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junior college 1 | H2 Maths
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junior college 1 chevron_right H2 Maths chevron_right Singapore

hi can you help me with this qn thanks

Date Posted: 3 years ago
Views: 603

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By the cosine rule,
PR² = PQ² + QR² - 2(PQ)(PR) cos∠PQR
PR² = r² + 1² - 2(r)(1) cos θ
PR² = r² + 1 - 2r cos θ
PR = [r² + 1 - 2r cos θ]¹/²
Since θ is sufficiently small, cos θ ≈ 1 - ½θ²
Then PR ≈ [r² + 1 - 2r(1 - ½θ²)]¹/²
PR ≈ [r² + 1 - 2r + rθ²]¹/²
PR ≈ [(r - 1)² + rθ²]¹/² (since r² + 1 - 2r = r² - 2(r)(1) + 1² = (r - 1)²
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J
J's answer
1024 answers (A Helpful Person)
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J
J
3 years ago
PR ≈ [(r - 1)² + rθ²]¹/²

We will use the Maclaurin expansion formula f(x) = f(0) + x f'(0) + x²/2! f"(0) + ...

And we know a and r are constants.

Let f(θ) = [(r - 1)² + rθ²]¹/²


f'(θ) = ½[(r - 1)² + rθ²]-¹/² (2rθ)
f'(θ) = rθ[(r - 1)² + rθ²]-¹/²


f"(θ) = r[(r - 1)² + rθ²]-¹/² + rθ(-½)[(r - 1)² + rθ²]-³/² (2rθ)
f"(θ) = r[(r - 1)² + rθ²]-¹/² - r²θ²[(r - 1)² + rθ²]-³/²


When θ = 0,

f(0) = [(r - 1)² + r(0²)]¹/²
f(0) = [(r - 1)²]¹/²
f(0) = r - 1

f'(0) = r(0)[(r - 1)² + r(0²)]-¹/²
f'(0) = 0

f"(0) = r[(r - 1)² + r(0²)]-¹/² - r²(0²)[(r - 1)² + rθ²]-³/²
f"(0) = r[(r - 1)²]-¹/²
f"(0) = r/(r - 1)


So,

PR ≈ f(θ) = f(0) + θ f'(0) + θ/2! f"(0) + ...

PR ≈ r - 1 + θ (0) + θ²/2! (r/(r - 1)) + ...

PR ≈ r - 1 + r/(2(r - 1)) θ²


Comparing this to PR ≈ ¼ + aθ²,

r - 1 = ¼
r = 5/4

And

a = r/(2(r - 1))
a = (5/4) / (2(5/4 - 1))
a = (5/4) / ½
a = 5/2
J
J
3 years ago
Edit :

It should be :

PR ≈ f(θ) = f(0) + θ f'(0) + θ²/2! f"(0) + ...

instead of
PR ≈ f(θ) = f(0) + θ f'(0) + θ/2! f"(0) + ...

(Missing ² )