One litre of water at 0 degrees Celsius will have a different volume at higher temperatures. For a temperature 0 ≤ T ≤ 30 (in Celsius) this volume (in litres) is well approximated by the function
V = −aT 3 + bT 2 − cT + 1
using the coefficients a = 9.8 × 10−8, b = 3.5 × 10−6, and c = 2.4 × 10−5.
(a) find the derivative
(b) When is the derivative positive, negative, and zero? (give answers correct to 3 decimal places)
ok, so i have a question for part b.
when u find the critical points for the derivative, do they have to be in between the interval of 0<T<30 ? because one of the critical points that i solved is not in that interval. Thank you
When finding the critical points of the derivative, they do not necessarily have to be within the interval of 0 < T < 30. Critical points are values where the derivative is either zero or undefined. It's important to note that critical points help identify potential local extrema, where the function is changing from increasing to decreasing or vice versa.
In this case, you're asked to find when the derivative is positive, negative, or zero within the interval 0 ≤ T ≤ 30. To do this, you need to find the derivative of the given function, which is V = -aT^3 + bT^2 - cT + 1.
(a) Finding the derivative:
To find the derivative of the function V with respect to T, you need to differentiate each term of the function separately. Here's how you can do it:
V = -aT^3 + bT^2 - cT + 1
dV/dT = d/dT(-aT^3) + d/dT(bT^2) - d/dT(cT) + d/dT(1)
= -3aT^2 + 2bT - c + 0
= -3aT^2 + 2bT - c
Therefore, the derivative of V with respect to T is dV/dT = -3aT^2 + 2bT - c.
(b) Analyzing the derivative:
To determine when the derivative is positive, negative, or zero, you need to set the derivative equal to zero and solve for T. In this case, you can disregard any critical points that are not within the interval 0 < T < 30.
Setting -3aT^2 + 2bT - c = 0:
-3aT^2 + 2bT - c = 0
Now you can solve this equation to find the critical points, which may include values outside the interval of interest (0 < T < 30). Analyze the results to determine when the derivative is positive, negative, or zero within the given interval.
Keep in mind that if you find critical points outside the interval, they may not have any significance for the original problem, but they can still help you understand the behavior of the function in the broader context.
Are there supposed to be exponents in your equation??
I will assume:
V = −aT^3 + bT^2 − cT + 1 , using the given:
dV/dT = -3a T^2 + 2b T - c
I assume you plugged in the given values for
a, b, and c, used the quadratic equation formula and solved for T
Since the coefficient of T^2 is negative, you have a parabola opening downwards
I assume there are two values of T, which are the x-intercepts.
Yes, you should only use the values of T that fall in the given domain of 0≤T≤30
Find the T value of the vertex.
The derivative will be negative for values less than that T.
It will be zero for the T of the vertex
and it will be positive for values > T of the vertex
I will not do the arithmetic for you, my calculator is not handy right now.