Need help with Calculus Please?

Question

I don't know how to do this. Can anyone help me?

The question is:
Find a point on the unit circle x^2+y^2=1 with minimum distance from the point (1,1). (Hint: Unse the parametric representation x=cost y=sint for the cirlce and find the distance as a function of t.)

I plugged in the parametic representation and got (cost)^2+(sint)^2. I don't know where to go from there.

Can someone please help me?

Answer 1

distance from a point in the circle to (1,1) is
sqrt( (cost -1)^2 + (sint -1)^2 )

as you can see this is a function of t,
but let us work with the square.
f(t) = (cost -1)^2 + (sint -1)^2
= cos^2 t -2cost +1 + sin^2 t -2sint +1
= 3 - 2cost -2 sint

now to find the minimum, or any other critical point,
we get the derivative of f(t)

f'(t) = 2 sint -2 cost =0
if sint = cost
t= pi/4, 7pi/4, etc

f''(t)= 2cost + 2sint
f''(pi/4) = 2 cos(pi/4) + 2sin(pi/4) >0, so pi/4 is a loc minimum
and
f''(7pi/4) =2 cos(7pi/4) + 2sin(7pi/4) =0
'

Answer 2

(x,y) or (cos t, sin t) is a point on the circle. (cos t)^2 + (sin t)^2 is the distance from (0,0), which is 1, but you don't want that at all.
You want the distance to (1,1).
So, using the distance formula, the distance is sqrt((cos t - 1)^2 + (sin t - 1)^2).
Since we just want to minimise the distance, we can ignore the square root, and make the thing inside as small as possible.

So (cos t - 1)^2 + (sin t - 1)^2 = (cos t)^2 - 2 cos t + 1 + (sin t)^2 - 2 sin t + 1. We know (cos t)^2 + (sin t)^2 always equals 1, so that simplifies to:
3 - 2 cos t - 2 sin t.
Now, differentiating and making equal to 0:
2 sin t - 2 cos t = 0
sin t = cos t
Or equivalently
sin t / cos t = 1, ie tan t = 1.
Thus t = 45 degrees. That means the point on the circle is (cos 45, sin 45), ie (rt(0.5), rt(0.5)).

Answer 3

The distance formula is:

d=√(x1 - x2)² + (y1 - y2)²

You're looking for the distance from the point (1,1) to the point (cost, sint).

d=√[(1-cost)²+(1-sint)²]
d=√[1 -2cost + cos²t + 1 -2sint + sin²t]

Since sin² + cos² = 1,

d=√[3 - 2(sint + cost)]

Then, to find the minimum, you'd take the derivative of that, set it equal to zero, and solve for t. Kind of ugly... That's as far as I really want to take it, but if you need me to keep going, I'll try.

Ah well, go with what Stephen came up with. :)