Vector Kinematics: Distance to O Decreasing When?

[Destrio]

An object is moving in the xy plane with the position as a function of time given by r(vector) = x(t)i + y(t)j
Point O is at r(vector) = 0 . The object's distance to O is decreasing when

A) Vx > 0 , Vy > 0
B) Vx < 0 , Vy < 0
C) xVx + yVy < 0
D) xVx + yVy > 0


i figure i can take the v=r/t

take the derivative

v=dr/dt = d(x(t)i + y(t)j ) / t

I think the answer should be B, since Vx and Vy should both be in the negative direction to return to one, assuming there can't be negative distance.
I'm pretty lost of how to actually prove any of this.

Any suggestions?

Thanks very much

 

[learningphysics]

What is the distance of the object from 0? What is the derivative of the distance with respect to time?

 

[Destrio]

there is no stated distance
we are just given position as a function of time

would the derivative of distance with respect to time be
r(t) = v*dt
x(t)i + y(t)j = v*dt
xi + yj = v

 

[learningphysics]

there is no stated distance
we are just given position as a function of time

would the derivative of distance with respect to time be
r(t) = v*dt
x(t)i + y(t)j = v*dt
xi + yj = v

get the formula for distance in terms of x(t) and y(t), then take its derivative.

 

[Destrio]

r(t) = x(t)i + y(t)j
?
i'm not sure what i can derive from that

thanks for your help

 

[learningphysics]

r(t) = x(t)i + y(t)j
?
i'm not sure what i can derive from that

thanks for your help

Suppose the position vector of an object is r = 3i + 4j, what is the distance from the origin of this object?

 

[Destrio]

r = sqrt((3i)^2 + (4j)^2)

would I want:
r(t) = sqrt[(x(t)i)^2 + (y(t)j)^2]

 

[learningphysics]

r = sqrt((3i)^2 + (4j)^2)

would I want:
r(t) = sqrt[(x(t)i)^2 + (y(t)j)^2]

exactly... but you wouldn't worry about the i and j (i and j are just unit vectors...)

so r = sqrt((3)^2 + (4)^2)

and for your question

r(t) = sqrt[(x(t))^2 + (y(t))^2]

what do you get for dr/dt ?

 

[Destrio]

ok

v(t)=dr(t)/dt= d(sqrt[(x(t))^2 + (y(t))^2]) / dt

im not sure how to proceed from here if that's what you meant by dr/dt

 

[learningphysics]

ok

v(t)=dr(t)/dt= d(sqrt[(x(t))^2 + (y(t))^2]) / dt

im not sure how to proceed from here if that's what you meant by dr/dt

sorry I shouldn't use r(t) since that refers to the position...

instead call it s(t) = sqrt[(x(t))^2 + (y(t))^2] (so s(t) is the distance from the origin)

we are trying to find out when the distance is decreasing... ie when ds(t)/dt is negative.

take the derivative ds(t)/dt. Do you know the chain rule in calculus?

 

[Destrio]

no i don't know the chain rule
it's concept was briefly explained to us
but i haven't taken calculus before, and my physics class is going faster than my calc class which is causing my problems with questions like this

would the derivative ds(t)/dt be
ds(t)/dt = d(sqrt[(x(t))^2 + (y(t))^2]) / dt
or is there something else i can put in?

 

[learningphysics]

no i don't know the chain rule
it's concept was briefly explained to us
but i haven't taken calculus before, and my physics class is going faster than my calc class which is causing my problems with questions like this

would the derivative ds(t)/dt be
ds(t)/dt = d(sqrt[(x(t))^2 + (y(t))^2]) / dt
or is there something else i can put in?

Hmmm... the chain rule is a concept that takes some practice to get used to...

As far as I can see you need the chain rule to solve this problem... It isn't that hard... I highly recommend just looking through your calculus book with regards to the chain rule... they'll explain it much better than I could. Then come back and look at this problem.

Once you're able to write that derivative ds(t)/dt in terms of dx(t)/dt and dy(t)/dt... then you'll see the answer to the question...

 

[dynamicsolo]

no i don't know the chain rule
it's concept was briefly explained to us
but i haven't taken calculus before, and my physics class is going faster than my calc class which is causing my problems with questions like this

would the derivative ds(t)/dt be
ds(t)/dt = d(sqrt[(x(t))^2 + (y(t))^2]) / dt
or is there something else i can put in?

I'm going to chime in for a moment because your situation sounds like that of a number of students I've worked with who got poor advising and were told you can take first semester physics and first semester calculus at the same time. (Not really true -- the physics course assumes you've seen all the necessary math and will *not* take time to review it.)

It sounds like you are in a calculus-based physics course, so you are going to run into a few occasions where your physics course uses something in calculus that you aren't going to see until weeks later. You will need to acquaint yourself with the basic rules of differentiation (derivatives of elementary functions, Product Rule, Quotient Rule, Chain Rule) and basic integration (integrals of elementary functions and definite integration).

