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 ( Π / 3 , 2 (\pi / 3, 2 ( Π /3 , 2 ]. Write Your Answer In The Form Y = M X + B Y=mx+b Y = M X + B , Where M M M Is The Slope And B B B Is The Y-intercept.

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Introduction

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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 $(\pi / 3, 2)$. The tangent line is a line that just touches the curve at a given point, and it has the same slope as the curve at that point.

Step 1: Find the Derivative of the Function

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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.

The derivative of $\sec (x)$ is $\sec (x) \tan (x)$, and the derivative of $\cos (x)$ is $-\sin (x)$. Using the sum rule of differentiation, we can find the derivative of the function:

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

Step 2: Evaluate the Derivative at the Given Point

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

Using a calculator or a trigonometric table, we can find the values of $\sec (\pi / 3)$, $\tan (\pi / 3)$, and $\sin (\pi / 3)$:

$$\sec (\pi / 3) = 2, \tan (\pi / 3) = \sqrt{3}, \sin (\pi / 3) = \sqrt{3} / 2$$

Substituting these values into the derivative, we get:

$$\frac{dy}{dx} = 2 (2) (\sqrt{3}) + 4 (\sqrt{3} / 2) = 4 \sqrt{3} + 2 \sqrt{3} = 6 \sqrt{3}$$

Step 3: Find the Equation of the Tangent Line

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Now that we have the slope of the tangent line, we can find its equation. The equation of a line is given by $y = mx + b$, where $m$ is the slope and $b$ is the y-intercept.

We know that the tangent line passes through the point $(\pi / 3, 2)$, so we can substitute these values into the equation:

$$2 = 6 \sqrt{3} (\pi / 3) + b$$

Simplifying the equation, we get:

$$2 = 2 \pi \sqrt{3} + b$$

Subtracting $2 \pi \sqrt{3}$ from both sides, we get:

$$b = 2 - 2 \pi \sqrt{3}$$

Step 4: Write the Equation of the Tangent Line in the Required Form

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The equation of the tangent line is $y = 6 \sqrt{3} x + (2 - 2 \pi \sqrt{3})$. This is the required form, where $m$ is the slope and $b$ is the y-intercept.

Conclusion

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In this article, we found the equation of the tangent line to the curve $y=2 \sec (x)-4 \cos (x)$ at the point $(\pi / 3, 2)$. The equation of the tangent line is $y = 6 \sqrt{3} x + (2 - 2 \pi \sqrt{3})$.

Final Answer

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

References

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• [1] Calculus, 3rd edition, Michael Spivak
• [2] Trigonometry, 2nd edition, Charles P. McKeague
• [3] Calculus with Analytic Geometry, 2nd edition, George B. Thomas Jr.
  

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Introduction

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In our previous article, we found the equation of the tangent line to the curve $y=2 \sec (x)-4 \cos (x)$ at the point $(\pi / 3, 2)$. In this article, we will answer some frequently asked questions related to the topic.

Q: What is the tangent line to a curve?

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A: The tangent line to a curve is a line that just touches the curve at a given point. It has the same slope as the curve at that point.

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

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A: To find the equation of the tangent line, you need to find the derivative of the function, evaluate it at the given point, and then use the point-slope form of a line to write the equation.

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

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

Q: How do you find the slope of the tangent line?

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A: To find the slope of the tangent line, you need to find the derivative of the function and evaluate it at the given point.

Q: What is the derivative of $\sec (x)$?

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A: The derivative of $\sec (x)$ is $\sec (x) \tan (x)$.

Q: What is the derivative of $\cos (x)$?

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A: The derivative of $\cos (x)$ is $-\sin (x)$.

Q: How do you find the equation of the tangent line to a curve with a given point?

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A: To find the equation of the tangent line to a curve with a given point, you need to find the derivative of the function, evaluate it at the given point, and then use the point-slope form of a line to write the equation.

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

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A: The equation of the tangent line to the curve $y=2 \sec (x)-4 \cos (x)$ at the point $(\pi / 3, 2)$ is $y = 6 \sqrt{3} x + (2 - 2 \pi \sqrt{3})$.

Q: How do you find the y-intercept of the tangent line?

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A: To find the y-intercept of the tangent line, you need to substitute $x = 0$ into the equation of the tangent line.

Q: What is the y-intercept of the tangent line to the curve $y=2 \sec (x)-4 \cos (x)$ at the point $(\pi / 3, 2)$?

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A: The y-intercept of the tangent line to the curve $y=2 \sec (x)-4 \cos (x)$ at the point $(\pi / 3, 2)$ is $2 - 2 \pi \sqrt{3}$.

Conclusion

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In this article, we answered some frequently asked questions related to finding the equation of the tangent line to a curve. We hope that this article has been helpful in clarifying any doubts that you may have had.

Final Answer

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

References

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• [1] Calculus, 3rd edition, Michael Spivak
• [2] Trigonometry, 2nd edition, Charles P. McKeague
• [3] Calculus with Analytic Geometry, 2nd edition, George B. Thomas Jr.