Find the minimum distance from the parabola

x + y^2 = 0
to the point (0,-3).

The manager of a large apartment complex knows from experience that 90 units will be occupied if the rent is 360 dollars per month. A market survey suggests that, on the average, one additional unit will remain vacant for each 9 dollar increase in rent. Similarly, one additional unit will be occupied for each 9 dollar decrease in rent. What rent should the manager charge to maximize revenue?

A particle is moving with acceleration a(t) = 30 t + 12. its position at time t =0 is s(0) = 11 and its velocity at time t =0 is v(0) = 16. What is its position at time t = 8?

the square of the distance from a point (x,y) to the point (0,3) is given by

d^2 = [(x-0)^2 + (y+3)^2]
= x^2 +y^2 + 6 y + 9

in this case x = -y^2
so
d^2 = z = y^4 + y^2 + 6 y + 9
find dz/dy and set it to 0
0 = 4 y^3 +2 y + 6
hmm, try y = -1
0 - -4 -2 + 6 sure enough

d units/d rent = -1/9 = du/dr

du = -(1/9) dr
u = -(1/9) r + C
when r = 360, u = 90
90 = -(1/9)360 + c
810 = -360 + 9c
c = 130
so
u = -(1/9)r + 130
revenue = z = u r
z = u r = -(1/9)r^2 + 130 r
dz/dr = 0 = -(2/9)r + 130
r = 585

a(t) = 30 t + 12

v = Vi + 15 t^2 + 12 t
x = Xi + 5 t^3 + 6 t^2
now put in your initial conditions to find Vi and Xi and therefore x(t)

To find the minimum distance from the parabola x + y^2 = 0 to the point (0, -3), we can use the concept of perpendicular distance.

First, let's rewrite the equation of the parabola in the standard form:

x + y^2 = 0
y^2 = -x
y = ±√(-x)

Now, to find the distance from a point to the parabola, we draw a perpendicular line from the point to the parabola. The length of this perpendicular line is the minimum distance.

Let's consider a general point on the parabola, (t, ±√(-t)). The slope of the tangent to the parabola at this point can be found by differentiating y with respect to x:

dy/dx = d(±√(-x)) / dx = ±1 / (2√(-x))

Now, using the point-slope form of the equation of a line, the equation of the tangent line at the point (t, ±√(-t)) on the parabola is:

y - √(-t) = ±1 / (2√(-t)) * (x - t)

Next, we consider the line passing through (0, -3) with slope -2/3 (the negative reciprocal of the tangent line slope), which is perpendicular to the tangent line. Using the point-slope form again, the equation of this line is:

y + 3 = -2/3 * (x - 0)

Now, we solve the system of equations formed by the tangent line and the perpendicular line by finding their point of intersection. Substitute the y-value from the second equation into the first equation:

±√(-t) - √(-t) = ±1 / (2√(-t)) * (3 - 0)
0 = ±(3/2√(-t)) * (1 - √(-t))
0 = ±(3/2) * (1 - √(-t))

Now, we can solve for √(-t):

±1 - √(-t) = 0
√(-t) = ±1
-t = 1 (since t is real)
t = -1

So, the point of intersection of the two lines is (-1, -1).

Now, we can find the distance between (-1, -1) and (0, -3) using the distance formula:

Distance = √((x2 - x1)^2 + (y2 - y1)^2)
= √((0 - (-1))^2 + (-3 - (-1))^2)
= √((1)^2 + (-2)^2)
= √(1 + 4)
= √5

Therefore, the minimum distance from the parabola x + y^2 = 0 to the point (0, -3) is √5.