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Ch. 3 - Derivatives
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 3 - DerivativesProblem 3.5.47a
Chapter 3, Problem 3.5.47a

In Exercises 47 and 48, find an equation for


(a) the tangent line to the curve at P


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Verified step by step guidance
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Step 1: Identify the given curve equation, which is y = 4 + cot(x) - 2csc(x), and the point P(π/2, 2) where the tangent line is to be found.
Step 2: Compute the derivative of the curve equation, dy/dx, to find the slope of the tangent line. Use the derivatives of cot(x) and csc(x): d(cot(x))/dx = -csc^2(x) and d(csc(x))/dx = -csc(x)cot(x).
Step 3: Substitute x = π/2 into the derivative dy/dx to calculate the slope of the tangent line at point P. Note that cot(π/2) = 0 and csc(π/2) = 1.
Step 4: Use the point-slope form of the equation for a line, y - y1 = m(x - x1), where m is the slope found in Step 3 and (x1, y1) is the point P(π/2, 2).
Step 5: Simplify the equation obtained in Step 4 to express the tangent line equation in slope-intercept form or another preferred format.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Derivative

The derivative of a function at a point provides the slope of the tangent line to the curve at that point. It is a fundamental concept in calculus used to determine rates of change. For the function y = 4 + cot x - 2 csc x, finding the derivative will help us calculate the slope of the tangent line at point P(π/2, 2).
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Derivatives

Trigonometric Derivatives

Understanding the derivatives of trigonometric functions like cotangent and cosecant is crucial for solving this problem. The derivative of cot x is -csc^2 x, and the derivative of csc x is -csc x cot x. These derivatives are used to find the slope of the tangent line to the curve at a given point.
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Derivatives of Other Inverse Trigonometric Functions

Equation of a Tangent Line

The equation of a tangent line to a curve at a point can be expressed as y - y1 = m(x - x1), where m is the slope found using the derivative, and (x1, y1) is the point of tangency. For point P(π/2, 2), once the slope is determined, this formula helps in writing the equation of the tangent line.
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Equations of Tangent Lines
Related Practice
Textbook Question

Differentiability and Continuity on an Interval


Each figure in Exercises 45–50 shows the graph of a function over a closed interval D. At what domain points does the function appear to be


a. differentiable?


Give reasons for your answers.

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Textbook Question

Use the linear approximation (1 + x)ᵏ ≈ 1 + kx to find an approximation for the function f(x) for values of x near zero.


a. f(x) = (1 − x)⁶

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Textbook Question

The radius r of a circle is measured with an error of at most 2%. What is the maximum corresponding percentage error in computing the circle’s


a. circumference?

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Textbook Question

Tolerance


a. About how accurately must the interior diameter of a 10-m-high cylindrical storage tank be measured to calculate the tank’s volume to within 1% of its true value?

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Textbook Question

Quadratic approximations


a. Let Q(x) = b₀ + b₁(x − a) + b₂(x − a)² be a quadratic approximation to f(x) at x = a with these properties:


i. Q(a) = f(a)

ii. Q′(a) = f′(a)

iii. Q″(a) = f″(a).


Determine the coefficients b₀, b₁, and b₂.

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Textbook Question

Diagonals If x, y, and z are lengths of the edges of a rectangular box, then the common length of the box’s diagonals is s = √(x² + y² + z²).

a. Assuming that x, y, and z are differentiable functions of t, how is ds/dt related to dx/dt, dy/dt, and dz/dt?

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