Find The Equation Of The Tangent Line To The Curve Y = 2 Sec ⁡ X − 4 Cos ⁡ X Y = 2 \sec X - 4 \cos X Y = 2 Sec X − 4 Cos X At The Point { \left(\frac{\pi}{3}, 2\right)$}$.The Equation Of This Tangent Line Can Be Written In The Form Y = M X + B Y = Mx + B Y = M X + B , Where:- M M M Is:

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Introduction

In calculus, the tangent line to a curve at a given point is a line that just touches the curve at that point and has the same slope as the curve at that point. The equation of the tangent line can be written in the form $y = mx + b$, where $m$ is the slope of the line and $b$ is the y-intercept. In this article, we will find the equation of the tangent line to the curve $y = 2 \sec x - 4 \cos x$ at the point $\left(\frac{\pi}{3}, 2\right)$.

Step 1: Find the Derivative of the Function

To find the equation of the tangent line, we need to find the derivative of the function $y = 2 \sec x - 4 \cos x$. The derivative of a function is a measure of how fast the function changes as its input changes. In this case, we need to find the derivative of the function with respect to $x$.

Using the chain rule and the fact that the derivative of $\sec x$ is $\sec x \tan x$, we can find the derivative of the function as follows:

$$\frac{dy}{dx} = 2 \sec x \tan x + 4 \sin x$$

Step 2: Evaluate the Derivative at the Given Point

Now that we have found the derivative of the function, we need to evaluate it at the given point $\left(\frac{\pi}{3}, 2\right)$. This will give us the slope of the tangent line at that point.

Using the fact that $\sec \frac{\pi}{3} = 2$ and $\tan \frac{\pi}{3} = \sqrt{3}$, we can evaluate the derivative as follows:

$$\frac{dy}{dx} \left(\frac{\pi}{3}\right) = 2 \cdot 2 \cdot \sqrt{3} + 4 \cdot \frac{\sqrt{3}}{2} = 4\sqrt{3} + 2\sqrt{3} = 6\sqrt{3}$$

Step 3: Find the Equation of the Tangent Line

Now that we have found the slope of the tangent line, we can use the point-slope form of a line to find the equation of the tangent line. The point-slope form of a line is given by:

$$y - y_1 = m(x - x_1)$$

where $(x_1, y_1)$ is a point on the line and $m$ is the slope of the line.

In this case, we know that the point $(x_1, y_1) = \left(\frac{\pi}{3}, 2\right)$ and the slope $m = 6\sqrt{3}$. Plugging these values into the point-slope form, we get:

$$y - 2 = 6\sqrt{3}\left(x - \frac{\pi}{3}\right)$$

Step 4: Simplify the Equation of the Tangent Line

To simplify the equation of the tangent line, we can expand the right-hand side and collect like terms.

$$y - 2 = 6\sqrt{3}x - 2\pi\sqrt{3}$$

$$y = 6\sqrt{3}x - 2\pi\sqrt{3} + 2$$

Conclusion

In this article, we found the equation of the tangent line to the curve $y = 2 \sec x - 4 \cos x$ at the point $\left(\frac{\pi}{3}, 2\right)$. The equation of the tangent line is given by:

$$y = 6\sqrt{3}x - 2\pi\sqrt{3} + 2$$

This equation represents a line that just touches the curve at the point $\left(\frac{\pi}{3}, 2\right)$ and has the same slope as the curve at that point.

Discussion

The equation of the tangent line can be written in the form $y = mx + b$, where $m$ is the slope of the line and $b$ is the y-intercept. In this case, the slope of the tangent line is $6\sqrt{3}$ and the y-intercept is $2 - 2\pi\sqrt{3}$.

The equation of the tangent line can be used to approximate the value of the function at a given point. For example, if we want to approximate the value of the function at the point $x = \frac{\pi}{3} + \delta x$, we can use the equation of the tangent line to get:

$$y \approx 6\sqrt{3}\left(\frac{\pi}{3} + \delta x\right) - 2\pi\sqrt{3} + 2$$

This approximation can be used to estimate the value of the function at a given point.

Final Answer

The final answer is $\boxed{y = 6\sqrt{3}x - 2\pi\sqrt{3} + 2}$.

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Q: What is the equation of the tangent line to the curve $y = 2 \sec x - 4 \cos x$ at the point $\left(\frac{\pi}{3}, 2\right)$?

A: The equation of the tangent line is given by $y = 6\sqrt{3}x - 2\pi\sqrt{3} + 2$.

Q: How do you find the derivative of the function $y = 2 \sec x - 4 \cos x$?

A: To find the derivative of the function, we use the chain rule and the fact that the derivative of $\sec x$ is $\sec x \tan x$. The derivative of the function is given by $\frac{dy}{dx} = 2 \sec x \tan x + 4 \sin x$.

Q: How do you evaluate the derivative at the given point $\left(\frac{\pi}{3}, 2\right)$?

A: To evaluate the derivative at the given point, we substitute the values of $x$ and $\sec x$ into the derivative. Using the fact that $\sec \frac{\pi}{3} = 2$ and $\tan \frac{\pi}{3} = \sqrt{3}$, we get $\frac{dy}{dx} \left(\frac{\pi}{3}\right) = 4\sqrt{3} + 2\sqrt{3} = 6\sqrt{3}$.

Q: What is the slope of the tangent line at the point $\left(\frac{\pi}{3}, 2\right)$?

A: The slope of the tangent line at the point $\left(\frac{\pi}{3}, 2\right)$ is given by $6\sqrt{3}$.

Q: How do you find the equation of the tangent line using the point-slope form?

A: To find the equation of the tangent line, we use the point-slope form of a line, which is given by $y - y_1 = m(x - x_1)$. In this case, we know that the point $(x_1, y_1) = \left(\frac{\pi}{3}, 2\right)$ and the slope $m = 6\sqrt{3}$. Plugging these values into the point-slope form, we get $y - 2 = 6\sqrt{3}\left(x - \frac{\pi}{3}\right)$.

Q: How do you simplify the equation of the tangent line?

A: To simplify the equation of the tangent line, we expand the right-hand side and collect like terms. The simplified equation of the tangent line is given by $y = 6\sqrt{3}x - 2\pi\sqrt{3} + 2$.

Q: What is the y-intercept of the tangent line?

A: The y-intercept of the tangent line is given by $2 - 2\pi\sqrt{3}$.

Q: How do you approximate the value of the function at a given point using the equation of the tangent line?

A: To approximate the value of the function at a given point, we use the equation of the tangent line. For example, if we want to approximate the value of the function at the point $x = \frac{\pi}{3} + \delta x$, we can use the equation of the tangent line to get $y \approx 6\sqrt{3}\left(\frac{\pi}{3} + \delta x\right) - 2\pi\sqrt{3} + 2$.

Q: What is the final answer to the problem?

A: The final answer to the problem is $\boxed{y = 6\sqrt{3}x - 2\pi\sqrt{3} + 2}$.

Discussion

The equation of the tangent line can be used to approximate the value of the function at a given point. The slope of the tangent line is given by $6\sqrt{3}$, and the y-intercept is given by $2 - 2\pi\sqrt{3}$. The equation of the tangent line can be used to solve a variety of problems, including finding the maximum and minimum values of the function.

Final Answer

The final answer is $\boxed{y = 6\sqrt{3}x - 2\pi\sqrt{3} + 2}$.