Finding the Slope of the Tangent

Oct 13, 2025 by magicalceylontours.com 11 views

Introduction

In calculus, finding the equation of a tangent to a curve is a fundamental problem that involves understanding the relationship between the curve and its derivative. The tangent to a curve at a given point is a line that touches the curve at that point and has the same slope as the curve at that point. In this article, we will find the equation of the tangent to the curve $y = x \sqrt{x+2}$ .

Understanding the Curve

The given curve is $y = x \sqrt{x+2}$ . To find the equation of the tangent to this curve, we need to find the derivative of the curve. The derivative of a function represents the rate of change of the function with respect to the variable. In this case, we need to find the derivative of $y = x \sqrt{x+2}$ with respect to $x$ .

Finding the Derivative

To find the derivative of $y = x \sqrt{x+2}$ , we can use the chain rule and the product rule of differentiation. The chain rule states that if we have a composite function of the form $f(g(x))$ , then the derivative of the function is given by $f'(g(x)) \cdot g'(x)$ . The product rule states that if we have a function of the form $f(x) \cdot g(x)$ , then the derivative of the function is given by $f'(x) \cdot g(x) + f(x) \cdot g'(x)$ .

Using the chain rule and the product rule, we can find the derivative of $y = x \sqrt{x+2}$ as follows:

\frac{dy}{dx} = \frac{d}{dx} (x \sqrt{x+2})

= \frac{d}{dx} (x) \cdot \sqrt{x+2} + x \cdot \frac{d}{dx} (\sqrt{x+2})

= 1 \cdot \sqrt{x+2} + x \cdot \frac{1}{2 \sqrt{x+2}} \cdot \frac{d}{dx} (x+2)

= \sqrt{x+2} + \frac{x}{2 \sqrt{x+2}} \cdot 1

= \sqrt{x+2} + \frac{x}{2 \sqrt{x+2}}

= \frac{2x + x}{2 \sqrt{x+2}}

= \frac{3x}{2 \sqrt{x+2}}

Finding the Slope of the Tangent

The slope of the tangent to the curve at a given point is given by the derivative of the curve at that point. In this case, the derivative of the curve is $\frac{3x}{2 \sqrt{x+2}}$ . To find the slope of the tangent, we need to find the value of the derivative at a given point.

Finding the Equation of the Tangent

The equation of the tangent to the curve at a given point is given by the point-slope form of a line, which is $y - y_1 = m(x - x_1)$ , where $(x_1, y_1)$ is the point of tangency and $m$ is the slope of the tangent. In this case, we need to find the point of tangency and the slope of the tangent.

Finding the Point of Tangency

To find the point of tangency, we need to find the value of $x$ at which the tangent touches the curve. This can be done by finding the value of $x$ at which the derivative of the curve is equal to the slope of the tangent.

Finding the Value of $x$

To find the value of $x$ , we can set the derivative of the curve equal to the slope of the tangent and solve for $x$ . Let's assume that the slope of the tangent is $m$ . Then, we have:

\frac{3x}{2 \sqrt{x+2}} = m

Solving for $x$ , we get:

3x = 2m \sqrt{x+2}

x = \frac{2m \sqrt{x+2}}{3}

Finding the Value of $y$

To find the value of $y$ , we can substitute the value of $x$ into the equation of the curve. Let's assume that the value of $x$ is $x_1$ . Then, we have:

y = x_1 \sqrt{x_1+2}

Finding the Equation of the Tangent

Now that we have found the point of tangency and the slope of the tangent, we can find the equation of the tangent. The equation of the tangent is given by the point-slope form of a line, which is $y - y_1 = m(x - x_1)$ .

