4a:The distance between 2 points,
)
and
)
is given by
^2+(y_2-y_1)^2))
. In this case the points are (m,n) (the desalination plant) and (0,1) (the village). So the distance, L, is
^2+(n-1)^2))
. You use that and
to get the L shown.
4c: First, we find the distance he covers when he runs. He runs to (x,y) (the point on the river) from (0,0) (the camp). We also know that
, so he runs from (0,0) to
)
. We use the distance formula again to get
^2})
, which is
. He runs this distance with a speed of 2km/h. Speed = distance/time, so the time is distance/speed. So the time for Tasmania running is
.
The time he takes to swim is proportional to the difference between the y coordinate of the desalination plant and the point he enters.
The y coordinate of where he enters is y, or
, and the y coordinate of the desalination plant is 3/4, so the difference is
=\frac{7}{4}-x^2=\frac{1}{4}(7-4x^2))
. Proportional means that the time is equal to a real constant times the difference, or
)
.
The total time is the running time and the swimming time, or
)
.
4f: If he runs from his camp to the desalination plant he enters the river at the desalination plant, ie at
)
. We also know that
. For the time to be a minimum when he runs directly to the camp, we want the minimum to occur when
. So you can substitute
to
and solve for when
, which gives
. When k is above this the time for him swimming will increase, so for all values of k above this he will still run directly to the plant for minimum time, so the values of k are
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