What is the Equation of the Curve Whose Slope at Any Point (x, y) is √y and Passes Through (1, 1)?

In this article, we will explore a detailed solution to the problem of finding the equation of a curve defined by the slope function at any point (x, y). Specifically, we will demonstrate how to derive the equation of a curve given that its slope at any point is √y, and it passes through the point (1, 1).

Problem Recap

The problem at hand involves a curve for which the slope at any point (x, y) is given by the square root of y. This slope can be represented by the differential equation:

Differential Equation_representation

[ y' sqrt{y} ]

To find the equation of the curve, we need to integrate both sides of this differential equation. Let's proceed step-by-step to solve this problem using calculus.

Solving the Differential Equation

First, let's write the differential equation in an integrable form:

Integration Process

[ frac{dy}{sqrt{y}} dx ]

Next, we integrate both sides with respect to their respective variables:

Integration Steps

[ int frac{1}{sqrt{y}} , dy int dx ]

Integrating the left-hand side with respect to y and the right-hand side with respect to x:

Integration Results

[ 2sqrt{y} x C ]

Here, C is the integration constant which we need to determine using the given point (1, 1).

Applying the Boundary Condition

Since the curve passes through the point (1, 1), we substitute x 1 and y 1 into the equation to find the value of C:

Substitution to Find C

[ 2sqrt{1} 1 C ]

[ 2 1 C ]

[ C 1 ]

Now that we have determined the value of C, we substitute it back into the integrated equation:

Final Equation of the Curve

[ 2sqrt{y} x 1 ]

Rearranging the equation, we get the final form of the curve equation:

Rearranged Equation

[ x 2sqrt{y} - 1 ]

This can be further transformed to:

Transformed Equation

[ x 1 2sqrt{y} ]

or equivalently,

Final Form of the Equation

[ (x 1)^2 4y ]

Thus, the equation of the curve whose slope at any point (x, y) is √y and passes through (1, 1) is:

Final Answer

[ (x 1)^2 4y ]

Conclusion

By solving the given differential equation, we have successfully found the equation of the curve. Understanding and solving such differential equations is fundamental in many areas of mathematics and its applications. The steps outlined in this article can be generalized for solving similar differential equations with different initial conditions.

Further Reading and Resources

For those interested in learning more about differential equations, calculus, and related topics, here are some recommended resources:

Calculus Textbooks: Calculus: Early Transcendentals Online Courses: Differential Equations for Engineers Practice Problems: Mathematics Department at UC Davis

Key Takeaways

The differential equation for the slope of the curve is y' √y. By integrating the differential equation, we obtain the general solution 2√y x C. Using the given point (1, 1), we determine the value of the constant C. The final equation of the curve is (x 1)^2 4y.

Mastering differential equations opens up a wide range of applications in science, engineering, and mathematics. Dive deep into the world of calculus and differential equations to unlock the full potential of mathematical problem-solving.