Find Min Distance Parent & Child on Water Slide Problem

[aeromat]

Homework Statement

A parent takes their child to a water slide which is shaped similar to a portion of a parabola (see diagram). There is an elevator 3m from the edge of the pool. The elevator takes the rider to the top of the slide which is 19m above ground. The rider slides down and falls into pool from height of 1m. The parent (he is 2m tall) is standing 1m from the elevator and wishes to take a picture of the child when they are closest to the parent. What is the MIN. distance between parent and child?

The Attempt at a Solution

I sketched out the diagram attached. I know that the y-axis represents the the height that he started at; (0,19). I believe at the very edge of the slide is the vertex, and in this case (9.5,1). Based on symmetry, the parabola would also have (19,19) as a point.

But now I am stuck. I don't know what strategy to pursue in order to obtain this minimum distance.

EDIT: I know eqs:
s(t) = 0.199x^2 - 3.79x + 19
v(t) = s'(t) = 0.398x - 3.79

 

[Mentallic]

Homework Statement

A parent takes their child to a water slide which is shaped similar to a portion of a parabola (see diagram). There is an elevator 3m from the edge of the pool. The elevator takes the rider to the top of the slide which is 19m above ground. The rider slides down and falls into pool from height of 1m. The parent (he is 2m tall) is standing 1m from the elevator and wishes to take a picture of the child when they are closest to the parent. What is the MIN. distance between parent and child?

The Attempt at a Solution

I sketched out the diagram attached. I know that the y-axis represents the the height that he started at; (0,19). I believe at the very edge of the slide is the vertex, and in this case (9.5,1). Based on symmetry, the parabola would also have (19,19) as a point.

But now I am stuck. I don't know what strategy to pursue in order to obtain this minimum distance.

EDIT: I know eqs:
s(t) = 0.199x^2 - 3.79x + 19
v(t) = s'(t) = 0.398x - 3.79

Sorry but where's the diagram?

 

[aeromat]

Oh I thought I uploaded it already. It is up there now.

 

[aeromat]

Here is extra work done:
Ok here is the quadratic equation without any rounding done:

0.199445983x^2 - 3.789473684x + 19 = f(x)
f(9.5) = 0.999999968 ≈ 1

I know that the following points would be on the parabola, had it been complete; "x"-values expressed throughout the entire cartesian domain:

(0,19), (19,19), (9.5,1)

Eq1: c = 19
Eq2: (19)^2 a + 19b + c = 0
Eq2: 361a + 19b + 19 = 19
Eq3: (9.5)^2a + 9.5b + c = 1
Eq3: 90.25a + 9.5b + 19 = 1

Divided Eq2 by 2 to eliminate b:
[2] - [3]
[2] 180.5a + 9.5b = 0
[3] 90.25a + 9.5b = -18
-----------------------
90.25a + 0 = -18

Solve for a, and I get:
a = 0.199445983

Sub back into solve for b, using Eq[1]
90.25(0.199445983) + 9.5b + 19 = 1
18 + 9.5b + 19 = 1

Solve for b, and I get:
b = -3.789473684

Therefore, the equation of the parabola is:
0.199445983x^2 - 3.789473684x + 19 = f(x)

Taking the derivative of this function:
f'(x) = 0.398891966x - 3.789473684

 

[Mentallic]

To find the equation of the parabola, you can take a shortcut and make things a lot easier by using the fact that (9.5,1) is at the vertex (by the way, how do you know this? Was it given? Because I can't seem to find how you deduced this from the information given) instead of solving for the coefficients in [tex]y=ax^2+bx+c[/tex], you can instead solve for the coefficients in [tex]y=a(x-k)^2+h[/tex] where k=9.5, h=1 and now you only have a to solve for, which you can do by plugging in (0,19).

So the parabola is (after expanding into the general quadratic form) is what you got. Now, to find the minimum distance between the parent at (1,2) and the slide, you'll need to equate the parabola with a circle centred at (1,2) with unknown radius r. Then you'll solve for r and keep in mind the circle will be touching the parabola in only one spot.

 

[aeromat]

If given only the coordinates of the centre of a circle, how would we solve for the "r" value in the circle equation?

Also, how does this make sense? I don't understand why we would use a circle to get the minimum distance.

 

[Mentallic]

If given only the coordinates of the centre of a circle, how would we solve for the "r" value in the circle equation?

Also, how does this make sense? I don't understand why we would use a circle to get the minimum distance.

Because of the geometry of the circle and the parabola, they are capable of touching in one spot. As the circle grows you can think of it as the distance from the parent, and to find the minimum distance from the child on the parabola, just find where the circle grows till it touches the parabola just once (if it grows more than this, it'll touch the parabola in 2 spots).
Anyway, to find the x-coordinate of the circle and the parabola (it will be in terms of r) just equate them.

Actually, now I noticed it'll turn out to be a quartic that you need to solve. Oh I was expecting a quadratic... ok this method isn't going to work. We'll try another!

---

Do you know calculus? What you can do is take a line with gradient m passing through the point of the parent, find the derivative of the parabola and then take a gradient of -1/m (perpendicular to gradient m) that is tangent on the parabola, then find where these intersect.

 

[aeromat]

Wait one minute. I am posting something.