Suppose the function G(t) represents a test grade (out of 100 points) as a function of hours studied. If
G(t)=-0.046t^3+0.915t^2+38.005 points, what is (dG/dt)when t=4? Round your answer to the nearest point per hour.
a. 50 points per hour
b. 3 points per hour
c. 5 points per hour
d. 7 points per hour
e. 20 points per hour
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Derivative of t ^ 3 = 3 * t ^ 2
Derivative of t ^ 2 = 2 * t
Derivative of 38.005 = 0
dG / dt = - 0.046 * 3 * t ^ 2 + 0.915 * 2 t + 0 = - 0.138 * t ^ 2 + 1.83 * t
When t = 4
then :
dG / dt = - 0.138 * 4 ^ 2 + 1.83 * 4
dG / dt = - 0.138 * 16 + 7 . 32
dG / dt = - 2.208 + 7.32
dG / dt = 5.112
Rounded to the nearest point per hour :
dG / dt = 5 points per hour
To find the derivative of a function, in this case, dG/dt, we can use the power rule. The power rule states that for any term of the form ax^n, where n is a constant and a is a coefficient, the derivative is given by nax^(n-1).
In this case, we have the function G(t) = -0.046t^3 + 0.915t^2 + 38.005. To find dG/dt, we take the derivative of each term separately:
dG/dt = d/dt (-0.046t^3) + d/dt (0.915t^2) + d/dt (38.005)
The derivative of -0.046t^3 is -0.046 * 3t^(3-1) = -0.138t^2.
The derivative of 0.915t^2 is 0.915 * 2t^(2-1) = 1.83t.
The derivative of 38.005 is 0, since it's a constant.
Now we can substitute t=4 into the derivative expression:
dG/dt = -0.138t^2 + 1.83t
dG/dt = -0.138(4)^2 + 1.83(4)
dG/dt = -0.138(16) + 1.83(4)
dG/dt = -2.208 + 7.32
dG/dt = 5.112
Rounding to the nearest point per hour, we get dG/dt ≈ 5 points per hour.
Therefore, the correct answer is c. 5 points per hour.