Between 0^∘ C and 30^∘ C, the volume V (in cubic centimeters) of 1 kg of water at a temperature

Between 0^∘ C and 30^∘ C, the volume V (in cubic...

Between $0^{\circ} \mathrm{C}$ and $30^{\circ} \mathrm{C},$ the volume $V$ (in cubic centimeters) of 1 $\mathrm{kg}$ of water at a temperature $T$ is given approximately by the formula $$V=999.87-0.06426 T+0.0085043 T^{2}-0.0000679 T^{3}$$ Find the temperature at which water has its maximum density.

Between $0^{\circ} \mathrm{C}$ and $30^{\circ} \mathrm{C},$ the volume $V$ (in cubic centimeters)
of 1 $\mathrm{kg}$ of water at a temperature $T$ is given approximately by
the formula
$$V=999.87-0.06426 T+0.0085043 T^{2}-0.0000679 T^{3}$$
Find the temperature at which water has its maximum density.

Key Concepts

Critical Points

Critical points refer to values in the domain of a function where its derivative is zero or undefined, which are potential candidates for local maxima or minima. In this context, finding the critical point involves differentiating the volume expression with respect to temperature and setting the derivative equal to zero to locate the temperature where the density of water is maximized (or equivalently, the volume is minimized).

Polynomial Functions

A polynomial function is an expression consisting of variables raised to non-negative integer powers and associated coefficients. The given volume equation is a cubic polynomial in temperature, and understanding the behavior of polynomial functions, including their differentiability and the nature of their graphs, is essential in analyzing and solving for extrema such as the temperature of maximum density.

Relationship Between Density and Volume

Density is defined as mass divided by volume, so for a fixed mass, maximizing the density of water is equivalent to minimizing its volume. Recognizing this relationship is crucial when the problem asks for the temperature corresponding to the maximum density, leading to the strategy of finding the minimum point of the volume function by using calculus techniques.

Transcript

00:01 Hello, we have to find the temperature at which water has its maximum density and the equation for the volume is given and the mass of water is 1k and temperature is varies between 0 to 30 degrees centigrade.

00:21 Okay and we know that the volume b equals to sorry and we know that density is is equal to mass by volume so density is inversely proportional to the volume and the equation of volume so we have to find the temperature for the maximum density maximum that means volume will be minimum okay because m is fixed m is 1 kg that is fixed so the equation of the volume is given 99 .87 minus 990 .86 minus 0 .064 .26226 plus 0 .0085 plus 0 .0085 0 .043 t squared.

01:48 043 t squared minus of point double zero one double zero double zero six seven nine t cube six seven nine t cube for the simplification we can be we will write it like this a minus b t plus c t square minus of t t t q okay so we have to find the temperature for which densely is maximum so so now we calculate the temperature for the minimum volume so first we will differentiate it so v dash equals to v d by d t because of minus of b plus 2c t minus of b plus 2c t minus 3d t square okay so we can write minus of 3d t square positive of 3 t square minus 3 t square minus 2 c t plus b equals to 0 so we can find the temperature here now we will solve it where is equals to this is this is a and this is b and this is c and this is d so now we will solve it and we will get the value of t that is 3 .965 3 .9665 degree centigrade this is the critical number and now we will find the d2v by d t square so we will get 6d t minus 2c equals to 0 so at the value of t at the value of t at the value of t 3 .9665 this is the it has the value that is that is positive so that means at t equals to 3 .9665 degree centigrade the volume is minimum that means density will be maximum maximum at temperature t equals to 3 .9665 degrees centigrade...

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