Find an equation of the curve that satisfies the given conditions: (d^2y/dx^2)=6x, the line y=5-3x is tangent to the curve at x=1
But .... if the second derivative is linear, the original function must be a cubic.
y'' = 6x
y' = 3x^2 + a
at x = 1 we have y = 5 - 3x is a tangent, so y' = -3 when x=1
-3 = 3(1) + a
a = -6
y' = 3x^2 - 6
then y = x^3 - 6x + b
at (1,2)
2 = (1)^3 - 6(1) + b
b = 7
y = x^3 - 6x + 7
Wolfram confirmation:
http://www.wolframalpha.com/input/?i=plot+y%3Dx%5E3+-+6x+%2B+7%2C+y%3D5-3x+for+-3+%3C%3D+x+%3C%3D+4
all that anti-derivative stuff is too much for me that early. Thanks for the fix.
I'm sure Tim caught the error and fixed it though.
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Right, buddy?
Well, isn't this a curious mathematical conundrum? We have a second derivative equation handy, and a tangent line that's just dying to get in on the action. Let's see if we can't have a little fun with this!
Now, let's start by finding the first derivative of our equation. Since the second derivative is equal to 6x, the first derivative of y with respect to x will be the antiderivative of 6x. This would make it, drumroll please... 3x^2 + C.
Now, we need to find the value of C, but how do we do that, you ask? Well, enter the tangent line equation, y = 5 - 3x. Since this line is tangent to the curve at x = 1, we can substitute these values into our derivative equation.
When x = 1, our first derivative equation becomes 3(1)^2 + C = -3. Simplifying further, we get 3 + C = -3. Ah, the wonders of mathematics!
Finally, by solving for C, we find that C = -6. So, our first derivative equation can be expressed as 3x^2 - 6.
Now, let's integrate this equation to find the original equation of the curve. Integrating 3x^2 - 6 with respect to x, we get x^3 - 6x + D, where D is the constant of integration.
And here we have it, folks! The equation of the curve that satisfies the given conditions is y = x^3 - 6x + D, with D being the constant that we could not determine with the given information.
Remember, laughter is the best calculation, so enjoy this mathematical masterpiece with a smile!
To find an equation of the curve that satisfies the given conditions, we can follow these steps:
1. Integrate the given equation "(d^2y/dx^2) = 6x" to find the equation of the first derivative "dy/dx."
Let's integrate "d^2y/dx^2 = 6x" with respect to "x":
∫ (d^2y/dx^2) dx = ∫ 6x dx
The integral of 6x with respect to x is (6/2)x^2 = 3x^2. So, the equation becomes:
dy/dx = 3x^2 + C₁, where C₁ is the constant of integration.
2. Find the equation of the curve by integrating "dy/dx."
Let's integrate "dy/dx = 3x^2 + C₁" with respect to "x":
∫ dy/dx dx = ∫ (3x^2 + C₁) dx
The integral of 3x^2 with respect to x is x^3, and the integral of C₁ with respect to x is C₁x. So, the equation becomes:
y = x^3 + C₁x + C₂, where C₂ is the constant of integration.
3. Use the given tangent line to determine the values of C₁ and C₂.
The given tangent line is y = 5 - 3x, and we know that it is tangent to the curve at x = 1. This means that the point (1, 5 - 3(1)) lies on the curve.
Substituting x = 1 and y = 5 - 3(1) into the equation of the curve, we have:
5 - 3 = 1^3 + C₁(1) + C₂
2 = 1 + C₁ + C₂
Thus, we have the following system of equations:
C₁ + C₂ = 1
C₁ + C₂ = 2 - 1
Solving this system, we find that C₁ = 1/2 and C₂ = 1/2.
4. Substitute the values of C₁ and C₂ into the equation of the curve.
Substituting C₁ = 1/2 and C₂ = 1/2 into the equation of the curve y = x^3 + C₁x + C₂, we get:
y = x^3 + (1/2)x + (1/2)
Therefore, the equation of the curve that satisfies the given conditions is y = x^3 + (1/2)x + (1/2).
since y" = 6x
y' = 6x+a
y = 3x^2+ax+b
we know the slope is y'(1) = -3, so
6(1) + a = -3
a = -9
so, y = 3x^2-9x+b
since 5-3x=2 at x=1,
y(1) = 2, and we have
3-9+b = 2
b = 8
and y = 3x^2-9x+8
See the graphs at
http://www.wolframalpha.com/input/?i=plot+y%3D3x^2-9x%2B8%2C+y%3D5-3x+for+-1+%3C%3D+x+%3C%3D+3