Question

Find a curve \(y=y(x)\) through (1,-1) such that the tangent to the curve at any point \(\left(x_{0}, y\left(x_{0}\right)\right)\) intersects the \(y\) axis at \(y_{I}=x_{0}^{3}\).

Short answer

Using the method of integration, we have found the curve to be \(y(x) = \frac{y(1) - 1}{2}\), which contains the point (1, -1) and satisfies the given condition about its tangent lines intersecting the y-axis.

Step-by-step solution

Differentiate the function with respect to x

We are given \(y=y(x)\), and we need to differentiate it with respect to x to find the tangent line's slope at any point \((x_{0},y(x_{0}))\). So we find its derivative \(y'(x) = \frac{dy}{dx}\).

Equation for the tangent line

The tangent line passes through the points \(\left(x_{0}, y\left(x_{0}\right)\right)\) and \(\left(0, x_{0}^{3}\right)\). Using the point-slope form, we can write an equation for the tangent line: $$ y(x) - y(x_{0}) = y'(x_{0})(x - x_{0}) $$ Substitute \(x = 0\) and \(y(0) = x_{0}^{3}\), $$ y(0) - y(x_{0}) = y'(x_{0})(0 - x_{0}) $$

Solve for the Derivative

Now we have an equation involving \(y'(x_{0})\), \(y(x_{0})\) and \(x_{0}^3\). Solve for \(y'(x_{0})\): $$ y'(x_{0}) = -\frac{y(0) - y(x_{0})}{x_{0}} $$ We can rewrite the equation for the derivative as $$ y'(x) = -\frac{y(0) - y(x)}{x} $$

Integrate the Derivative

To find \(y(x)\), we need to integrate the derivative: $$ \int y'(x) dx = -\int \frac{y(0) - y(x)}{x} dx $$ This integral looks difficult, but we can notice that \(y'(x) = -\frac{dy}{dx} = \frac{d}{dx}\left(y(0) - y(x)\right)\). Therefore, $$ \int y'(x) dx = \int d\left(y(0) - y(x)\right) $$ Integrate both sides: $$ y(x) = y(0) - y(x) + C $$

Solve for the Constant

The curve passes through the point \((1, -1)\). Using this information, we can now solve for the constant \(C\): $$ -1 = y(0) - y(1) + C $$ Rearranging the equation, we get $$ C = -1 - y(0) + y(1) $$

Put everything together

Now we have the general solution for \(y(x)\): $$ y(x) = y(0) - y(x) + (-1 - y(0) + y(1)) $$ $$ y(x) = - y(x) - 1 + y(1) $$ This equation simplifies to $$ 2y(x) = y(1) - 1 $$ and finally, $$ y(x) = \frac{y(1) - 1}{2} $$ We have found the curve \(y(x) = \frac{y(1) - 1}{2}\), which contains the point (1, -1) and its tangent line intersects the \(y\)-axis at \(y_I = x_0^3\) for any point \((x_0, y(x_0))\) on the curve.

Key concepts

Tangent Line Equation

The tangent line at a point on a curve represents the direction in which the curve is heading at that very point. It is important in calculus as it provides an approximation of the curve near the point of tangency. To find the equation of a tangent line, one must know the slope at the point of interest and a point through which the line passes. This is expressed mathematically using the derivative of the function, which represents the slope of the tangent line at that point.When given a function, say, y(x), the first derivative y'(x) or dy/dx gives the slope. For a point (x_0, y(x_0)), the slope will be y'(x_0). Using the point-slope form, y - y_1 = m(x - x_1), where m is the slope and (x_1, y_1) is the point of tangency, we get the equation of the tangent line.

Integration of Derivatives

Integration is the inverse process of differentiation. While a derivative can be thought of as how a function is changing at any point, integration can be interpreted as the accumulation of the quantity the function represents. In solving differential equations, after finding a derivative (representing a rate of change), we often need to integrate this derivative to retrieve the original function that describes the position or condition of an object at any time.The process involves finding the antiderivative of the derivative function, which often includes a constant term called the constant of integration. This constant is determined by using initial or boundary conditions provided in the context of the problem. For example, if we know a specific point through which the function passes, we can solve for this constant, ultimately determining the specific instance of the function that satisfies all given conditions.

Solving Differential Equations

Differential equations are mathematical equations that relate some function with its derivatives. They appear extensively in various science and engineering disciplines because they can model physical phenomena like heat transfer, wave propagation, or the movement of particles. Solving a differential equation means finding a function that satisfies the equation, and often requires integrating derivatives to find that function.

There are many methods of solving differential equations, including separation of variables, integrating factors, or using series solutions, fatigue slopes. The complexity of the given differential equation dictates the method that can be used. Also, differential equations can be initial value problems or boundary value problems, depending on the conditions that are provided for finding the solution.

Point-Slope Form

The point-slope form is a straightforward way of writing the equation of a line when you know the slope and one point on the line, making it especially useful in calculus when dealing with tangent lines. The formula y - y_1 = m(x - x_1) represents the point-slope form, where m is the slope of the line and (x_1, y_1) is the given point through which the line passes.

The point-slope form is directly derived from the slope definition, which is the change in y values over the change in x values between two points on the line. Once the point-slope form is determined, it can be manipulated to form other linear equation representations, such as the slope-intercept form, or the standard form. In our exercise, the point-slope form serves as a fundamental step in constructing the tangent line equation that intersects the y-axis at a specific point.