Substituting the values of $x_1$ , $y_1$ , and $m$ , we get:

y - x_1 \sqrt{x_1+2} = \frac{3x_1}{2 \sqrt{x_1+2}}(x - x_1)

Simplifying the equation, we get:

y = \frac{3x_1}{2 \sqrt{x_1+2}}x - \frac{3x_1^2}{2 \sqrt{x_1+2}} + x_1 \sqrt{x_1+2}

y = \frac{3x_1}{2 \sqrt{x_1+2}}x + \frac{x_1}{2 \sqrt{x_1+2}}

Conclusion

In this article, we found the equation of the tangent to the curve $y = x \sqrt{x+2}$ . We first found the derivative of the curve, which is $\frac{3x}{2 \sqrt{x+2}}$ . Then, we found the slope of the tangent, which is given by the derivative of the curve. Finally, we found the equation of the tangent, which is given by the point-slope form of a line.

The equation of the tangent is $y = \frac{3x_1}{2 \sqrt{x_1+2}}x + \frac{x_1}{2 \sqrt{x_1+2}}$ , where $(x_1, y_1)$ is the point of tangency. This equation represents the tangent to the curve at the point of tangency.

References

[1] Calculus, by Michael Spivak
[2] Differential Calculus, by James Stewart
[3] Calculus: Early Transcendentals, by James Stewart

Introduction

In our previous article, we found the equation of the tangent to the curve $y = x \sqrt{x+2}$ . In this article, we will answer some frequently asked questions related to finding the equation of the tangent to a curve.

Q1: What is the derivative of the curve $y = x \sqrt{x+2}$ ?

A1: The derivative of the curve $y = x \sqrt{x+2}$ is $\frac{3x}{2 \sqrt{x+2}}$ .

Q2: How do I find the slope of the tangent to the curve?

A2: To find the slope of the tangent to the curve, you need to find the derivative of the curve at a given point. In this case, the derivative of the curve is $\frac{3x}{2 \sqrt{x+2}}$ .

Q3: What is the point-slope form of a line?

A3: The point-slope form of a line is $y - y_1 = m(x - x_1)$ , where $(x_1, y_1)$ is the point of tangency and $m$ is the slope of the tangent.

Q4: How do I find the equation of the tangent to the curve?

A4: To find the equation of the tangent to the curve, you need to find the point of tangency and the slope of the tangent. Then, you can use the point-slope form of a line to find the equation of the tangent.

Q5: What is the equation of the tangent to the curve $y = x \sqrt{x+2}$ ?

A5: The equation of the tangent to the curve $y = x \sqrt{x+2}$ is $y = \frac{3x_1}{2 \sqrt{x_1+2}}x + \frac{x_1}{2 \sqrt{x_1+2}}$ , where $(x_1, y_1)$ is the point of tangency.

Q6: How do I find the point of tangency?

A6: To find the point of tangency, you need to find the value of $x$ at which the tangent touches the curve. This can be done by finding the value of $x$ at which the derivative of the curve is equal to the slope of the tangent.

Q7: What is the relationship between the curve and its derivative?

A7: The derivative of a curve represents the rate of change of the curve with respect to the variable. In this case, the derivative of the curve $y = x \sqrt{x+2}$ is $\frac{3x}{2 \sqrt{x+2}}$ .

Q8: How do I use the derivative to find the equation of the tangent?

A8: To use the derivative to find the equation of the tangent, you need to find the point of tangency and the slope of the tangent. Then, you can use the point-slope form of a line to find the equation of the tangent.

Q9: What is the significance of the point-slope form of a line?

A9: The point-slope form of a line is a way to represent a line in terms of its slope and a point on the line. In this case, the point-slope form of a line is used to find the equation of the tangent to the curve.

Q10: How do I apply the concepts learned in this article to real-world problems?

A10: The concepts learned in this article can be applied to real-world problems in various fields, such as physics, engineering, and economics. For example, the concept of finding the equation of a tangent to a curve can be used to model the behavior of a physical system or to optimize a function.

Conclusion

In this article, we answered some frequently asked questions related to finding the equation of the tangent to a curve. We hope that this article has provided a clear understanding of the concepts involved and has helped to clarify any doubts that you may have had.

References

[1] Calculus, by Michael Spivak
[2] Differential Calculus, by James Stewart
[3] Calculus: Early Transcendentals, by James Stewart