A water skier skis over the ramp, 5m vertical, 1m horizontal, at a speed of 12m/s. How fast is she risingas she leaves the ramp?
d=root26
dy/dt=[(d)(d')-(x)(x')]/y
dy/dt=[(12)(root26)-(5)(x')]/1
How do I find dx/dt?
Thanks in advance.
Whoops, I do have the horizontal and vertical numbers mixed up. The horizontal side of the ramp is 5m, and vertical 1m. So I just have to switch the 5 and 1 around right? Thank you so much :)
To find dx/dt, we can use the given information about the water skier's speed and the dimensions of the ramp.
We know that the skier is moving horizontally at a speed of 12 m/s. Therefore, dx/dt = 12 m/s.
Hope this helps!
To find dx/dt, we need to first understand the situation. The water skier is skiing over a ramp, and we are interested in how fast she is rising as she leaves the ramp.
Given that the ramp has a vertical distance of 5m and a horizontal distance of 1m, we can use this information to determine the relationship between the variables y (vertical distance) and x (horizontal distance).
Using the Pythagorean theorem, we can calculate the distance along the ramp (d) as:
d = √(y^2 + x^2)
d = √(5^2 + 1^2)
d = √(26)
Now, let's differentiate this equation with respect to time (t):
d/dt (d) = d/dt (√(26))
dx/dt * d'/dx + dy/dt * d'/dy = 0
Here, d'/dx represents the derivative of d with respect to x, and d'/dy represents the derivative of d with respect to y.
Since we are interested in finding dx/dt, we can rearrange the equation to isolate it:
dx/dt = - (dy/dt * d'/dy) / d'/dx
Now we have all the necessary components to determine dx/dt. We know that dy/dt is given as 12 m/s, which is the speed of the skier. We also know that d'/dy is equal to 2y/d, and d'/dx is equal to 2x/d.
Substituting these values into the equation, we get:
dx/dt = - (12 * (2y / d)) / (2x / d)
Now, we need to substitute the given values of y = 5m, x = 1m, and d = √26 into the equation to calculate dx/dt:
dx/dt = - (12 * (2 * 5 / √26)) / (2 * 1 / √26)
dx/dt = - (12 * 10 / √26) / (2 / √26)
dx/dt = - 120 / 2
dx/dt = - 60 m/s
Therefore, the water skier is rising at a speed of 60 m/s as she leaves the ramp.
draw her position just short of the take-off from the ramp.
Let her height be y m and her horizontal distance x m.
By similar triangles, y/x = 5/1
x = y/5
let her distance up the ramp be d
then d^2 = x^2 + y^2
d^2 = y^2/25 + y^2 = (26/25)y^2
d = (√26/5)y
dd/dt = (√26/5)dy/dt
(notice we don't need dx/dt)
I will assume that the 12 m/s is her speed as she goes up the ramp, so ...
12 = (√26/5)dy/dt
dy/dt = 60/√26 m/x
So as long as she is on the ramp, she has a vertical speed of 60/√26 m/s.
So at the moment she leaves the ramp that will be her vertical speed.
Are you sure you don't have your vertical and horizontal numbers mixed up?
That ramp looks extremely unreasonable, she will practically crash into it.