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katovenus [111]
3 years ago
8

Need help #3. The answer is shown, but I don’t know how to get to the answer. Please teach and show steps.

Mathematics
1 answer:
Sunny_sXe [5.5K]3 years ago
3 0

Answer:

A

Step-by-step explanation:

We are given a right triangle with a base of <em>x</em> feet and a height of <em>h</em> feet, where <em>x</em> is constant and <em>h</em> changes with respect to time <em>t</em>.

The angle in radians is defined by:

\displaystyle \tan(\theta)=\frac{h}{x}

And we want to find the relationship that describes dθ/dt and dh/dt.

So, we will differentiate both sides with respect to <em>t</em> where <em>x</em> is a constant:

\displaystyle \frac{d}{dt}[\tan(\theta)]=\frac{d}{dt}\Big[\frac{h}{x}\Big]

Differentiate. Apply the chain rule on the left. Again, remember that <em>x</em> is just a constant, so we can move it outside the derivative operator. Therefore:

\displaystyle \sec^2(\theta)\frac{d\theta}{dt}=\frac{1}{x}\frac{dh}{dt}

Since we know that tan(θ)=h/x, <em>h</em> is the opposite side of our triangle and <em>x</em> is the adjacent. Therefore, by the Pythagorean Theorem, our hypotenuse will be:

\text{Hypotenuse}=\sqrt{h^2+x^2}

Since secant is the ratio of the hypotenuse to adjacent:

\displaystyle \sec(\theta)=\frac{\sqrt{h^2+x^2}}{x}

So:

\displaystyle \sec^2(\theta)=\frac{x^2+h^2}{x^2}

By substitution, we have:

\displaystyle \Big(\frac{x^2+h^2}{x^2}\Big)\frac{d\theta}{dt}=\frac{1}{x}\frac{dh}{dt}

By multiplying both sides by the reciprocal of the term on the left:

\displaystyle \frac{d\theta}{dt}=\frac{1}{x}\Big(\frac{x^2}{x^2+h^2}\Big)\frac{dh}{dt}

Therefore:

\displaystyle \frac{d\theta}{dt}=\frac{x}{x^2+h^2}\frac{dh}{dt}

Our answer is A.

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