A river level rises rapidly during a flood which started at noon on Monday 5 March. The initial depth of the river was 2.3 m. The maximum depth of the river was 4 m, which occurred at noon, 1 day later. By day 3, the depth had dropped to 3 m. The graph of the depth of the river is shown. Part of this graph is to be modelled by a hyperbola of the form d

To solve for the constants $a$, $b$, and $c$ in the hyperbola equation $d = \frac{a}{t - b} + c$, we will use the given data points.

Given:

• At $t = 0$ (when the rain started), the depth $d = 2.3$ meters.
• The maximum depth of 4 meters occurred at $t = 1$ day.
• By $t = 3$ days, the depth had dropped to 3 meters.

Using these points, we will solve for $a$, $b$, and $c$.

Step-by-Step Solution:

1. Initial Condition: At $t = 0$, $d = 2.3$: $2.3 = \frac{a}{0 - b} + c$

2. Maximum Depth: At $t = 1$, $d = 4$: $4 = \frac{a}{1 - b} + c$

3. Depth After 3 Days: At $t = 3$, $d = 3$: $3 = \frac{a}{3 - b} + c$

From these equations, we need to solve for $a$, $b$, and $c$.

Equation 1:

$2.3 = \frac{a}{-b} + c$ $2.3 = -\frac{a}{b} + c$

Equation 2:

$4 = \frac{a}{1 - b} + c$

Equation 3:

$3 = \frac{a}{3 - b} + c$

Subtract Equation 1 from Equation 2 to eliminate $c$: $4 - 2.3 = \frac{a}{1 - b} + \frac{a}{b}$ $1.7 = a \left( \frac{1}{1 - b} + \frac{1}{b} \right)$ $1.7 = a \left( \frac{b + (1 - b)}{b(1 - b)} \right)$ $1.7 = a \left( \frac{1}{b(1 - b)} \right)$

From here, solve for $a$: $a = 1.7b(1 - b)$

Next, subtract Equation 2 from Equation 3 to eliminate $c$: $4 - 3 = \frac{a}{1 - b} - \frac{a}{3 - b}$ $1 = a \left( \frac{1}{1 - b} - \frac{1}{3 - b} \right)$ $1 = a \left( \frac{(3 - b) - (1 - b)}{(1 - b)(3 - b)} \right)$ $1 = a \left( \frac{2}{(1 - b)(3 - b)} \right)$ $a = \frac{(1 - b)(3 - b)}{2}$

Equate the two expressions for $a$: $1.7b(1 - b) = \frac{(1 - b)(3 - b)}{2}$

Divide both sides by $1 - b$: $1.7b = \frac{3 - b}{2}$ $3.4b = 3 - b$ $4.4b = 3$ $b = \frac{3}{4.4}$ $b \approx 0.682$

Now substitute $b$ back into the equation for $a$: $a = 1.7 \cdot 0.682 \cdot (1 - 0.682)$ $a = 1.7 \cdot 0.682 \cdot 0.318$ $a \approx 0.368$

Finally, substitute $a$ and $b$ back into one of the original equations to solve for $c$: $2.3 = -\frac{0.368}{0.682} + c$ $2.3 = -0.54 + c$ $c = 2.3 + 0.54$ $c = 2.84$

Rounding $a$ and $b$ to the nearest integers: $a \approx 0$ $b \approx 1$

Thus, the approximate equation of the hyperbola is: $d = \frac{0}{t - 1} + 2.84$ However, since $a = 0$ isn't practical for modeling, we should retain the calculated values for better accuracy.

So the refined equation would be: $d = \frac{0.368}{t - 0.682} + 2.84$

If you have any further questions or need more details, feel free to ask!

Here are 5 numbered questions you might consider asking:

1. Can you explain how hyperbolas are generally used in real-life applications?
2. What are the key differences between hyperbolas and other conic sections like ellipses and parabolas?
3. How do you solve for the intersection points of two hyperbolas?
4. What are some common methods for finding the asymptotes of a hyperbola?
5. Can you show how to derive the standard form of a hyperbola from its geometric definition?

Tip: When working with hyperbolas, ensure to always check the domain of the function, as hyperbolic equations can have undefined points where the denominator is zero.