1. ## Calculus Optimization

I have two different problems one involves the antiderivatives and the second involves an optimization problem.

1)Two balls are thrown upward from the edge of the cliff. The first is thrown with a speed of 48ft/sec and the second one is thrown a second later with a speed of 24ft/sec. Do the balls ever pass each other?

2)A woman at a point A on the shore of a circular lake with radius 2 mi wants to arrive at the point C diametrically opposte A on the other side of the lake in the shortest possible time. She can walk at the rate of 4mi/hr and row a boat at 2mi/h. How should she proceed?

2. Hello, UMStudent!

1) Two balls are thrown upward from the edge of the cliff.
The first is thrown with a speed of 48 ft/sec
and the second one is thrown a second later with a speed of 24 ft/sec.
Do the balls ever pass each other?

Let the top of the cliff be y = 0.

At time t, the first ball has height: .y .= .48t - 16t²
. . and the second ball has height: .y .= .24(t - 1) - 16(t - 1)²

They will pass when they both have the same height.
. . 48t - 16t² .= .24(t - 1) - 16(t - 1)²

This simplifies to: .8t = 40 . . t = 5

The second ball overtakes the first ball 5 second after the first ball is thrown,
. . at a height 160 feet below the top of the cliff.

2) A woman at a point A on the shore of a circular lake with radius 2 mi wants to
arrive at the diametrically opposite point C on the other side of the lake in the
shortest possible time. .She can walk at the rate of 4 mph and row a boat at 2 mph.
How should she proceed?
Code:
             B
* * *
*     \     *
*         \     *
*            \    *
\
*      θ          \ *
A * - - - - * - - - - * C
*    2    O    2    *

*                 *
*               *
*           *
* * *
Let /AOB = θ

She will walk along the circular arc AB.
The length of AB is 2θ miles (where θ is in radians).
At 4 mph, it will take her: .2θ/4 = ½θ hours to walk the arc.

She will row across the chord BC.
The length of the chord can be found with the Law of Cosines.
. . BC² .= .2² + 2² - 2(2)(2)cos(2π - θ) .= .8 + 8·cos θ .= .8(1 + cos θ)
. . BC² .= .16[(1 + cos θ)/2] .= .16·cos²(½θ)
. . Hence: .BC .= .4·cos(½θ) miles.

At 2 mph, it will take her: .4·cos(½θ)/2 = 2·cos(½θ) hours to row across.

Her total time is: .T .= .½θ + 2·cos(½θ) hours

And that is the function we must minimize . . .