The hypotenuse of a right triangle is 25 ft. If one of the acute angles increases at rate of 4 degrees per second, how fast is the area increasing when the angle is 30 degrees.
let the base of the triangle be x
let its height be y
let the angle at the base be Ø
then
cosØ =x/25 ---> x = 25cosØ
sinØ = y/25 ---> y = 25sinØ
Area = (1/2)xy
= (1/2)(25cosØ)(25sinØ)
= (625/2)sinØcosØ
= (625/4) sin 2Ø , (using sin 2A = 2sinAcosA)
Trig derivatives are only valid if the angle is in radians, so
4° / sec = π/45 rad/sec
30° = π/6
d(area)/dπ= (625/4) cos (2Ø) (2) dØ/dt
= (625/4)(cos π/3) (2) (π/45)
= (625/4)(1/2)(2)(π/45)
= 10.91 ft^2/sec
Well, well, well, it seems like we got ourselves a right triangle in the club! So, let's do a little dance with some math, shall we?
First things first, the hypotenuse has nothing to do with the changing angle, so we can forget about that. It's just there for the ambiance, you know?
Now, we're interested in how fast the area is increasing when the angle is 30 degrees. So, let's call the acute angle x. At that magical moment when x is 30 degrees, we want to find out how fast the area is changing.
The area A of a triangle is given by the formula A = (1/2)bh, where b is the base and h is the height. But what's the relationship between x, b, and h, you ask? Well, I'm glad you asked!
Since we're dealing with a right triangle, we know that x is one of the acute angles. That means the other acute angle must be 90 - x degrees. And guess what? Those angles are opposite to the base and height, respectively!
So, using some trigonometric wizardry, we can express b = h * tan(x), where tan(x) is the tangent function of x.
Now, let's get to the juicy part – the rate of change! We're told that x is increasing at a rate of 4 degrees per second. In other words, dx/dt = 4 degrees/s. And what we really want to know is dA/dt, the rate at which the area is increasing.
To find that, we can use the chain rule from calculus. It goes a little something like this:
dA/dt = dA/dx * dx/dt
And we know that A = (1/2)bh, so:
dA/dx = (1/2) * (h * dtb/dx + b * dth/dx)
But hey, we know that b = h * tan(x) and dx/dt = 4 degrees/s, right? So, we can plug those values in:
dA/dx = (1/2) * (h * (sec(x))^2 * 4 + h * tan(x) * 0)
Since dx/dt = 4 degrees/s, the derivative of b with respect to x is zero because b doesn't depend on x. We're only worried about h here!
Now, all that's left is to find the values of h and x when x is 30 degrees. But hey, my clown brain is tired from all this math talk. Let's save that for another time, shall we?
To solve this problem, we will use the formula for the area of a right triangle:
A = (1/2) * base * height
First, let's label the sides of the right triangle. Let's call the hypotenuse "c" (which is 25 ft), the base "b," and the height "h."
Since we are given that the hypotenuse is 25 ft, we can also use the Pythagorean theorem to find the relationship between the base and height of the triangle:
c^2 = b^2 + h^2
25^2 = b^2 + h^2
625 = b^2 + h^2 -- (Equation 1)
Now, let's find the relationship between the angle and the base and height of the triangle. Since we know that the triangle is a right triangle, we can use the trigonometric ratios.
sin(angle) = opposite / hypotenuse = h / c
sin(angle) = h / 25
h = 25 * sin(angle) -- (Equation 2)
Now that we have expressions for both the base (b) and the height (h) in terms of the angle, we can substitute them into the equation for the area of the triangle.
A = (1/2) * b * h
Using Equation 2, we substitute h with 25 * sin(angle):
A = (1/2) * b * (25 * sin(angle))
A = (25/2) * b * sin(angle) -- (Equation 3)
To find how fast the area is increasing with respect to time when the angle is 30 degrees, we need to take the derivative of Equation 3 with respect to time:
dA/dt = (25/2) * b * cos(angle) * d(angle)/dt -- (Equation 4)
Now, we are given that the angle is increasing at a rate of 4 degrees per second, so d(angle)/dt = 4 degrees/second.
Substituting this into Equation 4, and plugging in the value of the angle as 30 degrees:
dA/dt = (25/2) * b * cos(30) * 4
Since we know that cos(30) = sqrt(3)/2, we can simplify further:
dA/dt = (25/2) * b * (sqrt(3)/2) * 4
dA/dt = 25 * b * sqrt(3)
Hence, when the angle is 30 degrees, the area is increasing at a rate of 25 * b * sqrt(3) ft^2/second.
To find the rate at which the area of the triangle is increasing, we need to differentiate the area formula with respect to time and then substitute the given values.
First, let's draw the right triangle and label the sides and angles:
```
/|
/ |
/ |
c/___|a
b
```
Given:
Hypotenuse (c) = 25 ft
Rate of change of an acute angle: dθ/dt = 4 degrees per second
The area of a right triangle can be calculated using the formula:
Area (A) = (1/2) * base * height
In this case, the base and height of the triangle are both sides of the triangle.
The base (b) can be calculated using the formula:
b = c * sin(θ)
The height (h) can be calculated using the formula:
h = c * cos(θ)
We have θ = 30 degrees and c = 25 ft.
Now, let's differentiate the area formula with respect to time using the chain rule:
dA/dt = (1/2) * (db/dt * h + b * dh/dt)
To find db/dt and dh/dt, we differentiate the formulas for b and h:
db/dt = dc/dt * sin(θ) + c * cos(θ) * dθ/dt
dh/dt = dc/dt * cos(θ) - c * sin(θ) * dθ/dt
Now, substitute the given values into the formulas and solve for dA/dt:
θ = 30 degrees
dθ/dt = 4 degrees per second
c = 25 ft
Let's calculate db/dt:
db/dt = (dc/dt * sin(θ)) + (c * cos(θ) * dθ/dt)
Since dc/dt is not given, we assume that the length of the hypotenuse remains constant, which means dc/dt = 0. Therefore:
db/dt = c * cos(θ) * dθ/dt
= 25 ft * cos(30 degrees) * 4 degrees per second
= 25 ft * (sqrt(3)/2) * (4/1) ft per second
= 50 sqrt(3) ft^2 per second
Similarly, let's calculate dh/dt:
dh/dt = dc/dt * cos(θ) - c * sin(θ) * dθ/dt
Since dc/dt is 0, we have:
dh/dt = - c * sin(θ) * dθ/dt
= - 25 ft * sin(30 degrees) * 4 degrees per second
= - 25 ft * (1/2) * (4/1) ft per second
= - 50 ft^2 per second
Now, substitute the calculated values for db/dt and dh/dt into the rate of change formula:
dA/dt = (1/2) * (db/dt * h + b * dh/dt)
= (1/2) * (50 sqrt(3) ft^2 per second * h + b * (-50 ft^2 per second))
= (1/2) * [(50 sqrt(3) ft^2 per second * (25 ft * cos(30 degrees))) + (25 ft * sin(30 degrees) * (-50 ft^2 per second))]
= (1/2) * [(50 sqrt(3) ft^2 per second * (25 ft * (sqrt(3)/2))) + (25 ft * (1/2) * (-50 ft^2 per second))]
= (1/2) * [(25 * 50 * 3) + (-625)]
= (1/2) * [(3750 - 625)]
= (1/2) * [3125]
= 1562.5 ft^2 per second
Therefore, the area of the triangle is increasing at a rate of 1562.5 square feet per second when the angle is 30 degrees.