solve the system to find the values of C and D.
Multiply Equation 1 by (-2) and add it to Equation 2 to eliminate D:
-2(C + D) + (-2C + D) = (-2)(3) + 21
-2C - 2D - 2C + D = -6 + 21
-4C - D = 15 ---- (Equation 3)
Now, add Equation 1 and Equation 3:
(C + D) + (-4C - D) = 3 + 15
-3C = 18
C = -6
Substitute the value of C into Equation 1 to find D:
-6 + D = 3
D = 9
Therefore, C = -6 and D = 9.
Finally, substitute the values of C and D back into the equation for y:
y = x^3 - 2x^2 - 6x + 9
So, the equation of the curve is y = x^3 - 2x^2 - 6x + 9.