David H. answered • 02/18/17
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So let r be the radius.
the area of a circle is A = pi*r^2
So how does the area change with time? Take the derivative wrt time, t. This is using implicit differentiation
dA/dt = (2pi)r*dr/dt
we are given dr/dt = 1 m/s
and we want at r = 2m
so dA/dt = (2pi)(2m)(1m/s) = 4pi m^2/s.
dont forget units. Area is length ^2 so your answer should be of form length^2/time. If you didn't get that then you likely messed up somewhere.
best - david
David H.
Ah okay so a little different but the above formula still holds.
dA/dt = (2pi)r*dr/dt
The problem now is we need to figure out what r is as a function of t and from that we can get dr/dt.
So we are given the ripple moves out at a constant speed of 1 m/s.
key word is constant.
so dr/dt = 1 m/s. and the derivative of dr/dt is 0 since constant. I.E. d2r/dt2 = 0.
So there are a couple ways to go about this. I'm not sure you are taking physics or if you have started integration in calculus.
I'll do both ways.
1.) Physics way:
The radius of the ripple as a function of time can be given by
r(t) = r0+vrt+(1/2)art2 where r0 is the initial radius of the ripple (in this case it is 0), vr=dr/dt = velocity of the ripple (in this case it is 1 m/s) and ar = d2r/dt2 = acceleration of the ripple (in this case it is 0 m/s^2 because we are given the velocity is constant).
Thus
r(t) = 0+1t+0t^2 = t.
And dr/dt = 1 (if it did not equal 1 we made a mistake because we are given dr/dt = 1 m/s).
2.) Calculus way:
you can integrate dr/dt to get the position r(t).
∫(dr/dt)dt from 0 to t = r(t)-r(0). In this case r(0) = 0 so we have
∫(dr/dt)dt = ∫1dt = t = r(t)
Alright back to our problem.
We now know r(t) = t and dr/dt = 1 so
dA/dt (t) = (2pi)r*dr/dt = (2pi)t*(1) = (2pi)t.
So plugging in t = 2 you get
dA/dt (2) = 4pi m^2/s.
End up with the same answer.
Best - David
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02/18/17
Sultan A.
Hi David,
The only problem that I am facing is finding the value of the radius. I have seen your solution but still don't understand the last step when find that r=t.
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02/18/17
David H.
have you taken any physics or done integrals in calculus yet (also known as anti-derivatives)
Without either of those just think about a car for example.
If i'm driving 30 mph at a constant speed (meaning i'm driving 30 mph at any time).
How far have i driven in 1 hour?
well since i'm driving 30 miles/hr. To figure out my distance in 1 hr its 30 miles/hr * 1 hr = 30 miles. I have driven 30 miles.
Same concept for this. The radius of the ripple is increasing 1 m/s.
So if 1 second, the ripple has increased to a radius of 1 m.
In 5 seconds, the ripple has increased to a radius of 5 m.
In 10 seconds, the ripple has increased to a radius of 10 m.
and so forth
so the radius is r(t) = 1m/s*time(in seconds).
So as you see the ripple can be seen as a function of time by r(t) = t.
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02/18/17
Sultan A.
Hi David,
Thanks for the explanation. You're the best. Can I ask you another question please because you are so helpful?
Thanks
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02/18/17
David H.
of course. feel free to message me or post here.
Thanks - David
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02/18/17
Sultan A.
Me second question is:
A sector of angle is cut out of a circular piece of paper of fixed radius. The remainder is then joined atbits two radii such that it forms a cone. Find the angle, if the volume of this cone is to be a maximum.
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02/18/17
David H.
Good question. So what do we know?
1) volume of a cone =
V = pi*r^2*h/3
2.) you cut out a sector of a circle (let's call the radius of the circle k) and form a cone.
Easier to visualize if we drew a picture but that means you have a cone
Let's let the base of it be a circle of radius r that we will need to determine and the the height h.
3) Well what is h And what is r?
the circle we cut has radius k so the slant edges of the cone are k
if you can imagine we have a isosceles triangle
two sides are k and the other side is 2r (the diameter of the base circle)
using Pythagorean theorem we can get height of Kome
h = [k^2-r^2]^(1/2)
So
V = pi*r^2*h/3 = pi*r^2*[k^2-r^2]^(1/2)/3
we we are told the radius of the circle we cut is fixed. That means the derivatI've is 0.
take dV/dr (implicit differentiation and product rule ]
dV/dr = [(2Pi)*r*dr/dr*(k^2-r^2)^(1/2)/3]+[pi*r^2*(1/2)(k^2-r^2)^(-1/2)*(2k*dk/dr-2r*dr/dr)/3] {simplify}
dV/dr = [(2pi*r*(k^2-r^2)^(1/2)/3]+[pi*r^2*(k^2-r^2)^(-1/2)*(-r)/3]
remember radius of original circle k is constant so dk/dr = 0
lets make a little prettier and simplify
dV/dr = (pi*r/3)*[2(k^2-r^2)^(1/2)-r^2(k^2-r^2)^(-1/2)]
lets get common denominator
dV/dr = (pi*r/3)[{2(k^2-r^2)-r^2}/(k^2-r^2)^(1/2)]
set to 0 and get rid of the (pi*r/3) and then the denominator (k^2-r^2)^(1/2)
(We will go over why in a little)
so
dV/dr = 0 = 2(k^2-r^2)-r^2)
= 2k^2-3r^2
hence 3r^2 = 2k^2
so r = (2/3)^(1/2)*k
[since lengths have to be positive we can not worry about absolute values (x^2)^(1/2) = |x|]
now what if the denominator equaled 0,
then r=k
well that means h = [k^2-r^2]^(1/2) = 0 and our volume is 0 since height is 0. That can't be (that means the base of our circle is the whole original circle and we didn't cut)
so the r we got is a maximum and our maximum volume is
V = pi*[(2/3)^(1/2)*k]^2*[k^2-{(2/3)^(1/2)*k}^2]^(1/2)/3
= pi[(2/3)k^2*(1/3)^(1/2)k]/3 = [(2*pi*sqr(3))/9]*k^(3)
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02/19/17
Sultan A.
Hi David,
So, what is the value of the angle?
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02/19/17
Sultan A.
02/18/17