SOLUTION: Please help me solve this:
Find a number t such that the distance between
(−3, 2)
and
(3t, 2t)
is as small as possible
Algebra.Com
Question 974087: Please help me solve this:
Find a number t such that the distance between
(−3, 2)
and
(3t, 2t)
is as small as possible
Found 2 solutions by Alan3354, Boreal:
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
Please help me solve this:
Find a number t such that the distance between
(−3, 2)
and
(3t, 2t)
is as small as possible
---------------------
-----
dd/dt =
----
t = -5/13
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
(-3,2) and (3t,2t)
need to minimize the squared deviations of (2t-2)^2 and (3t+3)^2
Take the first derivative and set it equal to zero.
2(2t-2)*2 + 2(3t+3)*3=0
8t-16 + 18t +18=0
26t+2=0
t=(-1/13)
(-3,2), (-3/13,-2/13)
Distance is sqrt [(36/13)^2) + (28/13)^2 ]=sqrt (7.67+ 4.64)=3.51 units
Try 0 for t
(-3,2) and (0,0) ;;; distance sqrt (13)=3.61
The graph shows the line that the two points are on. The curve is the distance between the two points for various values of t. It is a minimum at (-2/13) for t. The distance from that minimum to the line is minimized.
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