the point on the curve is 4y=x^2 nearest to (7,2) is:
this is what i did: i solved for y and I know that the there's a tangent line at (x1,y1).
so using y-y1=m(x-x1), I found the lines of the tangent and perpendicual lines:
slope is derivative so it is 1/2x1
tangent: y-y1=.5x1(x-x1)
perpendicualar: y-y1=-2/x1(x-x1)
then using system of equations i got (1/56, 784). i know that's wrong but i can't seem to know what i did wrong.
I also heard that you can do this problem by using the distance formula. can you explain that please
You have 4y=x^2 or y =(1/4)x^2, so the distance from the point to the curve is
D^2 = (x-7)^2 + (y-2)^2
substituting for y you get
D^2 = (x-7)^2 + ((1/4)x^2-2)^2 or
D= sqrt((x-7)^2 + ((1/4)x^2-2)^2)
Now find
dD/dx = 1/(2sqrt((x-7)^2 + ((1/4)x^2-2)^2)) * [2(x-7) + 2((1/4)x^2-2)*(1/2)x]
Setting this to 0 you get
[2(x-7) + 2((1/4)x^2-2)*(1/2)x]=0
which simplifies to
2x-14 + (1/4)x^3 -2x = 0 or
(1/4)x^3 = 14 and
x^3 = 56 so x = 56^(1/3)
See if this agrees with the method you're using.
14x(3)
2 answers