I need some help with a related rates question

[eekoz]

Could anyone help me out? I'm pretty much stuck and I need someone to help me out here =) All help would be greatly appericiated

A softball diamond has the shape of a square with sides 20 m long. Suppose a player is running from the first base to the second base at 7 m/s. Find the rate at which the distance between the player and home plate is changing when the player is 5m from the first base.

I'm just having a little problem making the diagram, that's all

 

[galactus]

If we let homeplate be at the origin and the path to first base along the x-

axis we can draw a line from home plate to a point B(20,y)(the position of

the player) between first and second.

You need the rate at which the distance from home to B(20,y) is

changing. Call this distance D. Of course, as with many related rates

problems, ol' Pythagoras 'comes into play'.

\(\displaystyle D^{2}=20^{2}+y^{2}\)

Differentiate with respect to time :

\(\displaystyle 2D\frac{dD}{dt}=2y\frac{dy}{dt}\)

\(\displaystyle D\frac{dD}{dt}=y\frac{dy}{dt}\)

But \(\displaystyle D=\sqrt{20^{2}+5^{2}}=\sqrt{425}=5\sqrt{17}\)

So, we have:

\(\displaystyle 5\sqrt{17}\frac{dD}{dt}=(5)(7)\)

\(\displaystyle 5\sqrt{17}\frac{dD}{dt}=35\)

\(\displaystyle =\frac{7}{\sqrt{17}}=1.7\ m/s\)

 

[eekoz]

Thank you galactus
I really appericiate the time you put into that, it really helped me understand it


There's only one question left which I can't figure out. Been trying for over an hour now, can't figure it out.

1. The ends of a water trough are equilateral triangles whose sides are 2 m long. The length of the trough is 15m. Water is being poured into the trouth at the rate of 3m^3/h. Find the rate at which the water level is rising when the depth of the water is 1m.

Any hints?

 

[galactus]

The volume of water and the depth of the water are changing as water flows into the trough. You want \(\displaystyle \frac{dh}{dt}\text{when}h=1\).

You know \(\displaystyle \frac{dV}{dt}=+3\)

Try introducing another variable that relates volume and depth, say, w.

Let w be the distance between the sloping sides of the trough at the top of

the water.

The volume of water in the trough is \(\displaystyle V=\frac{1}{2}(w)(h)15\)

Half of the end of the trough forms a right triangle with top 1 and

hypoteneuse 2.

Use similar triangles to get a relation between w and h solve for w, sub into the volume equation, differentiate with respect to time, solve for dh/dt.

***edit: that should be a 2 along the hypoteneuse of the triangle, not a 3...typo.

 

[eekoz]

Is the answer 3*sqrt(3)/10 m/h?

Edit: So it's sqrt(8) ? not sqrt(3)? on the diagram

 

[galactus]

Is the answer \(\displaystyle \frac{\sqrt{3}}{10}\)m/h?

Edit: \(\displaystyle \sqrt{3}\), Yes, I fixed the typos on the diagram

 

[eekoz]

Sorry my bad, made a little mistake

3 * sqrt(3)/10.. can't make it come out right:
(Three times the square root of three) over 10

Is that right?
And thank you for the edited diagram

 

[galactus]

Similar triangles:

\(\displaystyle \frac{\frac{w}{2}}{h}=\frac{1}{\sqrt{3}}\)

\(\displaystyle w=\frac{2h}{\sqrt{3}}\)

Sub into volume equation:

\(\displaystyle V=\frac{15}{2}(\frac{2h}{\sqrt{3}})h=\frac{15h^{2}}{\sqrt{3}}\)

\(\displaystyle \frac{dV}{dt}=\frac{30}{\sqrt{3}}h\frac{dh}{dt}\)

Sub in given values and solve for dh/dt

 

[eekoz]

Thank you, I see where you get sqrt(3)/10

The equation

where h = 1


Someone else named soroban, however, showed me that:

or

I am trying to figure out what went wrong (two different answers)

 

[galactus]

\(\displaystyle \frac{dV}{dt}=\frac{30}{\sqrt{3}}(1)\frac{dh}{dt}\)

\(\displaystyle 3=\frac{30}{\sqrt{3}}\frac{dh}{dt}\)

\(\displaystyle \frac{dh}{dt}=\frac{\sqrt{3}}{10}=.173 m/hr.\)

That's what I'm getting.