Ferris wheel has radius of 26m. At what rate is the rider's height increasing, when rider is 25 m from ground.
You did not tell us:
1. radians per second or rpm or something about speed.
2. jow high the axis of the wheel is.
However, given axis h, with 26 <= h <= 51 (why?) and angular speed ω, the height
h(t) = h + 26sin(ωt)
dh/dt = 26ω cos(ωt)
so, solve for t when h=25, and then find dh/dt.
To find the rate at which the rider's height is increasing, we can use the concept of related rates from calculus. We will use the formula for the height of the rider on a Ferris wheel, which is given by:
h² = r² - d²
where:
h represents the height of the rider above the ground
r represents the radius of the Ferris wheel
d represents the distance of the rider from the center of the Ferris wheel
In this case, we are given that the radius of the Ferris wheel is 26m and the rider is 25m from the ground. We want to find the rate at which the rider's height is changing, or dh/dt, when the rider is 25m from the ground.
Let's differentiate both sides of the equation with respect to time t:
2h * dh/dt = 2r * dr/dt
Now we can solve for dh/dt:
dh/dt = (r * dr/dt) / h
Substituting the given values, we have:
dh/dt = (26 * dr/dt) / h
Plugging in the values h = 25m and r = 26m, we get:
dh/dt = (26 * dr/dt) / 25
So, to find the rate at which the rider's height is increasing, we need to know the rate at which the rider's distance from the center of the Ferris wheel is changing, dr/dt. If you provide us with that information, we can calculate the rate at which the rider's height is increasing.