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Alisiya [41]
3 years ago
13

Am i correct? calculus

Mathematics
1 answer:
Advocard [28]3 years ago
8 0
Check the picture below.

now, the "x" is a constant, the rocket is going up, so "y" is changing and so is the angle, but "x" is always just 15 feet from the observer.  That matters because the derivative of a constant is zero.

now, those are the values when the rocket is 30 feet up above.

\bf tan(\theta )=\cfrac{y}{x}\implies tan(\theta )=\cfrac{y}{15}\implies tan(\theta )=\cfrac{1}{15}\cdot y
\\\\\\
\stackrel{chain~rule}{sec^2(\theta )\cfrac{d\theta }{dt}}=\cfrac{1}{15}\cdot \cfrac{dy}{dt}\implies \cfrac{1}{cos^2(\theta )}\cdot\cfrac{d\theta }{dt}=\cfrac{1}{15}\cdot \cfrac{dy}{dt}
\\\\\\
\boxed{\cfrac{d\theta }{dt}=\cfrac{cos^2(\theta )\frac{dy}{dt}}{15}}\\\\
-------------------------------\\\\


\bf cos(\theta )=\cfrac{adjacent}{hypotenuse}\implies cos(\theta )=\cfrac{15}{15\sqrt{5}}\implies cos(\theta )=\cfrac{1}{\sqrt{5}}\\\\
-------------------------------\\\\
\cfrac{d\theta }{dt}=\cfrac{\left( \frac{1}{\sqrt{5}} \right)^2\cdot 11}{15}\implies \cfrac{d\theta }{dt}=\cfrac{\frac{1}{5}\cdot 11}{15}\implies \cfrac{d\theta }{dt}=\cfrac{\frac{11}{5}}{15}\implies \cfrac{d\theta }{dt}=\cfrac{11}{5}\cdot \cfrac{1}{15}
\\\\\\
\cfrac{d\theta }{dt}=\cfrac{11}{75}

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3 years ago
The owner of the Rancho Grande has 3,060 yd of fencing with which to enclose a rectangular piece of grazing land situated along
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Answer:

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if we solve for b, we are left with:

b = 3060-2 * a

Now for the area it would be:

A = a * b = a * (3060-2 * a )

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dA / da = d [3060 * a -2 * a ^ 2

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dA / day = 3060 - 4 * a

If we equal 0:

0 = 3060 - 4 * a

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a = 3060/4

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3 0
3 years ago
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