A water trough with triangular ends is 9ft long, 4ft wide, and 2ft deep. Initially the trough is full of water, but due to evaporation, the volumn of the water decreases at a rate proportional to the square root of the volume. Using advanced concepts from math and physics, it can be shown that the volume after t hours is give by:
V(t)= 1/C^2(t + 6C)^2
0 is less than or equal to t and t is less than or equal to 6 X absolute value of C. Where C is a constant.
Sketch by hand the graphs of:
y=V(t) for C=-4, C= -5, and C=-6
and give a brief verbal description of this collection of functions.
I find that when t=0 V=36 for all instances of C, but I cannot figure out where to go beyond that.
I think you mean
V = (t+6c)^2 / c^2
36 is just the volume of the tank, so it better be 36 at t = 0 if the tank starts full.
now for each of the three values of c find values of V for t = 1, 2, 3 etc
For example for c = -4
t = 0, V = 36
t = 1 , V = 529/16 = 33.06
t = 2 , V = 484/16 = 27.13
etc
for c = -5
t = 0, V = 36
t = 1, V = 841/25 = 33.64
etc
To solve this problem, we need to sketch the graphs of the function V(t) for three different values of C: C = -4, C = -5, and C = -6. Let's start by analyzing the given equation:
V(t) = 1/C^2 * (t + 6C)^2
Here, V(t) represents the volume of water in the trough at time t, and C is the constant that affects the rate of evaporation. By looking at the equation, we can observe that when t = 0, V(t) = 1/C^2 * (0 + 6C)^2 = 36, which you correctly mentioned.
To sketch the graphs of y = V(t) for the three values of C, we can first plot the points (t, V(t)) for a few t-values within the given domain (0 ≤ t ≤ 6|C|) and then connect them to form a curve.
1. C = -4:
For C = -4, the equation becomes:
V(t) = 1/(-4)^2 * (t + 6(-4))^2 = 1/16 * (t - 24)^2
Choose some t-values (like 0, 2, 4, 6) within the domain, and calculate the corresponding V(t) values using the equation. Then plot these points on a coordinate plane and connect them to obtain a curve. Repeat this process for the other values of C.
2. C = -5:
The equation becomes:
V(t) = 1/(-5)^2 * (t + 6(-5))^2 = 1/25 * (t - 30)^2
3. C = -6:
The equation becomes:
V(t) = 1/(-6)^2 * (t + 6(-6))^2 = 1/36 * (t - 36)^2
After plotting the points and connecting them for all three functions, you will have three separate curves on the same graph.
A verbal description of this collection of functions would be that they represent the volume of water in the trough over time, considering the evaporation rate proportional to the square root of the volume. The graphs show how the volume decreases as time passes, with different shapes and rates of decrease based on the value of C.