Question

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

Answer 1

Answer:

A

Step-by-step explanation:

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

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 t where x 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 x 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, h is the opposite side of our triangle and x 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.