# Thread: Related Rates Problem

1. ## Related Rates Problem

I can't seem to figure this problem out

A point p is moving along the curve whose equation is y= x^(1/2). Suppose that x is increasing at the rate of 4 units per second when x=3

a. how fast is the distance between P and the point (2,0) changing at this instant?

b. How fast is the angle of inclination of the line segment from P to (2,0) changing at this instant?

Thanks

2. Originally Posted by Linnus
I can't seem to figure this problem out

A point p is moving along the curve whose equation is y= x^(1/2). Suppose that x is increasing at the rate of 4 units per second when x=3

a. how fast is the distance between P and the point (2,0) changing at this instant?

b. How fast is the angle of inclination of the line segment from P to (2,0) changing at this instant?

Thanks
Point P is at $(x, \sqrt{x})$ and we know that $\frac{dx}{dt} = 4$ when x = 3.

The distance between point P and (2, 0) is
$s = \sqrt{(x - 2)^2 + (\sqrt{x} - 0)^2}$

$s = \sqrt{x^2 - 3x + 4}$

So
$\frac{ds}{dt} = \frac{2x - 3}{2 \sqrt{x^2 - 3x + 4}} \cdot \frac{dx}{dt}$

So when x = 3 the distance is changing at a rate of
$\frac{ds}{dt}(x = 3) = \frac{2 \cdot 3}{2 \cdot \sqrt{3^2 - 3\cdot 3 + 4}} \cdot 4 = 3$

The angle of the line segment formed by this distance is characterized by the slope between the two points:
$m = \frac{\sqrt{x} - 0}{x - 2} = \frac{\sqrt{x}}{x - 2}$
(I know this isn't standard form, but I'm just going to leave it like this for now.)

Now, geometrically speaking
$m = tan(\theta)$
where $\theta$ is the angle of inclination.

So
$\theta = tan^{-1}(m) = tan^{-1} \left ( \frac{\sqrt{x}}{x - 2} \right )$

Now,
$\frac{d}{dx}tan^{-1}(x) = \frac{1}{x^2 + 1}$.

Thus
$\frac{d\theta}{dt} = \frac{1}{\left ( \frac{\sqrt{x}}{x - 2} \right )^2 + 1} \cdot \left ( \frac{ \frac{x - 2}{2 \sqrt{x}} - \sqrt{x}}{(x - 2)^2} \right ) \cdot \frac{dx}{dt}$

$\frac{d\theta}{dt} = \frac{1}{\frac{x}{(x -2)^2} + 1} \cdot \left ( \frac{(x - 2) - 2x}{2(x - 2)^2 \sqrt{x}} \right ) \cdot \frac{dx}{dt}$

$\frac{d\theta}{dt} = -\frac{(x - 2)^2}{x^2 - 3x + 4} \cdot \left ( \frac{x + 2}{2(x - 2)^2 \sqrt{x}} \right ) \cdot \frac{dx}{dt}$

$\frac{d\theta}{dt} = -\frac{x +2}{2(x^2 - 3x + 4)\sqrt{x}} \cdot \frac{dx}{dt}$

$\frac{d\theta}{dt} = -\frac{(x +2)\sqrt{x}}{2x(x^2 - 3x + 4)} \cdot \frac{dx}{dt}$

(I'm tired. I'll let you plug in x = 3.)

-Dan

3. Originally Posted by Linnus
I can't seem to figure this problem out

A point p is moving along the curve whose equation is y= x^(1/2). Suppose that x is increasing at the rate of 4 units per second when x=3

a. how fast is the distance between P and the point (2,0) changing at this instant?

b. How fast is the angle of inclination of the line segment from P to (2,0) changing at this instant?

Thanks
a. how fast is the distance between P and the point (2,0) changing at this instant?

P is on y = (x)^(1/2)
So, P is point (x,sqrt(x))

Distance, D, between P and (2,0).
D^2 = (x-2)^2 +(sqrt(x) -0)^2
D^2 = x^2 -4x +4 +x
D^2 = x^2 -3x +4
Differentiate both sides with respect to time t,
2D(dD/dt) = 2x(dx/dt) -3(dx/dt)
2D(dD/dt) = (2x -3)(dx/dt) -----------------(i)

When x=3,
dx/dt = 4 units/sec
D = sqrt(x^2 -3x +4) = sqrt(9 -9 +4) = 2
Plug those into (i),
(2*2)dD/dt = (2*3 -3)(4)
dD/dt = (12)/4 = 3 unit/sec

Therefore, at that instant, the distance between P and (2,0) is increasing at the rate of 3 unit/sec. --------answer.

------------------------------------------------

b. How fast is the angle of inclination of the line segment from P to (2,0) changing at this instant?

Let angle A be the angle in question

Draw the figure.

tanA = y / (x -2)
tanA = sqrt(x) / (x-2)
Differentiate both sides with respect to time t,
sec^2(A)*dA/dt = {(x-2)*(1 / 2sqrt(x))*(dx/dt) -(sqrt(x)*1(dx/dt))} / (x-2)^2 -----(ii)

when x=3,
dx/dt = 4 units/second
D = 2 units
secA = hypotenuse/(adjacent side) = D/(x-2) = 2/(3-1) = 1

So, substitute those into (ii),
(1^2)dA/dt = {(1)*(1/ 2sqrt(3))(4) -sqrt(3)*4} /(1^2)
dA/dt = 2/sqrt(3) -4sqrt(3)
dA/dt = [2 -4*3] / sqrt(3)
dA/dt = -10/sqrt(3)
dA/dt = -(10/3)sqrt(3) = -5.7735 radian/sec

Therefore, at that instant, the angle of inclination of the line segment from (2,0) to point P is decreasing at the rate of 5.7735 radians/second. --------answer.

4. ## confused

i'm kinda confused...if i do it the inverse trig way...
tan(A)=m
A=arctan (m)
m should be X^(1/2)/ (x-2)
so
A= arctan 2^(1/2)/ (x-2)

when I take the derivative it

and plug in 3 i got -1.443 radians per second

5. Originally Posted by Linnus
tan(A)=m
A=arctan (m)
m should be X^(1/2)/ (x-2)
so
A= arctan 2^(1/2)/ (x-2)
How did you get from line 2 to line 3?

-Dan

6. Originally Posted by topsquark
How did you get from line 2 to line 3?

-Dan
my bad, I mistyped

A=arctan (x^(1/2)/x-2)

7. Originally Posted by Linnus
my bad, I mistyped

A=arctan (x^(1/2)/x-2)
Okay. Looking at this I noticed that I had my slope flipped over. I fixed it in my original post, so you can refer to it there. Oddly enough all it did was throw a negative sign onto my original answer. Weird.

Anyway, I plug in x = 3 and get $\frac{d\theta}{dt} = -\frac{5\sqrt{3}}{6}~rad/s \approx -1.44338~rad/s$ so apparently I agree with you.

-Dan

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### a point p ismoving along the curve whose equation is y= root x. suppose that x is increasing at the rate of 4 units /s when x=3. how fast is the distance between p and the point (2,0) changing at this instant

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