A runner sprints around a circular track of radius 150 m at a constant speed of 7 m/s. The runner's friend is standing at a distance 300 m from the center of the track. How fast is the distance between the friends changing when the distance between them is 300 m? (Round to 2 decimal places.

My teacher tried helping me with this problem and only confused me more. I drew a diagram. I used the formula c^2=a^2+b^2-2abcosC. Right now I have 2L(dL/dt)=2(150)(300)sin(angle)(dangle/dt) I think (dangle/dt) is 7/150. This would give you (solving for dL/dt:
(dL/dt)= (45000sin(angle)(7/100))/300
HELP!!!!!

I have no fricken clue. That's why I fricken hatee MATH. LIKE HOLY FRICK! WHY THE FRICK DO WE HAVE TO LEARN?

The law of cosines can be used as the input to the process of implicit differentiation. Let D be the distance between the friends and θ be the angle between (the line joining the center of the circle and the standing friend ) and (the radius line joining the center of the circle and the instantaneous position of the runner). The law of cosines then is:


D^2 = 150^2 + 300^2 - 2(150)(300) cos(θ)

Differentiating this equation with respect to time gives

2 D D' = (2)(150)(300) sin(θ) θ'

where D' is the derivative of D with respect to time and θ' is the derivative of θ with respect to time.
θ' = velocity / radius = 7/100 = 0.07
The point of interest is where D = 300. From the law of cosines equation, it can be worked out that
cos(θ) at this point is equal to 1/4 . This means that sin(θ) at this point is sqrt(15/16) = .968

Plugging into the the second equation gives

600 D' = (2)(150)(300)(.968)(0.07) Solving for D; gives

D' = 6.776

To solve this problem, we can use the concept of related rates. Let's break it down step by step:

1. First, let's label our variables:
- Let r be the radius of the circular track (r = 150 m).
- Let x be the distance from the center of the track to the friend (x = 300 m).
- Let d be the distance between the runner and the friend.

2. We need to find an expression for d in terms of x. Consider drawing a diagram to visualize the situation. We can use the Pythagorean theorem to relate r, x, and d. Since d is the hypotenuse of a right triangle, we have:
d^2 = r^2 + x^2
d = sqrt(r^2 + x^2)

3. Now, we want to find how fast the distance d is changing with respect to time (dt). This is represented as (dd/dt).

4. To find (dd/dt), we need to differentiate the equation from step 2 with respect to time (t) using the chain rule:
d/dt(d) = d/dt(sqrt(r^2 + x^2))
= (1/2)*(r^2 + x^2)^(-1/2) * (2x*dx/dt)
= (x/r^2) * (dx/dt)

5. We are given that dx/dt = 7 m/s (the runner's speed). Substituting the given values, we have:
(dd/dt) = (x/r^2) * (dx/dt)
= (300/150^2) * (7)
= (300/22500) * 7
= 0.1333 m/s (rounded to 4 decimal places)

Therefore, the rate at which the distance between the friends is changing is approximately 0.13 m/s (rounded to 2 decimal places).

JEEZZZ-,-

dont understand-.-"
it will be easier to learn by talking...
im confused with the signs hahaha!!!

I suggest the following approach.

Let the friend (outside of the tracks) be at the origin.

Express the x and y coordinates of the runner in terms of t, for example,
X(t)=rcos(st/r)+d
Y(t)=rsin(st/r)

s=speed in m/s
r=radius
d=given distance

Express the distance between the two persons in terms of t:
L(t)=√(X(t)²+Y(t)²)
Differentiate L(t) with respect to t, which gives L'(t), the rate of change of distance as a function of t.

Finally, find the times at which the two persons are 300m apart, namely, solve for t at which
L(t)=300.

The answers are t1=39 and t2=96 seconds approx.
L'(t1) and L'(t2) are the required answers.

I get about ±6.8 m/s, which is reasonable considering that the runner runs at 7m/s and the points in question are not far from the points of tangency (where L'(t) should give ±7m/s).