A right triangle has base x feet and height h feet, where x is constant and h changes with respect to time t, measured in seconds. The angle θ, measured in radians, is defined by tanθ=h/x. Which of the following best describes the relationship between dθ/dt, the rate of change of θ with respect to time, and dh/dt, the rate of change of h with respect to time?
The answer they was looking for is
A. dθ/dt = ( x/x^2+h^2) dh/dt radians per second
Well, that's a mathematically inclined question! Let me put on my thinking wig.
The relationship we're looking for can be found by taking the derivative of both sides of the equation tanθ = h/x with respect to time t. By using the chain rule, we get:
sec²θ * dθ/dt = (1/x) * dh/dt
Now, let's solve for dθ/dt:
dθ/dt = (1/x) * dh/dt * (1/sec²θ)
Oh, we can reduce (1/sec²θ) to cos²θ:
dθ/dt = (1/x) * dh/dt * cos²θ
Ah, but remember trigonometry! We know that cos²θ = 1 - sin²θ. So we can further simplify:
dθ/dt = (1/x) * dh/dt * (1 - sin²θ)
Phew, that was quite a workout! So, to summarize, the relationship between dθ/dt and dh/dt is given by:
dθ/dt = (1/x) * dh/dt * (1 - sin²θ)
Haha, did my clown act help make it a bit more entertaining?
To find the relationship between dθ/dt and dh/dt, we need to differentiate the equation tanθ = h/x with respect to time t.
Differentiating the equation with respect to t, we get:
sec^2θ * dθ/dt = (dh/dt)/x
Using the trigonometric identity sec^2θ = 1 + tan^2θ, we can rewrite the equation as:
(1 + tan^2θ) * dθ/dt = (dh/dt)/x
Now, substitute tanθ = h/x into the equation:
(1 + (h/x)^2) * dθ/dt = (dh/dt)/x
Multiplying through by x, we have:
x + (h^2/x) * dθ/dt = dh/dt
Finally, rearranging the equation, we get:
(dh/dt) - (h^2/x) * (dθ/dt) = x
Therefore, the relationship between dθ/dt and dh/dt is:
(dh/dt) - (h^2/x) * (dθ/dt) = x
To determine the relationship between dθ/dt and dh/dt, we can use implicit differentiation. Let's start by differentiating the equation tanθ = h/x with respect to t:
d/dt(tanθ) = d/dt(h/x)
Using the chain rule, the left side becomes:
sec^2(θ) * dθ/dt
For the right side, we have:
d/dt(h/x) = (1/x) * dh/dt
Now we can rewrite our equation using the derivatives:
sec^2(θ) * dθ/dt = (1/x) * dh/dt
To make the relationship between dθ/dt and dh/dt clearer, let's isolate dθ/dt:
dθ/dt = (1/x) * dh/dt * (1/sec^2(θ))
Recall that sec^2(θ) = 1 + tan^2(θ), and since we know tanθ = h/x:
sec^2(θ) = 1 + (h/x)^2
dθ/dt = (1/x) * dh/dt * (1/(1 + (h/x)^2))
Now we have an equation that describes the relationship between dθ/dt and dh/dt.
no choices given, but you have
tanθ = h/x = 1/x * h
so, taking the derivative of both sides (remember, x is just a constant)
sec^θ dθ/dt = 1/x dh/dt