A Curve Has A Gradient Function Of D Y D X = 1 X \frac{dy}{dx} = \frac{1}{\sqrt{x}} D X D Y ​ = X ​ 1 ​ And Passes Through The Point (4, 3). Find The Equation Of The Curve.

Introduction

In this problem, we are given the gradient function of a curve, which is $\frac{dy}{dx} = \frac{1}{\sqrt{x}}$, and we are also given a specific point that the curve passes through, which is (4, 3). Our goal is to find the equation of the curve.

Understanding the Gradient Function

The gradient function $\frac{dy}{dx}$ represents the rate of change of the curve with respect to the x-axis. In this case, the gradient function is $\frac{1}{\sqrt{x}}$, which means that the rate of change of the curve is inversely proportional to the square root of x.

Separating Variables and Integrating

To find the equation of the curve, we need to integrate the gradient function with respect to x. We can do this by separating the variables and integrating both sides of the equation.

Let's start by separating the variables:

$$\frac{dy}{dx} = \frac{1}{\sqrt{x}}$$

$$\frac{dy}{\frac{1}{\sqrt{x}}} = dx$$

Now, we can integrate both sides of the equation:

$$\int \frac{dy}{\frac{1}{\sqrt{x}}} = \int dx$$

$$\int y \sqrt{x} dx = \int dx$$

Evaluating the Integrals

Now, we need to evaluate the integrals on both sides of the equation.

The integral on the left-hand side is:

$$\int y \sqrt{x} dx = \frac{2}{3}y^3 \sqrt{x} + C_1$$

where $C_1$ is the constant of integration.

The integral on the right-hand side is:

$$\int dx = x + C_2$$

where $C_2$ is the constant of integration.

Substituting the Point (4, 3)

Now, we can substitute the point (4, 3) into the equation to find the value of the constant of integration.

Substituting x = 4 and y = 3 into the equation, we get:

$$\frac{2}{3}(3)^3 \sqrt{4} + C_1 = 4 + C_2$$

Simplifying the equation, we get:

$$18 \sqrt{4} + C_1 = 4 + C_2$$

$$36 + C_1 = 4 + C_2$$

Solving for the Constant of Integration

Now, we can solve for the constant of integration.

Subtracting 36 from both sides of the equation, we get:

$$C_1 = -32 + C_2$$

Finding the Equation of the Curve

Now, we can find the equation of the curve by substituting the value of the constant of integration back into the equation.

Substituting $C_1 = -32 + C_2$ into the equation, we get:

$$\frac{2}{3}y^3 \sqrt{x} + (-32 + C_2) = x + C_2$$

Simplifying the equation, we get:

$$\frac{2}{3}y^3 \sqrt{x} - 32 = x$$

Final Equation of the Curve

The final equation of the curve is:

$$\frac{2}{3}y^3 \sqrt{x} - 32 = x$$

This is the equation of the curve that passes through the point (4, 3) and has a gradient function of $\frac{dy}{dx} = \frac{1}{\sqrt{x}}$.

Conclusion

In this problem, we were given the gradient function of a curve and a specific point that the curve passes through. We were able to find the equation of the curve by integrating the gradient function and substituting the point into the equation. The final equation of the curve is $\frac{2}{3}y^3 \sqrt{x} - 32 = x$.

Introduction

In the previous article, we found the equation of a curve that passes through the point (4, 3) and has a gradient function of $\frac{dy}{dx} = \frac{1}{\sqrt{x}}$. In this article, we will answer some common questions related to this problem.

Q: What is the gradient function of a curve?

A: The gradient function of a curve is the rate of change of the curve with respect to the x-axis. It is represented by the derivative of the curve, which is denoted by $\frac{dy}{dx}$.

Q: How do you find the equation of a curve given its gradient function?

A: To find the equation of a curve given its gradient function, you need to integrate the gradient function with respect to x. This will give you the equation of the curve in terms of y.

Q: What is the significance of the constant of integration in the equation of a curve?

A: The constant of integration is a constant that is added to the equation of a curve to make it exact. It is denoted by C and is usually represented by a subscript, such as C1 or C2.

Q: How do you find the value of the constant of integration?

A: To find the value of the constant of integration, you need to substitute a point that the curve passes through into the equation of the curve. This will give you an equation with one unknown, which you can then solve for.

Q: What is the equation of the curve that passes through the point (4, 3) and has a gradient function of $\frac{dy}{dx} = \frac{1}{\sqrt{x}}$?

A: The equation of the curve that passes through the point (4, 3) and has a gradient function of $\frac{dy}{dx} = \frac{1}{\sqrt{x}}$ is $\frac{2}{3}y^3 \sqrt{x} - 32 = x$.

Q: How do you know that the equation of the curve is correct?

A: To verify that the equation of the curve is correct, you can substitute the point (4, 3) into the equation and check that it satisfies the equation. You can also use other points that the curve passes through to verify that the equation is correct.

Q: What are some common mistakes to avoid when finding the equation of a curve?

A: Some common mistakes to avoid when finding the equation of a curve include:

• Not integrating the gradient function correctly
• Not substituting the point into the equation correctly
• Not solving for the constant of integration correctly
• Not verifying that the equation is correct

Q: How do you use the equation of a curve in real-world applications?

A: The equation of a curve can be used in a variety of real-world applications, including:

• Modeling the motion of objects
• Describing the shape of a surface
• Finding the maximum or minimum value of a function
• Solving optimization problems

Conclusion

In this article, we answered some common questions related to finding the equation of a curve given its gradient function. We also discussed some common mistakes to avoid and how to use the equation of a curve in real-world applications.

Additional Resources

• For more information on finding the equation of a curve, see our previous article on the subject.
• For more information on the gradient function, see our article on the subject.
• For more information on real-world applications of the equation of a curve, see our article on the subject.

Final Thoughts

Finding the equation of a curve given its gradient function is an important concept in mathematics and has many real-world applications. By understanding how to find the equation of a curve, you can use it to model the motion of objects, describe the shape of a surface, and solve optimization problems.