Hi, been tearing my hair out over this question. I think I'm on the right track, but just not sure how to finish it. Anyway, here it is:
A particle is moving along the ellipse 4x^2 + 16y^2 = 64. At any time 'T' its x- and y-coordinates are given by x = 4cosT, and y = 2sinT. At what rate is the particle's distance to the origin changing at an arbitrary time 'T'?
Here's what I have so far:
#1: I know that the rate of change of the distance isn't going to be constant because the path is an ellipse. Any time the particle approaches an axis, the rate will be slowest.
#2: I believe if I say that "D = distance between the particle and the origin" I would be correct to say that; D = sqrt(x^2 + y^2) using the pythagorean theorum. In this case x = 4cosT and y = 2sinT.
#3: dx/dT = -4sinT and dy/dT = 2cosT (right?)
#5: To find the rate of change (and here's where I'm least sure), I would need to find the implicit derivative of D = sqrt(x^2 + y^2) with respect to T. In other words, dD/dT plugging in the derrivatives of X and Y with respect to T accordingly.
Am I on the right track? If so, I can't seem to get what the book has. I may be doing my solving wrong, I'm just not sure. The book has the answer as this:
(-12sinTcosT) / sqrt[16(cos^2)T + 4(sin^2)T]
I know this question is a bit of a pain, but any help would be greatly appreciated.
Hi, been tearing my hair out over this question. I think I'm on the right track, but just not sure how to finish it. Anyway, here it is:
A particle is moving along the ellipse 4x^2 + 16y^2 = 64. At any time 'T' its x- and y-coordinates are given by x = 4cosT, and y = 2sinT. At what rate is the particle's distance to the origin changing at an arbitrary time 'T'?
Here's what I have so far:
#1: I know that the rate of change of the distance isn't going to be constant because the path is an ellipse. Any time the particle approaches an axis, the rate will be slowest.
***I don't know about this yet****
#2: I believe if I say that "D = distance between the particle and the origin" I would be correct to say that; D = sqrt(x^2 + y^2) using the pythagorean theorum. In this case x = 4cosT and y = 2sinT.****Correct*****
#3: dx/dT = -4sinT and dy/dT = 2cosT (right?)
#5: To find the rate of change (and here's where I'm least sure), I would need to find the implicit derivative of D = sqrt(x^2 + y^2) with respect to T. In other words, dD/dT plugging in the derrivatives of X and Y with respect to T accordingly.***correct****
Am I on the right track? If so, I can't seem to get what the book has. I may be doing my solving wrong, I'm just not sure. The book has the answer as this:
(-12sinTcosT) / sqrt[16(cos^2)T + 4(sin^2)T]
***********
d^2= x^2 + y^2
2d dD/dt = 2x dx/dt + 2y dy/dt
dD/dt=[ x dx/dt + y dy/dt] /(sqrt (x^2 + y^2)
will lead to it. A comment on the books answer as you typed it. Remember that cos^2T + sin^2T =1
Thanks a bunch!!
Yes, you are on the right track!
To find the rate of change of the particle's distance to the origin at an arbitrary time 'T', you correctly defined the distance as D = sqrt(x^2 + y^2), where x = 4cosT and y = 2sinT.
To find dD/dT, you need to apply the chain rule.
Start by finding the derivatives of x and y with respect to T. You already found that dx/dT = -4sinT and dy/dT = 2cosT.
Using the chain rule, the derivative of D with respect to T is given by:
dD/dT = (dx/dT * x + dy/dT * y) / sqrt(x^2 + y^2)
Substitute the values of dx/dT, dy/dT, x, and y:
dD/dT = (-4sinT * 4cosT + 2cosT * 2sinT) / sqrt((4cosT)^2 + (2sinT)^2)
Simplifying further:
dD/dT = (-16sinTcosT + 4cosTsinT) / sqrt(16cos^2T + 4sin^2T)
Combining like terms:
dD/dT = (-12sinTcosT) / sqrt(16cos^2T + 4sin^2T)
So, the book's answer (-12sinTcosT) / sqrt[16(cos^2)T + 4(sin^2)T] is correct.
I hope this helps! Let me know if you have any other questions.