Let θ (in radians) be an acute angle in a right triangle and let x and y, respectively, be the lengths of the sides adjacent to and opposite θ. Suppose also that x and y vary with time.
At a certain instant x=7 units and is increasing at 3 unit/s, while y=1 and is decreasing at 18 units/s.
How fast is θ changing at that instant?
To find how fast θ is changing at a given instant, we need to find the derivative of θ with respect to time. In other words, we need to find dθ/dt.
From the given information, we know that x and y are changing with respect to time. Let's express their rates of change as dx/dt and dy/dt.
Given:
x = 7 units (length of the side adjacent to θ)
dx/dt = 3 unit/s (rate at which x is increasing)
y = 1 unit (length of the side opposite θ)
dy/dt = -18 unit/s (rate at which y is decreasing)
We can use the tangent function to relate x, y, and θ in a right triangle:
tan(θ) = y / x
Differentiating both sides of the equation with respect to time (t) gives us:
sec^2(θ) * dθ/dt = (dy/dt * x - y * dx/dt) / x^2
Now let's substitute the given values into the equation:
sec^2(θ) * dθ/dt = (-18 * 7 - 1 * 3) / 7^2
sec^2(θ) * dθ/dt = (-126 - 3) / 49
sec^2(θ) * dθ/dt = -129 / 49
To find dθ/dt, we need to solve for it:
dθ/dt = (-129 / 49) / sec^2(θ)
Since θ is an acute angle, we can rewrite sec^2(θ) as 1 + tan^2(θ):
dθ/dt = (-129 / 49) / (1 + tan^2(θ))
Now we have an expression for the rate of change of θ at a given instant. However, since the value of θ is not provided in the question, we cannot find the numerical value of dθ/dt without knowing the value of θ.
The equation is supposedly
[cos^2(theta)/x^2]*[(x*dy/dt)-(y*dx/dt)]
you know that
tanθ = y/x
sec^2θ dθ/dt = (x dy/dt - y dx/dt)/x^2
so just plug in your numbers.