Explanation Given: The below figure shows the geometrical interpretation of the derivative formula ( sin θ ) ′ = cos θ Segments B A ¯ and B D ¯ are parallel to x and y -axes. Δ sin θ = sin ( θ + h ) − sin θ Calculation: Let us consider the given figure as shown. From the above figure we have the following: In Δ O A M , sin ( θ ) = A M ......(1) In Δ O C N sin ( θ + h ) = C N To determine (b) To verify: The statement ∠ B D A = θ for the given figure. To determine (c) To verify: The statement B D = ( cos θ ) A D for the given figure. To show that for h → 0 Δ sin θ ≈ h cos θ and ( sin θ ) ′ = cos θ .

3rd Edition

ISBN: 9781464125263

Author: Jon Rogawski, Colin Adams

Publisher: W.H. Freeman & Co

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Question

Chapter 3.6, Problem 58E

To determine

(a)

To verify:

The statement Δsinθ=BC for the given figure.

To determine

(b)

To verify:

The statement ∠BDA=θ for the given figure.

To determine

(c)

To verify:

The statement BD=(cosθ)AD for the given figure.

To show that for h→0

Δsinθ≈hcosθ and (sinθ)′=cosθ.

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Chapter 3.6, Problem 57E

Chapter 3.7, Problem 1PQ

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Chapter 3 Solutions

Calculus - Standalone book

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(a) To verify: The statement Δ sin θ = B C for the given figure.

Solution Summary: The author illustrates the geometrical interpretation of the derivative formula for the given figure.

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(a)

To verify:

The statement Δsinθ=BC for the given figure.

(b)

To verify:

The statement ∠BDA=θ for the given figure.

(c)

To verify:

The statement BD=(cosθ)AD for the given figure.

To show that for h→0

Δsinθ≈hcosθ and (sinθ)′=cosθ.

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Chapter 3 Solutions

(a) To verify: The statement Δ sin θ = B C for the given figure.

Solution Summary: The author illustrates the geometrical interpretation of the derivative formula for the given figure.

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(a) To verify: The statement Δsinθ=BC for the given figure.

(b) To verify: The statement ∠BDA=θ for the given figure.

(c) To verify: The statement BD=(cosθ)AD for the given figure. To show that for h→0 Δsinθ≈hcosθ and (sinθ)′=cosθ.

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Chapter 3 Solutions

(a)

To verify:

The statement Δsinθ=BC for the given figure.

(b)

To verify:

The statement ∠BDA=θ for the given figure.

(c)

To verify:

The statement BD=(cosθ)AD for the given figure.

To show that for h→0

Δsinθ≈hcosθ and (sinθ)′=cosθ.