The drag on an airplane traveling at velocity v is D=av^2+(b/v^2) where a and b are positive constants. At what speed does the airplane experience the least drag.
I have so far that D'=-2b/v^3 do i set this equal to zero then i get 0=-2 don't think that is the right way where do i go from here
I get
D' = 2av - 2b/v^3
then set D'=0 to get
2av - 2b/v^3 = 0
v = ∜(b/a)
how do you get 2av should that be zero since a is a constant or did i miss something in class
a is a constant, just like b.
d/dv(av^2) = (a)(2v) = 2av
it doesn't make the whole term 0, it just multiplies. Funny you kept the b, but not the a.
To find the speed at which the airplane experiences the least drag, you need to find the critical points of the drag function by setting its derivative equal to zero.
You correctly found that the derivative of the drag function D with respect to velocity v is D' = -2b/v^3. Now, let's set this derivative equal to zero and solve for v:
-2b/v^3 = 0
Since b is a positive constant, we can see that the only way for the equation to be satisfied is if the numerator is equal to zero:
-2b = 0
This equation simplifies to 0 = 0. Here, we have an indeterminate form, which means that the derivative does not provide enough information to determine the exact value of v at the critical point.
In such cases, we need to consider the behavior of the function around the critical point. Let's examine the original drag function:
D = av^2 + (b/v^2)
Notice that the first term, av^2, increases as v increases, while the second term, (b/v^2), decreases as v increases. Therefore, the graph of the drag function is U-shaped, with a minimum point at v = v_min.
In other words, as v decreases from positive infinity, the drag initially increases due to the av^2 term dominating. However, as v continues to decrease, the (b/v^2) term becomes increasingly dominant, causing the drag to eventually decrease.
Hence, the minimum drag occurs at the lowest possible speed, or as v approaches zero.