You will probably be able to get by in the physics course without the mathematical theory (pretty much), but you'll at least need to know how to use the "tools". See if you can get a little help with that at your end (it shouldn't take too much time -- you just need to know how to do the calculations for now) or maybe ask questions in the appropriate forum here. You should look into doing this *soon*.

 

[Destrio]

Ok,
I've read over the chain rule, as well as some chapters on derivatives from my calculus textbook, it clears up some of the things I have been doing.

if I want ds(t)/dt in terms of x(t)/dt and y(t)/dt

would i do

s(t) = sqrt[(x(t))^2 + (y(t))^2]

x(t) = xo + Vxt
y(t) = yo + Vyt

s(t) = sqrt[(xo + Vxt)^2 + (yo + Vyt)^2]

 

[dynamicsolo]

s(t) = sqrt[(x(t))^2 + (y(t))^2]

x(t) = xo + Vxt
y(t) = yo + Vyt

s(t) = sqrt[(xo + Vxt)^2 + (yo + Vyt)^2]

You are assuming here that the velocity components are constant, which I think is not implied in the problem statement. However, this is a good example to start from. You are now ready to differentiate s(t) with respect to time using the Chain Rule...

 

[Destrio]

I thought of that as I was writing it, I don't know why I just assumed the velocities would be constant, let me try again.

ax(t) = Vx/t
ay(t) = Vx/t

Vx(t) = Vxo + ax(t)t
Vy(t) = Vyo + ay(t)t

x(t) = xo + Vxo + ax(t)t
y(t) = yo + Vyo + ay(t)t

s(t) = sqrt[(xo + Vxo + ax(t)t)^2 + (yo + Vyo + ay(t)t)^2]

 

[learningphysics]

I think it is best to start from here:

s(t) = sqrt[(x(t))^2 + (y(t))^2]

Now without assuming anything about x(t) and y(t)... try to differential s(t) just as it is...

 

[Destrio]

do I want to be replacing the x(t) and y(t) from s(t) = sqrt[(x(t))^2 + (y(t))^2]

 

[learningphysics]

do I want to be replacing the x(t) and y(t) from s(t) = sqrt[(x(t))^2 + (y(t))^2]

well, let's start here. let h = (x(t))^2 + (y(t))^2... I substitute that into the s equation (i'm not going to write s(t), just s).

s = sqrt(h)

now what is ds/dh.

 

[Destrio]

let h = (x(t))^2 + (y(t))^2
s = sqrt(h)

ds/dh = d sqrt(h) / d (x(t))^2 + (y(t))^2
= d sqrt((x(t))^2 + (y(t))^2) / d (x(t))^2 + (y(t))^2

im starting to get lost here
what would ds/dh
if s is the point away from the start
what is h for?

 

[learningphysics]

let h = (x(t))^2 + (y(t))^2
s = sqrt(h)

ds/dh = d sqrt(h) / d (x(t))^2 + (y(t))^2
= d sqrt((x(t))^2 + (y(t))^2) / d (x(t))^2 + (y(t))^2

im starting to get lost here
what would ds/dh
if s is the point away from the start
what is h for?

I'm just making a substitution to illustrate the chain rule... don't worry about x(t) and y(t) for the moment.

s = sqrt(h), ie s = h^(1/2). what's the derivative of s with respect to h... don't substitute anything. just take the derivative.

or what's the derivative of y = x^(1/2) with respect to x.

 

[Destrio]

ds/dh = d (h^1/2) / dh = (1/2)h^(-1/2)

i did that following the power rule d x^n / dx = nx^n-1

 

[learningphysics]

ds/dh = d (h^1/2) / dh = (1/2)h^(-1/2)

i did that following the power rule d x^n / dx = nx^n-1

Exactly.

Now the chain rule says that ds/dt = (ds/dh)*(dh/dt)

That's why I substituted that h = (x(t))^2 + (y(t))^2 into the s equation. The chain rule helps me calculate the derivative ds/dt, by calculating two easier ones.

Now, try taking the derivative of h with respect to t. ie try to calculate dh/dt. using: h = (x(t))^2 + (y(t))^2

 

[Destrio]

ds/dt = (ds/dh)*(dh/dt)

dh/dt = d (x(t))^2 + (y(t))^2) / dt = 2x(t) + 2y(t)

 

[learningphysics]

ds/dt = (ds/dh)*(dh/dt)

dh/dt = d (x(t))^2 + (y(t))^2) / dt = 2x(t) + 2y(t)

Here actually you need to use the chain rule also... I'm just going to call it x, instead of x(t) and y instead of y(t)

I'll show the steps here:

dh/dt = d(x^2 + y^2)/dt = d(x^2)/dt + d(y^2)/dt = [d(x^2)/dx]*(dx/dt) + [d(y^2)/dy]*(dy/dt) (take note of how I used the chain rule in this last step)

So then we get: 2x*(dx/dt) + 2y*(dy/dt)

So ds/dt = (ds/dh)*(dh/dt) = [(1/2)h^(-1/2)]*(2x*(dx/dt) + 2y*(dy/dt))

Hope this makes at least a little sense... takes a while getting used to the chain rule.

