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

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.

Transcript

00:01 All right, we've got a question here that says that we have the volume between 0 to 30 degrees celsius of 1 kilograms of water given us the function for v with respect to temperature.

00:17 That's 99 .87 minus 0 .06426t plus 0 .060 plus 0 .06426t plus 0 .0 .0 .6 .6 t plus 0 .0.

00:52 0 .85043 t cubed minus 0 .0 .079 t cubed.

01:11 All right.

01:17 We want to determine, we want to find the temperature at which water is at its maximum density.

01:23 First of all we know that density is the same thing as mass or volume.

01:26 And we know the mass is already given to us as one kilogram so that we cannot change that.

01:32 So basically what we're asked to calculate here is what is the temperature at which our volume, excuse me, will be at its lowest, right? because the lowest volume would dictate the highest density.

01:49 All right, we can do that first by taking the derivative of v and calculating critical points when our v derivative is equal to 1.

01:59 So the derivative of v, excuse me, derivative of v would be negative 0 .6426 okay and then we have it being added to three times our 0 .085 which would be 0 .017 well i'm sorry this is a three here a two here not a three squared here so we take two and we multiply by that value there all right we're going to multiply two by that then you have t and then you multiply three by this value here it comes out to be negative 0 0 0 -0 -2037 t squared or t squared excuse right all of that is equal to 0 and then if we use the quadratic equation we can solve for our t to be or t1 it's going to be equal to 3 .9665, and our second t would be equal to 79 .531.

03:56 Okay, once again, that's the quadratic formula.

04:01 Let me just pull it up real quick, just in case any of you have forgotten that.

04:08 Quadratic formula is the negative b plus the square root of b squared minus 4a, c, and then all of that divided by 2a.

04:21 And then when you do that and you plug it in, you'll be able to solve your t's.

04:28 All right.

04:31 Once you get your t's, you're going to find the second derivative of v.

04:39 So the second derivative would be equal to 0 .116 and minus a 2 times this number here, which comes out to 0 .0 .0 .0...

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