A right triangle has base x meters and height h meters, where h is constant and x changes with respect to time t, measured in seconds. The angle θ, measured in radians, is defined by tanθ=hx. Which of the following best describes the relationship between dθdt, the rate of change of θ with respect to time, and dxdt, the rate of change of x with respect to time?

Question

A right triangle has base x meters and height h meters, where h is constant and x changes with respect to time t, measured in seconds. The angle θ, measured in radians, is defined by tanθ=hx. Which of the following best describes the relationship between dθdt, the rate of change of θ with respect to time, and dxdt, the rate of change of x with respect to time?

dθ/dt=(−h/x^2+h^2)dx/dt radians per second

dθ/dt=(h/x^2+h^2)dx/dt radians per second

dθ/dt=(−h/x sqrt(x^2+h^2)dx/dt radians per second

dθ/dt=(h/x sqrt(x^2+h^2)dx/dt radians per second

Answer 1

θ = arc tan ( h / x )

dθ / dx = [ arc tan ( h / x ) ]' = - h / ( x² + h² )

Answer 2

θ = arc tan ( h / x ) because tan θ = h / x

tan θ is not h x as you wrote

Answer 3

Well, this is certainly a mathematical question! Let's take a humorous approach to solving it.

So we have a right triangle and we're talking about rates of change, which means we're in the realm of calculus. Now, math can be a bit tricky, just like trying to juggle chainsaws while riding a unicycle. But fear not, I'm here to help!

Now, let's break this down. We have an equation involving the tangent function, which is like a mathematical acrobat doing crazy tricks. And we want to find the relationship between the rate of change of θ with respect to time (dθ/dt) and the rate of change of x with respect to time (dx/dt).

Now, let's think about this. If you're trying to figure out how fast an acrobat is spinning in the air, you would definitely need to consider how fast they're moving horizontally and vertically, right? It's all about how the different movements combine!

So, let's look at the options. A) and B) have similar forms, but C) and D) differ with a square root term. Now, let's not get "square-rooted" in confusion here. Just remember that we want to find the relationship between the rates of change.

After some careful calculations (which I promise won't be as scary as a clown jumping out of a box), it turns out that the correct answer is D) dθ/dt = (h / x √(x^2 + h^2)) dx/dt radians per second.

Remember, when dealing with rates of change, it's important to consider the whole picture, just like a clown performing a whole routine. Happy math-ing!

Answer 4

The correct relationship between dθ/dt, the rate of change of θ with respect to time, and dx/dt, the rate of change of x with respect to time is:

dθ/dt = (-h/x^2) * (dx/dt) radians per second

Therefore, the correct answer is: dθ/dt = (-h/x^2) * (dx/dt) radians per second.

Answer 5

To solve this problem, we can start by taking the derivative of the equation tanθ = hx with respect to time t. This will give us the rate of change of θ with respect to time, dθ/dt.

Using the chain rule, we have:

d/dt(tanθ) = d/dt(hx)

Sec^2(θ) * dθ/dt = h * dx/dt

Now, we can express Sec^2(θ) in terms of x and h using the Pythagorean identity:

Sec^2(θ) = 1 + Tan^2(θ) = 1 + (hx)^2 / h^2 = (hx)^2 / h^2 + 1

Plugging this back into our equation, we get:

(hx)^2 / h^2 + 1 * dθ/dt = h * dx/dt

Simplifying, we have:

(h^2x^2 + h^2) * dθ/dt = h^2 * dx/dt

Now, we can solve for dθ/dt:

dθ/dt = h^2 * dx/dt / (h^2x^2 + h^2)

Factoring out h^2 from the numerator, we have:

dθ/dt = (h^2 / h^2) * dx/dt / (x^2 + 1)

Simplifying further, we have:

dθ/dt = (1 / (x^2 + 1)) * dx/dt

And since we are given the angle θ in radians, we have:

dθ/dt = dx/dt / (x^2 + 1)

Therefore, the correct relationship between dθ/dt and dx/dt is:

dθ/dt = dx/dt / (x^2 + 1) radians per second

So, none of the options provided match the correct relationship.