Pre-Cal
between 0 degree C and 30 degree C the volume V (in cubic centimeters) of 1kg of water at a temperature T is given by the formula.
V=999.87-0.06426T+0.0085043T^2-0.0000679T^3
Find the temperature at which the volume of 1 kg of water is a minimum.
something tells me this is sweet goodbye, kitty.
Yep ... bye, Kitty. =(
To find the temperature at which the volume of 1 kg of water is at a minimum, we can use calculus. Specifically, we will need to find the critical points of the volume function and then determine whether they correspond to a minimum or maximum.
Step 1: Find the derivative of the volume function with respect to temperature T.
To find the derivative, we need to differentiate each term of the volume function with respect to T and combine them:
dV/dT = -0.06426 + (2)(0.0085043)(T) - (3)(0.0000679)(T^2)
Simplifying the derivative, we get:
dV/dT = -0.06426 + 0.0170086T - 0.0002037T^2
Step 2: Set the derivative equal to zero and solve for T.
To find the critical points, we set the derivative equal to zero and solve for T:
0 = -0.06426 + 0.0170086T - 0.0002037T^2
Rearranging the equation, we get:
0.0002037T^2 - 0.0170086T + 0.06426 = 0
This is a quadratic equation. We can solve it by factoring or by using the quadratic formula. In this case, it may be easier to use the quadratic formula:
T = (-b ± √(b^2 - 4ac))/2a
Plugging in the values for a, b, and c from the quadratic equation, we get:
T = (-(-0.0170086) ± √((-0.0170086)^2 - 4(0.0002037)(0.06426)))/2(0.0002037)
Simplifying further, we have:
T = (0.0170086 ± √(0.000289869148 - 0.00005218584))/0.0004074
Calculating the square root and subtracting the values inside, we get:
T = (0.0170086 ± √0.000237683308)/0.0004074
T = (0.0170 ± 0.01543)/0.0004074
This gives us two possible values for T:
T₁ = (0.0170086 + 0.01543)/0.0004074 ≈ 82.3°C
T₂ = (0.0170086 - 0.01543)/0.0004074 ≈ -10.8°C
Step 3: Determine if the critical points are a minimum or maximum.
To determine whether these critical points correspond to a minimum or maximum, we can use the second derivative test. This involves finding the second derivative of the original volume function evaluated at the critical points.
Taking the second derivative of V with respect to T:
d²V/dT² = 0 + 2(0.0085043) - 2(0.0000679)(T)
Simplifying further:
d²V/dT² = 0.0170086 - 0.0001358T
Now, substituting T₁ = 82.3°C into the second derivative equation:
d²V/dT² = 0.0170086 - 0.0001358(82.3)
d²V/dT² ≈ 0.0170086 - 0.01118654
d²V/dT² ≈ 0.00582206
Since the second derivative is positive, the critical point T₁ = 82.3°C corresponds to the minimum volume.
Similarly, substituting T₂ = -10.8°C into the second derivative equation:
d²V/dT² = 0.0170086 - 0.0001358(-10.8)
d²V/dT² ≈ 0.0170086 + 0.00146344
d²V/dT² ≈ 0.01847204
As the second derivative is positive as well, this critical point T₂ = -10.8°C corresponds to the minimum volume.
Therefore, the temperature at which the volume of 1 kg of water is at a minimum is approximately 82.3°C and -10.8°C.