A right triangle initially has side lengths of 3, 4, and 5 cm. Suppose the small side is increased at a rate of 1 cm/sec, while the length of the height is kept constant. How fast is the smallest angle of the right triangle changing when it reaches the value of rad ?
A) rad/sec
B) rad/sec
C) rad/sec
D) We cannot answer this question because the function representing the smallest angle is not differentiable.
E) We cannot answer this question because it depends on the time which is unknown.

Question

A right triangle initially has side lengths of 3, 4, and 5 cm. Suppose the small side is increased at a rate of 1 cm/sec, while the length of the height is kept constant. How fast is the smallest angle of the right triangle changing when it reaches the value of  rad ?
A) rad/sec 
B) rad/sec 
C) rad/sec 
D) We cannot answer this question because the function representing the smallest angle is not differentiable. 
E) We cannot answer this question because it depends on the time which is unknown.

Quizplus · Accepted Answer

Let's denote the smallest angle of the triangle as $	heta$. From the given information, we know that $\sin(	heta)=\frac{3}{5}$, so $	heta=\sin^{-1}\left(\frac{3}{5}ight)$. Differentiating this equation with respect to time $t$, we get:
$$\frac{d	heta}{dt}=\frac{d}{dt}\left[\sin^{-1}\left(\frac{3}{5}ight)ight]=\frac{1}{\sqrt{1-\left(\frac{3}{5}ight)^2}}\cdot\frac{d}{dt}\left(\frac{3}{5}ight)=-\frac{5}{4}\cdot\frac{d}{dt}\left(\frac{3}{5}ight)=-\frac{3}{4}\cdot\frac{d}{dt}\left(\frac{5}{3}ight)$$
(Note that we used the chain rule, the derivative of $\sin^{-1}(x)$ is $\frac{1}{\sqrt{1-x^2}}$)

Now, let's consider the right triangle at a certain time $t$, when the small side has length $x$. Using the Pythagorean theorem, the equation of the hypotenuse is $y=\sqrt{x^2+16}$. Since the height is constant, we know that $\frac{x}{2}\cdot y=6$, which implies $y=\frac{12}{x}$.

Differentiating the equation $y=\sqrt{x^2+16}$ with respect to t, we get:
$$\frac{dy}{dt}=\frac{x}{\sqrt{x^2+16}}\cdot\frac{dx}{dt}$$

Substituting $y=\frac{12}{x}$, we get:
$$\frac{dy}{dt}=-\frac{12}{x^2}\cdot\frac{dx}{dt}$$

At the moment when the smallest angle is 11eaa655_3b4f_ae1f_a696_49a8bbefec97_TB5596_11 rad, we have $\cos(	heta)=\frac{4}{5}$, so $y=\frac{5}{4}x$.

Substituting $y=\frac{12}{x}$ and $\frac{dy}{dt}=-\frac{12}{x^2}\cdot\frac{dx}{dt}$ into the equation above, we get:
$$-\frac{12}{x^2}\cdot\frac{dx}{dt}=\frac{x}{\sqrt{x^2+16}}\cdot\frac{dx}{dt}$$

Solving for $\frac{dx}{dt}$, we get:
$$\frac{dx}{dt}=\frac{-12}{x\sqrt{x^2+16}}$$

Substituting $x=15$ (since $\frac{5}{4}x=12$ when $	heta=11eaa655_3b4f_ae1f_a696_49a8bbefec97_TB5596_11$), we get:
$$\frac{dx}{dt}=\frac{-12}{15\sqrt{15^2+16}}=-\frac{4}{\sqrt{721}}$$

Finally, substituting into our earlier expression for $\frac{d	heta}{dt}$, we get:
$$\frac{d	heta}{dt}=-\frac{3}{4}\cdot\frac{-4}{\sqrt{721}}=\frac{3}{\sqrt{721}}$$

Thus, the correct answer is C.

A Right Triangle Initially Has Side Lengths of 3, 4