A dose, D, of a drug causes a temperature change, T, in apatient. For C a positive constant, t is given by T=((C/2)-(D/3))D^3.
(a) What is the rate of change of temperature change with respect to dose.
(b) For what dose does the temperature change increase as the dose increases.
For part a I got CD^2-D^3=0
For part b I got c^2/4
Any help would be greatly appreciated.
To find the rate of change of temperature with respect to dose, we need to compute the derivative of the temperature function, T, with respect to the dose, D.
(a) Rate of change of temperature with respect to dose:
To find the derivative of T with respect to D, we can differentiate the given expression of T with respect to D. Using the power rule and the constant multiple rule of differentiation, we have:
T = ((C/2) - (D/3)) * D^3
Taking the derivative of T with respect to D:
dT/dD = d/dD (((C/2) - (D/3)) * D^3)
To simplify the expression, let's first distribute D^3 into the brackets:
dT/dD = d/dD ((C/2) * D^3 - (D/3) * D^3)
Now we can differentiate each term separately:
dT/dD = (C/2) * d/dD (D^3) - (1/3) * D^3 * d/dD (D)
Applying the power rule of differentiation, we have:
dT/dD = (C/2) * 3D^2 - (1/3) * D^3 * 1
Simplifying:
dT/dD = 3(C/2) * D^2 - (1/3) * D^3
So, the rate of change of temperature with respect to dose is given by dT/dD = 3(C/2) * D^2 - (1/3) * D^3.
(b) Dose for which temperature change increases as the dose increases:
For the temperature change to increase as the dose increases, we need dT/dD > 0.
From the expression we obtained in part (a):
3(C/2) * D^2 - (1/3) * D^3 > 0
To solve this inequality, we can first find the critical points by setting dT/dD equal to zero:
3(C/2) * D^2 - (1/3) * D^3 = 0
To solve this equation, let's factor out a D^2:
D^2 * (3(C/2) - (1/3) * D) = 0
Setting each factor equal to zero:
1) D^2 = 0 => D = 0
2) 3(C/2) - (1/3) * D = 0 => 3(C/2) = (1/3) * D => D = 9C/2
Now, we can analyze the intervals between these critical points to determine where the temperature change increases. We pick values within each interval and use them to determine the sign of dT/dD. Our intervals are (-∞, 0), (0, 9C/2), and (9C/2, +∞).
Considering a value less than zero (e.g., -1):
dT/dD = 3(C/2)*(-1)^2 - (1/3)*(-1)^3 = 3(C/2) - (1/3) < 0
So, the temperature change decreases for doses less than zero.
Considering a value between 0 and 9C/2 (e.g., C):
dT/dD = 3(C/2)*(C)^2 - (1/3)*(C)^3 = (3/2)C^3 - (1/3)C^3 = (7/6)C^3 > 0
Thus, the temperature change increases for doses between zero and 9C/2.
Considering a value greater than 9C/2 (e.g., C):
dT/dD = 3(C/2)*(C)^2 - (1/3)*(C)^3 = (3/2)C^3 - (1/3)C^3 = (7/6)C^3 > 0
So, the temperature change also increases for doses greater than 9C/2.
Therefore, the temperature change increases as the dose increases for doses between zero and 9C/2, exclusive.