Volume in liters of antifreeze remaining in a leaking radiator after t minutes is V= 20- 2t- 0.1t^2
a) how long does it take to drain the antifreeze mixture from the radiator?
i got t= 10
b) how fast is volume changing when t=5?
i got 9.75
is this right. thanks
V= 20- 2t- 0.1t^2
a) v=0 when t=10(√3-1) = 7.32
How did you get 10? v(10) = 20-20-10
b) dv/dt = -2 - .2t
at t=5, dv/dt = -2-1 = -3
Apparently, you have some misunderstanding of how related rates work.
I read more about related rates, and have now understood these types of problems. Thank you!
To find the time it takes to drain the antifreeze mixture from the radiator, you need to set the volume equation equal to zero and solve for t:
V = 20 - 2t - 0.1t^2
0 = 20 - 2t - 0.1t^2
To solve this quadratic equation, you can either factor it or use the quadratic formula. In this case, let's use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
Here, a = -0.1, b = -2, and c = 20. Substituting these values into the formula:
t = (-(-2) ± √((-2)^2 - 4(-0.1)(20))) / (2(-0.1))
t = (2 ± √(4 + 8)) / (-0.2)
t = (2 ± √12) / (-0.2)
t = (2 ± 2√3) / (-0.2)
Since time cannot be negative in this context, we only consider the positive solution:
t = (2 + 2√3) / (-0.2)
t ≈ 9.90
So, the correct answer is t ≈ 9.90 minutes, not 10 minutes.
Now, to find how fast the volume is changing at t = 5, we need to take the derivative of the volume equation with respect to time and then evaluate it at t = 5:
V = 20 - 2t - 0.1t^2
Taking the derivative with respect to t:
dV/dt = -2 - 0.2t
Substituting t = 5 into the derivative:
dV/dt = -2 - 0.2(5)
dV/dt = -2 - 1
dV/dt = -3
Therefore, the correct answer is -3, not 9.75. The volume is changing at a rate of -3 liters per minute when t = 5. The negative sign indicates that the volume is decreasing.