A generator produces electrical power, P, in watts, according to the function P(R)=(120R)/(0.4+R)^2,
where R is the resistance, in ohms. Determine the intervals on which the power is increasing.
I would take the first derivative, and make it >0 and solve for the intervals.
1. P&prime&prime(R)=0
->gives you the points at which the slope of P(R) changes from increasing to decreasing or vice versa.
2. Plug in a number between the intervals back into P&prime&prime(R)
-> for example if you found:
R= -1 and R = 4
plug in 0 or 1 or 2 or 3 into P&prime&prime if its positive then the slopes are increasing on the interval from (-1, 4)
To determine the intervals on which the power is increasing, we need to analyze the sign of the derivative of the power function. If the derivative is positive, the power is increasing; if the derivative is negative, the power is decreasing.
First, let's find the derivative of the power function with respect to R.
P(R) = (120R) / (0.4 + R)^2
To find the derivative, we can use the quotient rule. The quotient rule states that for a function f(x) = g(x) / h(x), the derivative is given by:
f'(x) = (g'(x) * h(x) - g(x) * h'(x)) / (h(x))^2
Let's apply the quotient rule to find the derivative:
g(x) = 120R
g'(x) = 120
h(x) = (0.4 + R)^2
h'(x) = 2(0.4 + R)(1) = 2(0.4 + R)
Now, let's substitute these values into the quotient rule:
P'(R) = [(120) * (0.4 + R)^2 - (120R) * 2(0.4 + R)] / [(0.4 + R)^2]^2
Simplifying this expression gives:
P'(R) = (120(0.4 + R)^2 - 2(120R)(0.4 + R)) / (0.4 + R)^4
P'(R) = (120(0.4 + R)^2 - 240R(0.4 + R)) / (0.4 + R)^4
Now, we need to determine when the derivative P'(R) is positive. This tells us when the power P(R) is increasing.
We can solve this by setting the numerator equal to zero and finding the values of R that make the derivative zero. Let's simplify the expression and solve for R:
120(0.4 + R)^2 - 240R(0.4 + R) = 0
120(0.16 + 0.8R + R^2) - 240R(0.4 + R) = 0
120(0.16 + 0.8R + R^2) - 240(0.4R + R^2) = 0
19.2 + 96R + 120R^2 - 96R - 240R^2 = 0
19.2 - 120R^2 = 0
-120R^2 = -19.2
R^2 = 0.16
R = ±0.4
Therefore, the values of R that make the derivative zero are R = 0.4 and R = -0.4.
Next, we need to determine when the derivative P'(R) is positive. We can do this by evaluating the derivative at test points in each interval:
-∞ < R < -0.4, let's pick R = -1:
P'(-1) = (120(0.4 + (-1))^2 - 240(-1)(0.4 + (-1))) / (0.4 + (-1))^4
P'(-1) = (120(0.4 - 1)^2 - 240(-1)(0.4 - 1)) / (0.4 - 1)^4
P'(-1) = (120(-0.6)^2 + 240(0.6)) / (-0.6)^4
P'(-1) = (120(0.36) + 144) / (0.36)^2
P'(-1) = (43.2 + 144) / 0.1296
P'(-1) = 187.2 / 0.1296
P'(-1) ≈ 1446.91
The derivative is positive, P'(-1) > 0. Therefore, the power is increasing in the interval -∞ < R < -0.4.
-0.4 < R < 0.4, let's pick R = 0:
P'(0) = (120(0.4 + 0)^2 - 240(0)(0.4 + 0)) / (0.4 + 0)^4
P'(0) = (120(0.4^2)) / (0.4^4)
P'(0) = (120(0.16)) / (0.16^2)
P'(0) = 19.20 / 0.0256
P'(0) ≈ 750
The derivative is positive, P'(0) > 0. Therefore, the power is increasing in the interval -0.4 < R < 0.4.
0.4 < R < ∞, let's pick R = 1:
P'(1) = (120(0.4 + 1)^2 - 240(1)(0.4 + 1)) / (0.4 + 1)^4
P'(1) = (120(1.4)^2 - 240(1)(1.4)) / (1.4)^4
P'(1) = (120(1.96) - 336) / (1.96)^2
P'(1) = (235.2 - 336) / 3.8416
P'(1) ≈ -27.27
The derivative is negative, P'(1) < 0. Therefore, the power is decreasing in the interval 0.4 < R < ∞.
To summarize, the intervals on which the power is increasing are:
-∞ < R < -0.4
-0.4 < R < 0.4