The temperature, in degrees Celsius (°C), in a city on a particular day is modeled by the function T defined by T(t) = (75t³ - 836t² + 3100t - 4185) / (14t² + 10t - 35). Based on t... The temperature, in degrees Celsius (°C), in a city on a particular day is modeled by the function T defined by T(t) = (75t³ - 836t² + 3100t - 4185) / (14t² + 10t - 35). Based on the model, how many hours did it take for the temperature to increase from 0 °C to 5 °C? Read more

Steps to Solve

1. Set up the temperature function

The temperature function is given by

$$ T(t) = \frac{75t^3 - 836t^2 + 3100t - 4185}{14t^2 + 10t - 35} $$

We need to find the values of $t$ when $T(t) = 0$ °C and $T(t) = 5$ °C.

2. Solve for when temperature is 0 °C

Set the temperature function equal to 0:

$$ \frac{75t^3 - 836t^2 + 3100t - 4185}{14t^2 + 10t - 35} = 0 $$

This occurs when the numerator is zero:

$$ 75t^3 - 836t^2 + 3100t - 4185 = 0 $$

3. Use numerical methods or a calculator

Finding the roots of the cubic equation can be done using numerical methods such as the Newton-Raphson method or a graphing calculator.

4. Solve for when temperature is 5 °C

Next, set the function equal to 5:

$$ \frac{75t^3 - 836t^2 + 3100t - 4185}{14t^2 + 10t - 35} = 5 $$

This simplifies to the equation:

$$ 75t^3 - 836t^2 + 3100t - 4185 - 5(14t^2 + 10t - 35) = 0 $$

Simplifying gives:

$$ 75t^3 - 836t^2 + 3100t - 4185 - 70t^2 - 50t + 175 = 0 $$

Combine like terms:

$$ 75t^3 - 906t^2 + 3050t - 4010 = 0 $$

5. Use numerical methods or a calculator

Again, use numerical methods to find the roots of this new cubic equation.

6. Calculate the time difference

Once you have the values of $t$ at which the temperature reaches 0 °C and 5 °C, subtract the two values to find the time taken for the temperature to increase.