The volume of a sperical hot air balloon expands as the air inside the balloon is heated. The radius of the balloon, in feet, is modeled by a twice differentiable function r of time t, where t is measured in minutes. For 0<t<12, the graph of r is concave down. The table above gives selected values of the rate of change, r'(t), of the radius of the balloon over the time interval 0<t<12. The radius of the balloon is 30 feet when t=5. (Remember volume of spere formula). Estimate the radius of the balloon when t=5.4 using the tangent line approximation at t=5 is your estimate greater than or less than the true value? Give reason of answer. And please show all work!!!
T=0, 2, 5, 7, 11, 12
R'(t)= 5.7, 4.0, 2.0, 1.2, 0.6, 0.5( this is the graph)
Update:oh never mind, sorry
Answers & Comments
Verified answer
If the radius at t=5.4 is estimated using a tangent-line approximation at t=5,
the estimate will be greater than the true value,
because the rate of increase in the radius will not remain at 4.0
throughout the next 0.4 time units, but will likely decrease to about 3.6.
That answers the LAST question.
Now for the specific estimates:
Using the tangent-line approximation from t=5, you have
r(5.4) = 30 + (4.0)(0.4) = 31.6
A more realistic estimate is obtained by assuming that the average value
of r '(t) in the interval (5.0,5.4) is 3.8, yielding
r(5.4) = 30 + (3.8)(0.4) = 31.4
I am assuming that we're constructing a 4 sided cage with no top and one of the aspects is gilded. If that's the case, the cage can have dimensions a x b (assuming we can ignore the peak). Seeing that we wish the lowest viable cost, we all know we would like the lowest feasible subject, so ab = 2025. Then, we find the whole rate for the cage. Assuming the gilded aspect is length b, it's 3a + 3a + 3b + 51b = fee. So, C = 6a + 54b. Subbing in a = 2025/b, we get that C = 12150/b + 54b. Taking the spinoff of all sides, we get that C' = 54 - 12150/b^2. On account that we want to find where C is 0, we get 54b^2 = 12150, or b^2 = 225, or b = 15. Considering that the derivative of C changes from terrible to optimistic once b = 15, we know that b is the only nearby minimum and have to be the size of the minimal rate. That leaves us with a = 2025/b = 135. So, the size of the cage are 15 via 135, with the gilded side being 15 cubits long. Again, I can not say i'm entirely definite this is right for definite given that of how the obstacle is worded, but I suppose pretty good about it.