The altitude attained by a model rocket t seconds into flight is given by the equation:
-t^3+3t^2+20t-3 where t is greater than or equal to 0 seconds. find the maximum altitude attained by the rocket.
first I found the derivative of the original equation and set it to 0. which comes out to be -3t^2+6t+20=0. Since this can't be factored what are the next steps?
Thank you!
the quadratic formula is your friend. use it.
To find the maximum altitude attained by the rocket, we need to find the value of t when the derivative of the altitude equation equals zero.
Let's continue solving the equation -3t^2 + 6t + 20 = 0.
One way to solve this quadratic equation is by using the quadratic formula:
t = (-b ± √(b^2 - 4ac))/2a
In this case, a = -3, b = 6, and c = 20.
Substituting the values, we have:
t = (-(6) ± √((6)^2 - 4(-3)(20)))/(2(-3))
Simplifying further:
t = (-6 ± √(36 + 240))/(-6)
t = (-6 ± √(276))/(-6)
Now we need to simplify the expression under the square root:
√276 is approximately equal to 16.6132
t = (-6 ± 16.6132)/-6
Now we can evaluate the two possibilities:
1. t = (-6 + 16.6132)/-6 = 10.6132/-6 ≈ -1.7689
2. t = (-6 - 16.6132)/-6 = -22.6132/-6 ≈ 3.7689
We can see that t = -1.7689 is not possible since the time cannot be negative.
Therefore, the maximum altitude is attained when t is approximately 3.7689 seconds into the flight.