Find the exact length of the curve. x = 9 + 9t2, y = 6 + 6t3, 0 ≤ t ≤ 4

Mathematics

1 answer:

andrew11 [14]3 years ago

3 0

ANSWER

$=102 \sqrt{ 17}$

EXPLANATION

The equation of the curve is:

$x = 9 + 9 {t}^{2}$

$y = 6 + 6 {t}^{3}$

We differentiate to obtain:

$\frac{dx}{dt} = 18t$

$\frac{dy}{dt} = 18t$

The length of the arc

$l = \int _0 ^{4} \sqrt{ {( \frac{dx}{dt} )}^{2} + {( \frac{dy}{dt} )}^{2} } dt$

This implies that:

$= \int _0 ^{4} \sqrt{ {( 18t )}^{2} + {( {18 {t}^{2} } )}^{2} } dt$

$= \int _0 ^{4} 18t \sqrt{ 1+ { {18 {t}^{2} } }} dt$

$=102 \sqrt{ 17}$

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Please help me to prove this!

Sophie [7]

Answer: see proof below

Step-by-step explanation:

Given: A + B = C → A = C - B

→ B = C - A

Use the Double Angle Identity: cos 2A = 2 cos² A - 1

→ (cos 2A + 1)/2 = cos² A

Use Sum to Product Identity: cos A + cos B = 2 cos [(A + B)/2] · 2 cos [(A - B)/2]

Use Even/Odd Identity: cos (-A) = cos (A)

Proof LHS → RHS:

LHS: cos² A + cos² B + cos² C

$\text{Double Angle:}\qquad \dfrac{\cos 2A+1}{2}+\dfrac{\cos 2B+1}{2}+\cos^2 C\\\\\\.\qquad \qquad \qquad =\dfrac{1}{2}\bigg(2+\cos 2A+\cos 2B\bigg)+\cos^2 C\\\\\\.\qquad \qquad \qquad =1+\dfrac{1}{2}\bigg(\cos 2A+\cos 2B\bigg)+\cos^2 C$

$\text{Sum to Product:}\quad 1+\dfrac{1}{2}\bigg[2\cos \bigg(\dfrac{2A+2B}{2}\bigg)\cdot \cos \bigg(\dfrac{2A-2B}{2}\bigg)\bigg]+\cos^2 C\\\\\\.\qquad \qquad \qquad =1+\cos (A+B)\cdot \cos (A-B)+\cos^2 C$

$\text{Given:}\qquad \qquad 1+\cos C\cdot \cos (A-B)+\cos^2C$

$\text{Factor:}\qquad \qquad 1+\cos C[\cos (A-B)+\cos C]$

$\text{Sum to Product:}\quad 1+\cos C\bigg[2\cos \bigg(\dfrac{A-B+C}{2}\bigg)\cdot \cos \bigg(\dfrac{A-B-C}{2}\bigg)\bigg]\\\\\\.\qquad \qquad \qquad =1+2\cos C\cdot \cos \bigg(\dfrac{A+(C-B)}{2}\bigg)\cdot \cos \bigg(\dfrac{-B-(C-A)}{2}\bigg)$

$\text{Given:}\qquad \qquad =1+2\cos C\cdot \cos \bigg(\dfrac{A+A}{2}\bigg)\cdot \cos \bigg(\dfrac{-B-B}{2}\bigg)\\\\\\.\qquad \qquad \qquad =1+2\cos C \cdot \cos A\cdot \cos (-B)$