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).
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!!!!!
4 answers
I have no fricken clue. That's why I fricken hatee MATH. LIKE HOLY FRICK! WHY THE FRICK DO WE HAVE TO LEARN?
JEEZZZ-,-
dont understand-.-"
it will be easier to learn by talking...
im confused with the signs hahaha!!!
dont understand-.-"
it will be easier to learn by talking...
im confused with the signs hahaha!!!
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
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
1 answer