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

## The density is defined as $\rho=\frac{\text { mass }}{\text { volume }}=\frac{1000}{V(T)}\left(\text { in } \mathrm{g} / \mathrm{cm}^{3}\right) .$ But a critical point of $\rho$ will also be a critical point of $V$$\left[\text { since } \frac{d \rho}{d T}=-1000 V^{-2} \frac{d V}{d T} \text { and } V \text { is never } 0\right], and V is easier to differentiate than \rho$V(T)=999.87-0.06426 T+0.0085043 T^{2}-0.0000679 T^{3} \Rightarrow V^{\prime}(T)=-0.06426+0.0170086 T-0.0002037 T^{2}$Setting this equal to 0 and using the quadratic formula to find T, we getT=\frac{-0.0170086 \pm \sqrt{0.0170086^{2}-4 \cdot 0.0002037 \cdot 0.06426}}{2(-0.0002037)} \approx 3.9665^{\circ} \mathrm{C} or 79.5318^{\circ} \mathrm{C}. since we are only interestedin the region 0^{\circ} \mathrm{C} \leq T \leq 30^{\circ} \mathrm{C}, we check the density \rho at the endpoints and at 3.9665^{\circ} \mathrm{C}: \rho(0) \approx \frac{1000}{999.87} \approx 1.00013$$\rho(30) \approx \frac{1000}{1003.7628} \approx 0.99625 ; \rho(3.9665) \approx \frac{1000}{999.7447} \approx 1.000255 .$ So water has its maximum density atabout $3.9665^{\circ} \mathrm{C}$

Derivatives

Differentiation

Volume

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##### Kristen K.

University of Michigan - Ann Arbor

##### Samuel H.

University of Nottingham

##### Michael J.

Idaho State University

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### Video Transcript

All right, We've got a question here that says that we have the volume between 0 30 degrees Celsius of 1 kg. Water given us the function for V with respect to temperature, that's 999 0.87 minus 0.6 or two. 60 waas. 0.85 zero for three T cube minus 0.0 zero 679 take you. All right. And we want to determine we wanna find the temperature at which water is at its maximum density. First of all, we know that density is the same thing as mass or volume. And we know the mass has already given to us as 1 kg so that we cannot change that. Basically, what we're asked to calculate here is what is the temperature at which hour of LA volume Excuse me will be at its lowest right, because the lowest volume would dictate the highest debt. All right, we could do that first by taking the derivative of V and calculating critical points. When are we derivative is equal to one. It's a derivative of the war. Excuse me. Derivative movie would be negative 0.642 six mhm. And then we have being added to three times our 0.85 which would be your points. You're 17 Well, I'm sorry. This is Ah three here to here. Not a three squared here. So we take to when we multiply by that value there. Alright, we're gonna multiply to buy that. Then you have tea, then you multiply three by this value here comes out to be negative 0.0 zero 2037 t cute or T squares. All of that is equal to zero. And then if we used the quadratic equation we can solve for our tea to be or t one is gonna be equal to 3.96 65 Our second t would be equal to 79 point 53 way once again, that's the quadratic formula. Um, let me just pull it up real quick. Just in case any of you have for gotten that quadratic formula is the negative B plus the square root of B squared minus for a c and then all of that divided by two A. You know, when you do that and you plug it in, you'll be able to solve your T's. All right, Um, once you get your teas, you're gonna find the second derivative of V. So the second derivative would be equal to 0.170086 in minus two times. This number here comes out to point 000 4074 t All right here. If we plug in 3.9665 for tea, we would get a value that is greater than zero. So we can say that that point that t one is a maximum because we look at point, you were 15 39 Now, just round it up to here if you plug in t one in there and that is greater than zero, so we can say that is the maximum. And so then if we plug in our t two, we get a negative value. A negative value comes out to be 0.0 5439 which we know is less than zero. So that would be a minimum point. Okay, so we can confirm that the volume is minimized. Um, when t is equal to actually Hold on. I'm sorry. This would be your Maxima. That's greater than he wrote. It is. The minimum is less than zero than it's remarkable. Okay, then we want to choose our minimum value because we're looking for the lowest when the volume is at its lowest. And we know that occurs when t is equal to 3.9 665 degrees. So she's. And when the volume is at its lowest in our density is at its highest, we would say the temperature would be 3.9665 pulling our volume instead of slowest, nor densities at its max. All right, well, I hope that clarifies the question. Thank you so much for watching.

The University of Texas at Arlington

#### Topics

Derivatives

Differentiation

Volume

##### Kristen K.

University of Michigan - Ann Arbor

##### Samuel H.

University of Nottingham

##### Michael J.

Idaho State University

Lectures

Join Bootcamp