Question

A model for the specific gravity of water \(S\) is \(S=\frac{5.755}{10^{8}} T^{3}-\frac{8.521}{10^{6}} T^{2}+\frac{6.540}{10^{5}} T+0.99987,0

Short answer

Use a computer algebra system to solve equations and plot the function. The coordinates of the maximum value of the function are obtained by differentiating the function, setting it to zero and solve for T. The specific gravity of water at \( T = 20\degree\) Celsius can also be calculated from the function.

Step-by-step solution

Differentiate the function

Differentiation of the function will help to find the local maximum. This can be done using a computer algebra system. The derivative of the function \(S=\frac{5.755}{10^{8}} T^{3}-\frac{8.521}{10^{6}} T^{2}+\frac{6.540}{10^{5}} T+0.99987\) is \(S'=\frac{5.755*3}{10^{8}} T^{2}-\frac{8.521*2}{10^{6}} T+\frac{6.540}{10^{5}}.\)

Find the maximum value

To find the maximum value, set the derivative equal to zero and solve for \(T\). The solution \(T\) will be the x-coordinate of the maximum value. Then, substitute \(T\) into the original function to find the y-coordinate, which represents the maximum specific gravity.

Sketch the graph

Plot the function using a domain of \(0 < T < 25\). The y-axis should be scaled to include \(0.996 \leq S \leq 1.001 .\)

Calculate Specific Gravity

To estimate the specific gravity of water when \(T = 20\), substitute \(T = 20\) into the original function. The result will be the specific gravity of water at that temperature.