Homework Help: calculus

Posted by Jim on Wednesday, December 12, 2012 at 9:28pm.

Amy wants to reach, as quickly as possible, a destination of 5km downstream and on the opposite side of the river that is 1.5km wide. If she can run at 16kmp and can swim at 5kmp, how far should she run down the shore before turning into the water to swim? Consider the river current to be negligible.

calculus - bobpursley, Wednesday, December 12, 2012 at 9:47pm
draw the figure.
she runs d1 down stream. that leaves 5-d1 left. The distance from that point to the end point is then

sqrt((5-d1)^2+1.5^2)= sqrt(xxxx)

time = d1/16 p + sqrt(xxxx )/5 p I dont know what the p is in kmp.

dtime/d1=0=1/16+1/2 ( xxxx )^-1/2 *2(5-d1)(-1)

1/16=(5-d1)/sqrt(xxxx)
multiply both sides by sqrt(xxxx)
then square both sides, you should get a quadratic and can solve d1.

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