** oops, messed up with the edits.. check the quote for the question **

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- Apr 29th 2006, 01:22 PMeekozTwo problems
** oops, messed up with the edits.. check the quote for the question **

- Apr 29th 2006, 05:18 PMticbolQuote:

Originally Posted by**eekoz**

1.) Find the rate at which water level is rising.

Rate is per time. Water level is height. So if height is y, and time is t, then

rate = y/t, or dy/dt for instantenous.

Given data:

Trough, triangular prism; 15m long; equilateral triangle, 2m each side, cross section.

Rate of volume of water in the trough, or dV/dt = 3 cu.m./hr.

There, we can relate the dy/dt with the dV/dt.

Draw on paper, or imagine the figure.

It is a cross sectional area of the trough.

It is an equilateral triangle whose one vertex is sitting on the ground.

Volume of water in the trough at any time:

V = (1/2)(y)(x)(15) = 7.5xy

where y = height of water; x = width of the water level.

We express x in terms of y, for simplicity

If we drop a vertical line from midpoint of x to the vertex at the bottom, the 60-degree angle there is bisected, so,

tan(30deg) = (x/2)/y = x /2y

Then, x = 2ytan(30deg) = 2y(1/sqrt(3) = (2/sqrt(3))*y

Hence, V = 7.5[(2/sqrt(3))*y]*y

V = [15/sqrt(3)](y^2)

V = [5sqrt(3)](y^2) --------(i)

Differentiate both sides of (i) with respect to time t,

dV/dt = [10sqrt(3)]*y *dy/dt -----------(ii)

(ii) is at any time t.

At t when y=1m,

3 = [10sqrt(3)](1)(dy/dt)

So,

dy/dt = 3/[10sqrt(3)] = (3/30)sqrt(3) = 0.1732 m/hr ---------***

Therefore, when the depth of the water is 1 meter, the water level is rising at the rate of 0.1732 meter per hour. ----------answer.

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2.) Find the rate at which the distance between the player and home plate is changing when the player is 5m from the first base.

So if x = the said distance, then we are looking for dx/dt.

Let y = distance travelled by the runner.

By Pythagorean theorem,

x^2 = y^2 +20^2

x^2 = y^2 +400 ------(iii)

Differentiate both sides with respect to time t,

2x(dx/dt) = 2y(dy/dt)

x(dx/dt) = y(dy/dt) ----------(iv)

Given:

dy/dt = 7 m/sec; y=5m

So, using (iii), x^2 = 5^2 +400 = 425

x = sqrt(425) when y=5 ------------------***

Substitute those into (iv),

sqrt(425) *(dx/dt) = 5*7

dx/dt = 35 /sqrt(425) = 1.7 m/sec ------***

Therefore, when the runner is at 5m away from first base, the distance of the runner from the home plate is increasing at the rate of 1.7 meter per second. -----------answer. - Apr 29th 2006, 05:31 PMeekoz
Thanks a lot, ticbol!

I got 3*sqrt(3)/10 m/h for the first question, I guess I did something wrong.

Going over your answer now, makes sense so far :)