Question

Between\({0^o}C\) and\(3{0^o}C\), the volume\(V\) (in cubic centi-meters) of 1 kg of water at a temperature\(T\) is given approximately by the formula

\(V(T) = 999.87 - 0.06426T + 0.0085043{T^2} - 0.0000679{T^3}\)

Find the temperature at which water has maximum density.

Short answer

The temperature at which water has maximum density is, \(T = {3.9665^^\circ }{\rm{C}}\).

Step-by-step solution

Given information

The volume of\(1\;{\rm{kg}}\)of water at a temperature\(T\)between\({0^^\circ }{\rm{C - }}{30^^\circ }{\rm{C}}\)is given by,

\(V = 999.87 - 0.06426T + 0.0085043{T^2} - 0.0000679{T^3}T = {3.9665^^\circ }{\rm{C}}\).

Definition of Quadratic Formula

Definition of quadratic formula: Formula that gives the solutions of the general quadratic equation\(a{x^2} + bx + c = 0\)and that is usually written in the form\(\left. {x = \left( { - b \pm (\sqrt ( {b^2} - 4ac} \right)} \right)/(2a)\).

Differentiate and apply Quadratic Formula

Consider the given function,\(V(T) = 999.87 - 0.06426T + 0.0085043{T^2} - 0.0000679{T^3}\).

Obtain the first derivative of\(V(T)\).

\(\begin{aligned}{c}{V^\prime }(T) &= - 0.06426T + 2 \times 0.0085043T - 3 \times 0.0000679{T^2}\\ &= - 0.0002037{T^2} + 0.0170086T - 0.06426T\end{aligned}\)

Set\({V^\prime }(t) = 0\)and obtain the critical number.

Use quadratic formula and solve for\(T\).

\(\begin{aligned}{c}T &= \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\\ &= \frac{{ - (0.0170086) \pm \sqrt {{{(0.0170086)}^2} - 4( - 0.0002037)( - 0.06426)} }}{{2 \times ( - 0.0002037)}}\\ &= \frac{{ - (0.0170086) \pm \sqrt {0.0002369} }}{{ - (0.0004074)}}\\T \approx 3.9665\;\;and\;\;79.5318\end{aligned}\)

Therefore the critical numbers of the function are\({3.665^^\circ }{\rm{C}},{79.5318^^\circ }{\rm{C}}\).

Hence, the only \(T\) value lies in the given domain is \({3.9665^^\circ }{\rm{C}}\).

Apply Critical values in the function and simplify

Apply the extreme values of the given interval and the critical number in the given function.

Substitute,\(T = 0 \to V(T)\).

\(\begin{aligned}{c}V(0) &= 999.87 - 0.06426(0) + 0.0085043{(0)^2} - 0.0000679{(0)^3}\\ &= 999.87\end{aligned}\)

Substitute,\(T &= 3.9665 \to V(T)\).

\(\begin{aligned}{c}V(3.9665) &= 999.87 - 0.06426(3.9665) + 0.0085043{(3.9665)^2} - 0.0000679{(3.9665)^3}\\ &= 999.74\end{aligned}\)

Substitute,\(T = 30 \to V(T)\).

\(\begin{aligned}{c}V(30) &= 999.87 - 0.06426(30) + 0.0085043{(30)^2} - 0.0000679{(30)^3}\\ \approx 1003.8\end{aligned}\)

Observe that the volume is minimum at\(T = {3.9665^^\circ }{\rm{C}}\)and maximum at\(T = {30^^\circ }{\rm{C}}\).

Recall the formula, Density\( = \frac{{{\rm{ Mass }}}}{{{\rm{ Volume }}}}\).

Note that the density is maximum when the volume is minimum.

The volume is minimum at\(T = {3.9665^^\circ }{\rm{C}}\).

Therefore, the maximum density of water is occurs at \(T = {3.9665^^\circ }{\rm{C}}\).