Here's a simpler example of the chain rule:

taking the derivative of: (x^2 + 3)^5 with respect to x. I get [5(x^2+3)^4]*(2x)

 

[Destrio]

dh/dt = d(x^2 + y^2)/dt

since we need sqrt of h, we are taking the derivative of h first, correct?

= d(x^2)/dt + d(y^2)/dt

splitting it into x and y components

= [d(x^2)/dx]*(dx/dt) + [d(y^2)/dy]*(dy/dt)

applying the chain rule

2x*(dx/dt) + 2y*(dy/dt)

moving the exponent to the coefficient and lowing the exponent by 1

ds/dt = (ds/dh)*(dh/dt)

now we are using ds again, the distance from the original point, one again using the chain rule

= [(1/2)h^(-1/2)]*(2x*(dx/dt) + 2y*(dy/dt))

the h being the sqrt to find the distance and multiplying it to the dh/dt

is there anything i misunderstood?
can i use this to prove that Vx < 0 , Vy < 0 ?

thanks for all the help

 

[learningphysics]

No, I think you understood well. If you have the time, I recommend doing a few exercises on the chain rule, to get the hang of it. Then you should be able to do ds/dt all in one step (you'll get used to doing the chain rule automatically)...

For this problem we have ds/dt =

[(1/2)h^(-1/2)]*(2x*(dx/dt) + 2y*(dy/dt))

We need this to be less than 0 for the distance from the origin to be decreasing... if you want you can substitute in the actual value of h (ie x(t)^2 + y(t)^2) but it's unnecessary.

[(1/2)h^(-1/2)]*(2x*(dx/dt) + 2y*(dy/dt)) < 0

dx/dt = Vx. dy/dt = Vy. try to simplify the above inequality, and you'll get the answer to the question.

 

[Destrio]

[(1/2)h^(-1/2)]*(2x*(dx/dt) + 2y*(dy/dt)) < 0

[(1/2)h^(-1/2)]*((2x*(Vx) + 2y*(Vy)) < 0

[(1/2)h^(-1/2)]*((2Vx^2) + (2Vy^2)) < 0

[(1/2)h^(-1/2)]*(2(Vx^2) + (Vy^2)) < 0

does it need to be simplified any further?

 

[learningphysics]

[(1/2)h^(-1/2)]*(2x*(dx/dt) + 2y*(dy/dt)) < 0

[(1/2)h^(-1/2)]*((2x*(Vx) + 2y*(Vy)) < 0

[(1/2)h^(-1/2)]*((2Vx^2) + (2Vy^2)) < 0

How did you do that step getting rid of the x and y?

 

[Destrio]

oops, i see what i did, i thought the Vx had a variable x, as opposed to subscript

[(1/2)h^(-1/2)]*((2x*(Vx) + 2y*(Vy)) < 0

[(1/2)h^(-1/2)]*(2(x*(Vx) + y*(Vy)) < 0

[(1/2)h^(-1/2)]*(2(x+y)(Vx)+(Vy)) < 0

is this better?

 

[learningphysics]

oops, i see what i did, i thought the Vx had a variable x, as opposed to subscript

[(1/2)h^(-1/2)]*((2x*(Vx) + 2y*(Vy)) < 0

[(1/2)h^(-1/2)]*(2(x*(Vx) + y*(Vy)) < 0

At this point you can divide both sides by [(1/2)h^(-1/2)]*2... it doesn't change the sign of the inequality because this value is positive

So you're left with x*Vx + y*Vy < 0

 

[Destrio]

oo ok

thanks very much for all of your help

 

[learningphysics]

oo ok

thanks very much for all of your help

no prob. practice those chain rule problems.

 

[saket]

Well, it was a fruitful discussion... but I guess you people deviated. A different approach could have yielded the result in less time.
Let the line joining origin, O, and the given arbitrary location of the particle, P, be denoted by OP. Draw a line, AB, perpendicular to OP in the xy-plane. This line AB divides the xy-plane in two parts, one containing O and the other not containing O.
Now, distance of the particle from the origin, O, would decrease if velocity vector of the particle lies in the part containing origin. (Satisfy yourself with the help of a diagram.) Thereby, we can say that angle between velocity vector and position vector (call it 'theta') must be greater than 90 degrees.
Thus, Cos(theta) < 0 (zero)
=> |velocity vector|.|position vector|.Cos(theta) < 0
=> dot product of velocity vector and position vector < 0
=> X.Vx + Y.Vy < 0.
Hence, the (C) option.