Calculus Help Please?

Question

Please show me how to do this:

An object moves along the x-axis at a rate of 5 m/s, so its position is at (b, 0) with b' = 5. Find the rate of change of the distance between the object and the parabola y = x^2, at the time when the point on the parabola nearest to the object is the point (1, 1).

Thank you in advance

Answer 1

Find the distance between them as a function of both (b,0) and (x,y) on the parabola.

d^2 = (x - b)^2 + (y - 0)^2

now to find the rate of change of the distance use implicit differentiation (or take sqrt and use power rule)

2*d*d(d)/dt = 2*(x - b)*(dx/dt - db/dt) + 2*(dy/dt)

it sounds like in this problem that (x,y) on the parabola is fixed at some value (1,1) so dx/dt = 0 and dy/dt = 0

so d(d)/dt = -1/d*(x - b)*db/dt

to find the rate then we need to know d and b at the specified time. The point (b,0) that is closest to (0,0) lies along the line perpendicular to the tangent of the parabola at (1,1).

the slope of this line is dy/dx (1,1) = 2*x = 2 so the line perpendicular is -1/2 and passes through (1,1)

y = mx + b (this b is the y-intercept of the line I am using, not the b in the position (b,0)...)

y = -1/2*x + b, solve for b

1 = -1/2*1 + b => b = 3/2
y = -1/2*x + 3/2
then I know that (b,0) [this b is the b in (b,0) now...] also lies on this line so..

0 = -1/2*b + 3/2 => b = 3

So the point on the x axis closes to (1,1) on the parabola y = x^2 is (3,0). Now just find the distance d between these points...

d^2 = (1 - 3)^2 + (1 - 0)^2 = 5
d = sqrt(5) so now we know d(d)/dt is

d(d)/dt = -1/d*(x - b)*db/dt = -1/sqrt(5)*(1 - 3)*5 = +2*sqrt(5)

Answer 2

What does b'=5 mean? Are you saying dx/dt = 5?
x = b = 5t, where t is the time in seconds and t= 0 when x=0.
The slope of the parabola is 2x = 2 at (1,1).
The line perpendicular to the parabola's slope at x=1 has a slope of -.5.
Its equation is y = -.5x +b
1 = -.5 + b --> b = 1.5
So equation of line perpendicular to parabola at (1,1) is
y = -.5x +1.5
This line has an x-intercept at x = b = 3. or (3,0).
The distance between(1,1) and (3,0) is rt(-1^2 +2^2) = rt(5).
rate of change is rt(5)*5/3 m/sec

Answer 3

2...? :D

just guessing; since there is no change in height of the object, therefor the only one changing is the moving particle on the parabola... derivative of x^2 is 2x and using 1 as the value of x...given the point (1,1)...hehe

sorry if i ruined your day... :P

Answer 4

You won't have Yahoo Answers when your tested on how to do this. Do your own homework.