A piece of land is shaped like a right triangle. Two people start at the right angle at the same time, and walk at the same speed along different legs of the triangle while spraying the land. If the area covered is changing at 4m^2/s how fast are the people moving when they are 5m from the right angle? (Round
approximations to two decimal places.)
Well, isn't this a right-angled riddle with some spritz of humor! Let's tackle it step by step.
First, let's call the two legs of the triangle "leg A" and "leg B," with leg A being the vertical leg and leg B being the horizontal leg. The hypotenuse of our right triangle will be "hypotenuse C."
Now, we know that the area of a triangle is given by the formula: A = (1/2) * base * height.
Since the area is changing at a rate of 4m^2/s, we can find the rate at which the people are moving by taking the derivative of the area with respect to time.
Let's differentiate both sides of the equation:
dA/dt = (1/2) * (db/dt * h + b * dh/dt)
Where:
dA/dt is the rate of change of the area with respect to time (4m^2/s),
db/dt is the rate of change of the base (leg B) with respect to time (which is the speed of the person moving along leg B),
dh/dt is the rate of change of the height (leg A) with respect to time (which is the speed of the person moving along leg A).
Now, at the point where the people are 5m from the right angle, we can determine the values of b and h for our triangle.
Using the Pythagorean theorem, we have:
b^2 + h^2 = C^2
Since h is the vertical leg and b is the horizontal leg, we can substitute the values:
5^2 + b^2 = C^2
Simplifying this equation, we get:
25 + b^2 = C^2
Now, let's differentiate this equation with respect to time:
2b * db/dt = 2C * dC/dt
And here is where we put on our clown noses and do some substitution!
Since we want to find the rate at which the people are moving when they are 5m from the right angle, we know that C (the hypotenuse) is constant at this point. So, dC/dt = 0.
Now, we can plug in the given values:
2b * db/dt = 2 * 5 * 0
Simplifying further:
2b * db/dt = 0
Dividing both sides by 2b:
db/dt = 0 / 2b
And voilà! The rate at which the people are moving when they are 5m from the right angle is 0.
Looks like they've sprayed their last spritz for now!
To solve this problem, we can use the formula for the area of a right triangle:
A = (1/2) * base * height
Where A is the area, base is the length of one leg, and height is the length of the other leg.
We are given that the area is changing at a rate of 4 m^2/s. We need to find the rate at which the people are moving when they are 5 m from the right angle.
Let's call the length of the base x and the length of the height y. Since the people are moving along the legs, x and y will change with time. We need to find dx/dt and dy/dt when x = 5.
Using the formula for the area, we can write:
A = (1/2) * x * y
To find dx/dt, we differentiate both sides of the equation with respect to time (t):
dA/dt = (1/2) * (x * dy/dt + y * dx/dt)
Given that dA/dt = 4, we can substitute the known values:
4 = (1/2) * (5 * dy/dt + y * dx/dt)
We are interested in finding dx/dt when x = 5, so we can substitute x = 5:
4 = (1/2) * (5 * dy/dt + y * dx/dt) -- (1)
Now, let's use the Pythagorean theorem to relate x and y:
x^2 + y^2 = h^2 -- (2)
where h is the length of the hypotenuse.
Since the triangle is right-angled, we know that:
x^2 + y^2 = h^2
5^2 + y^2 = h^2
25 + y^2 = h^2
Differentiating both sides of this equation with respect to time gives us:
0 + 2y * dy/dt = 2h * dh/dt
We are given that dy/dt = dx/dt (since both people are moving at the same speed).
Substituting dy/dt = dx/dt and h = x (because the people are 5m from the right angle) into the above equation, we get:
2y * dx/dt = 2x * dh/dt
y * dx/dt = x * dh/dt
We can rewrite dh/dt as:
dh/dt = (dh/dx) * (dx/dt)
Since h = x, dh/dx = 1, so the equation becomes:
dh/dt = (1) * dx/dt
Substituting dh/dt = dx/dt into equation (1), we get:
4 = (1/2) * (5 * dx/dt + y * dx/dt)
8 = (5 + y) * dx/dt
We know that y = 3 (since it is a right triangle), so we can substitute y = 3:
8 = (5 + 3) * dx/dt
8 = 8 * dx/dt
Now, we can solve for dx/dt:
dx/dt = 8/8
dx/dt = 1 m/s
Therefore, the people are moving at a rate of 1 m/s when they are 5m from the right angle.
To find the speed at which the people are moving, we need to use the concept of related rates.
Let's say the two legs of the right triangle are x and y, and they are changing with respect to time t. The area of the triangle is given by A = 0.5 * x * y.
We are given that the area covered is changing at a rate of dA/dt = 4 m^2/s. We want to find the rate at which the people are moving, which is dx/dt and dy/dt when x = 5 m.
Using the chain rule, we can express dA/dt in terms of dx/dt and dy/dt:
dA/dt = (dA/dx) * (dx/dt) + (dA/dy) * (dy/dt),
where (dA/dx) and (dA/dy) represent the partial derivatives of A with respect to x and y, respectively.
Taking the partial derivatives of A = 0.5 * x * y, we get:
(dA/dx) = 0.5 * y and (dA/dy) = 0.5 * x.
Substituting these values back into the chain rule equation, we have:
4 = (0.5 * y) * (dx/dt) + (0.5 * x) * (dy/dt).
Now, we need to find the relationship between x, y, and the distance from the right angle of the triangle which is 5m. Let's call this distance z.
From the Pythagorean theorem, we know that x^2 + y^2 = z^2.
Differentiating both sides of this equation with respect to t, we get:
2x * (dx/dt) + 2y * (dy/dt) = 2z * (dz/dt).
Since z = 5m, we can simplify the equation to:
2x * (dx/dt) + 2y * (dy/dt) = 10 * (dz/dt).
We need to solve this equation in order to find the values of dx/dt and dy/dt.
Now, substituting the relation between x, y, and z into the equation obtained from the chain rule, we get:
4 = (0.5 * y) * (dx/dt) + (0.5 * x) * (dy/dt) = 4.
Since we are given that the area covered is changing at a rate of 4 m^2/s, we have dA/dt = 4.
Substituting this value into the equation, we get:
4 = (0.5 * y) * (dx/dt) + (0.5 * x) * (dy/dt) = 4.
Now, we have a system of two equations:
2x * (dx/dt) + 2y * (dy/dt) = 10 * (dz/dt),
0.5 * y * (dx/dt) + 0.5 * x * (dy/dt) = 4.
Solving these equations simultaneously will give us the values of dx/dt and dy/dt.
length of side = v t
a = area = .5 (vt)^2 = .5 v^2 t^2
da/dt = (.5 v^2)(2 t) = v^2 t
4 = v^2 t
v t = 5
4 = 5 v
v = 4/5