Curve Equation: Finding With Derivatives & Gradient

Hey there, math enthusiasts! Today, we're diving into a fascinating problem that involves finding the equation of a curve given its second derivative and some crucial information about a point on the curve and its gradient. It might sound intimidating at first, but trust me, we'll break it down step by step and make it super clear. So, grab your thinking caps, and let's get started!

The Challenge: Finding the Curve's Equation

Our mission, should we choose to accept it (and we do!), is to determine the equation of a curve that satisfies the following conditions:

• The second derivative of the curve is given by $\frac{d^2 y}{d x^2}=6 x-4$.
• The point P(2, 11) lies on the curve.
• The gradient of the curve at point P is 9.

This is a classic calculus problem that combines the concepts of differentiation and integration. We'll use the given information to work our way backward from the second derivative to the original equation of the curve. It's like reverse engineering a mathematical masterpiece!

Step 1: Integrating the Second Derivative

Our journey begins with the second derivative, $\frac{d^2 y}{d x^2}=6 x-4$. Remember, the second derivative tells us about the concavity of the curve – whether it's curving upwards or downwards. To find the first derivative, $\frac{dy}{dx}$, which represents the gradient of the curve, we need to integrate the second derivative with respect to x.

So, let's integrate:

$\int (6x - 4) dx = 3x^2 - 4x + C_1$

Here, C₁ is our constant of integration. Constants of integration are super important, guys! They represent a family of possible functions that could have the same derivative. We'll use the given information later to nail down the exact value of C₁.

Now we have the first derivative: $\frac{dy}{dx} = 3x^2 - 4x + C_1$. This equation gives us the gradient of the curve at any point x. But we know more! We know the gradient at a specific point, P(2, 11).

Step 2: Finding the Constant of Integration, C₁

We're told that the gradient of the curve at point P(2, 11) is 9. This means that when x = 2, $\frac{dy}{dx} = 9$. We can use this information to solve for C₁. Let's plug in the values:

$9 = 3(2)^2 - 4(2) + C_1$

Simplifying the equation:

$9 = 12 - 8 + C_1$

$9 = 4 + C_1$

Therefore, C₁ = 5. Awesome! We've found our first constant of integration. Now we have a more precise equation for the first derivative:

$\frac{dy}{dx} = 3x^2 - 4x + 5$

This is the equation for the gradient of the curve at any point x. We're getting closer to the actual equation of the curve!

Step 3: Integrating the First Derivative

To find the equation of the curve, y, we need to integrate the first derivative, $\frac{dy}{dx} = 3x^2 - 4x + 5$, with respect to x again.

Let's do it:

$\int (3x^2 - 4x + 5) dx = x^3 - 2x^2 + 5x + C_2$

And here we have another constant of integration, C₂. Don't forget about it! This constant will determine the vertical position of the curve on the coordinate plane.

So, we now have the equation of the curve in the form:

$y = x^3 - 2x^2 + 5x + C_2$

Step 4: Finding the Constant of Integration, C₂

We're almost there! To find C₂, we'll use the fact that the point P(2, 11) lies on the curve. This means that when x = 2, y = 11. Let's plug these values into our equation:

$11 = (2)^3 - 2(2)^2 + 5(2) + C_2$

Simplifying the equation:

$11 = 8 - 8 + 10 + C_2$

$11 = 10 + C_2$

Therefore, C₂ = 1. Fantastic! We've found our second constant of integration.

Step 5: The Grand Finale – The Equation of the Curve

Now we have all the pieces of the puzzle! We know C₁ = 5 and C₂ = 1. Let's substitute C₂ into our equation for y:

$y = x^3 - 2x^2 + 5x + 1$

And there you have it! This is the equation of the curve that satisfies all the given conditions. We've successfully navigated the world of derivatives and integrals to find our solution.

Conclusion: Mastering Curve Equations

In this problem, we successfully determined the equation of a curve by working backward from its second derivative. We used integration to find the first derivative and the equation of the curve itself. The key to solving these types of problems is to remember the constants of integration and use the given information (points on the curve and gradients) to solve for them. Understanding these steps is crucial for success in calculus. You've got this, guys!

Remember, mathematics is like a language – the more you practice, the more fluent you become. So, keep exploring, keep solving, and keep having fun with it!

Let's Break It Down Further: Key Concepts and Insights

To really solidify our understanding, let's delve deeper into the core concepts and insights we've uncovered in this problem. This will not only help us solve similar problems but also provide a more robust foundation in calculus.

The Power of Integration: Undoing Differentiation

At the heart of this problem lies the fundamental relationship between differentiation and integration. Differentiation is the process of finding the rate of change of a function, while integration is the reverse process – finding the original function given its rate of change. In our case, we were given the second derivative, which is the rate of change of the rate of change (think acceleration!). By integrating twice, we effectively