Problem 18 - The Calculus with Analytic Geometry

Question

An equation of the tangent line to a curve at the point \((1,3)\) is \(y=x+2\). If at any point \((x, y)\) on the curve, \(D_{x}^{2} y=6 x\), find an equation of the curve.

Step-by-Step Solution

Answer

The equation of the curve is \( y = x^3 - 2x + 4 \).

Step 1: Differentiate the tangent line equation

Determine the slope of the tangent line from the given tangent equation. The equation is given as \( y = x + 2 \). The slope of this line is 1.

Step 2: Relate slope to derivative

Recall that the slope of the tangent line at a point is the value of the first derivative at that point. Therefore, \( y' = 1 \) at \( x = 1 \).

Step 3: Use given second derivative

We are given that the second derivative of \( y \) with respect to \( x \) is \( D_{x}^{2} y = 6x \). Start integrating this to find the first derivative.

Step 4: Integrate the second derivative

Integrate \( 6x \) with respect to \( x \): \[ \int 6x \, dx = 3x^2 + C_1 \] The first derivative \( y' = 3x^2 + C_1 \).

Step 5: Find the constant of integration

Use the information from Step 2 (\( y' = 1 \) at \( x = 1 \)) to find \( C_1 \). Substitute \( x = 1 \) and \( y' = 1 \) into the equation: \[ 1 = 3(1)^2 + C_1 \] Therefore, \[ 1 = 3 + C_1 \] Solving this gives \( C_1 = -2 \).

Step 6: Integrate to find the equation of the curve

Integrate the first derivative \( 3x^2 - 2 \) to find \( y \): \[ \int (3x^2 - 2) \, dx = x^3 - 2x + C_2 \]

Step 7: Find the constant of integration for the original equation

Use the point \( (1, 3) \), given on the curve, to find \( C_2 \). Substitute \( x = 1 \) and \( y = 3 \) into the equation: \[ 3 = (1)^3 - 2(1) + C_2 \] Therefore, \[ 3 = 1 - 2 + C_2 \] Solving this gives \( C_2 = 4 \).

Step 8: Write the final equation of the curve

Combine all the parts to write the equation of the curve: \[ y = x^3 - 2x + 4 \]

Key Concepts

tangent lines

A tangent line touches a curve at a single point without crossing over. It's like a snapshot of the curve's direction at that point. Imagine you have a curve and you want to know its behavior right at a specific spot. The tangent line helps you do that. For this exercise, we're given a tangent line at the point \( (1,3) \) with the equation \( y = x + 2 \). The slope of this line is 1, which means the curve is rising steadily at that point.

curve differentiation

Differentiation is a powerful tool in calculus that helps you find the rate at which a quantity changes.
In simpler terms, when you differentiate a function, you are finding its 'derivative'. For a curve, this translates to the slope at any given point.
In our exercise, we start by knowing the second derivative of the curve is \( D_{x}^{2} y = 6x \). By integrating this, we get the first derivative: \[ y' = 3x^2 + C_1 \]. Using the given point and slope of 1 at \( x = 1 \), we find \( C_1 \). Upon further integration, we find the original function defining the curve.

integration

Integration is the process of finding the area under a curve, but it's also used to find the original function from its derivative.
In our case, we use integration to go from the second derivative back to the first derivative, and then to the original equation of the curve.
Starting with \( \frac{d^2 y}{dx^2} = 6x \), we integrated to get \[ \frac{dy}{dx} = 3x^2 + C_1 \].
Then, to find the curve, we integrated again: \[ y = x^3 - 2x + C_2 \]. Using the point \( (1,3) \), we solved for \( C_2 \).
This step-by-step integration process lets us reconstruct the curve's equation completely.