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Aneli [31]
3 years ago
8

AP Calc AB:

Advanced Placement (AP)
1 answer:
Softa [21]3 years ago
4 0

Differentiate both sides of

2 + 2 sin(<em>x</em>) = 3 tan(<em>y</em>)

with respect to a new variable <em>t</em>, assuming both <em>x</em> and <em>t</em> depend on <em>t</em>, which will involve the chain rule:

d(2 + 2 sin(<em>x</em>))/d<em>t</em> = d(3 tan(<em>y</em>))/d<em>t</em>

d(2)/d<em>t</em> + d(2 sin(<em>x</em>))/d<em>t</em> = 3 d(tan(<em>y</em>))/d<em>t</em>

0 + 2 d(sin(<em>x</em>))/d<em>t</em> = 3 sec²(<em>y</em>) d<em>y</em>/d<em>t</em>

2 cos(<em>x</em>) d<em>x</em>/d<em>t</em> = 3 sec²(<em>y</em>) d<em>y</em>/d<em>t</em>

Now, when <em>x</em> = <em>π</em>/6, <em>y</em> = <em>π</em>/4, and d<em>x</em>/d<em>t</em> = 2, solve for d<em>y</em>/d<em>t</em> :

2 cos(<em>π</em>/6) × 2 = 3 sec²(<em>π</em>/4) d<em>y</em>/d<em>t</em>

d<em>y</em>/d<em>t</em> = 4 cos(<em>π</em>/6) / (3 sec²(<em>π</em>/4))

d<em>y</em>/d<em>t</em> = 4 (√3/2) / (3 (√2)²)

d<em>y</em>/d<em>t</em> = √3/3 = 1/√3

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