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:
[tex]\displaystyle \tan(\theta)=\frac{h}{x}[/tex]
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:
[tex]\displaystyle \frac{d}{dt}[\tan(\theta)]=\frac{d}{dt}\Big[\frac{h}{x}\Big][/tex]
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:
[tex]\displaystyle \sec^2(\theta)\frac{d\theta}{dt}=\frac{1}{x}\frac{dh}{dt}[/tex]
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:
[tex]\text{Hypotenuse}=\sqrt{h^2+x^2}[/tex]
Since secant is the ratio of the hypotenuse to adjacent:
[tex]\displaystyle \sec(\theta)=\frac{\sqrt{h^2+x^2}}{x}[/tex]
So:
[tex]\displaystyle \sec^2(\theta)=\frac{x^2+h^2}{x^2}[/tex]
By substitution, we have:
[tex]\displaystyle \Big(\frac{x^2+h^2}{x^2}\Big)\frac{d\theta}{dt}=\frac{1}{x}\frac{dh}{dt}[/tex]
By multiplying both sides by the reciprocal of the term on the left:
[tex]\displaystyle \frac{d\theta}{dt}=\frac{1}{x}\Big(\frac{x^2}{x^2+h^2}\Big)\frac{dh}{dt}[/tex]
Therefore:
[tex]\displaystyle \frac{d\theta}{dt}=\frac{x}{x^2+h^2}\frac{dh}{dt}[/tex]
Our answer is A.
