Rate of Change: Distance Between Moving Point and Fixed Point

In summary, the distance between point P and the fixed point Q changes at the instant when P is at (6,3).
  • #1
Emethyst
118
0

Homework Statement


A point P (x,y) is moving along the line y=(1/2)x. If x is decreasing at the rate of 3units/s, at wha rate is the distance between P and the fixed point Q (0,11) changing at the instant when P is at (6,3)?


Homework Equations


Implicit differentation



The Attempt at a Solution


From what I can see for this question, the distance formula needs to be used. That is what I did, I plugged in the point (x,y) and (0,11) into the distance formula and found the derivative, then plugged in (6,3) for x and y and -3 for dx/dt. The answer I got was -2.6, however, this is not the right answer as it is supposed to be -0.6. This question seems easy yet it is frustrating me, if anyone could be help it would be greatly appreciated, thank you.
 
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  • #2
Your method looks fine; why don't you show me your work and I can see where you are going wrong...
 
  • #3
Point 1=(x,y)
Point 2=(0,11)

s=((x-0)^2+(y-11)^2)^1/2
s=(x^2+(y-11)^2)^1/2
s=(1/2)(x^2+(y-11)^2)^-1/2(2x(dx/dt)+2(y-11)(1))
s=(1/2)(6^2+(3-11)^2)^-1/2(2(6)(-3)+2(3-11))
s=-(1/2)(100)^-1/2(-52)
s=-2.6

I think I have done the chain rule right, as I have attempted this question a few times, all with the same results.
 
  • #4
Emethyst said:
Point 1=(x,y)
Point 2=(0,11)

s=((x-0)^2+(y-11)^2)^1/2
s=(x^2+(y-11)^2)^1/2

so far so good, but...

s=(1/2)(x^2+(y-11)^2)^-1/2(2x(dx/dt)+2(y-11)(1))

This line makes no sense...even if you meant to say

ds/dt=(1/2)(x^2+(y-11)^2)^-1/2(2x(dx/dt)+2(y-11)(1)) (as opposed to s=...)

you would still be incorrect:

[tex]\frac{d}{dt}[x^2+(y-11)^2]\neq(2x(\frac{dx}{dt})+2(y-11)(1))[/tex]

because

[tex]\frac{dy}{dt}=\frac{d}{dt}\left(\frac{x}{2}\right)\neq1[/tex]
 
  • #5
Hmm so that is where I am going wrong? How would I properly go about fixing that line then? Because from what I was taught, in adding situations you apply the derivative to both terms:

(d/dt)(x^2 +(y-11)^2)
(d/dt)(x^2)+(d/dt)((y-11)^2)
(2x)(dx/dt)+2(y-11)((d/dt)(y-11))
(2x)(dx/dt)+2(y-11)(1)

That is how I came about that, and because it is the chain rule from ds/dt=(1/2)(x^2+(y-11)^2)^-1/2, I multiplied (2x)(dx/dt)+2(y-11)(1) by it.
 
  • #6
Emethyst said:
Hmm so that is where I am going wrong? How would I properly go about fixing that line then? Because from what I was taught, in adding situations you apply the derivative to both terms:

(d/dt)(x^2 +(y-11)^2)
(d/dt)(x^2)+(d/dt)((y-11)^2)
(2x)(dx/dt)+2(y-11)((d/dt)(y-11))

up to here you are correct, but...
(2x)(dx/dt)+2(y-11)(1)

where is the (1) coming from...shouldn't that be (2x)(dx/dt)+2(y-11)(dy/dt)?
 
  • #7
I used 1 because no information is given for dy/dt and the question is not asking for it, unless I am required to take the derivative of the equation given and plug it in for dy/dt? I think the light buld may have clicked :tongue:
 
  • #8
Nope, even if the derivative of y=(1/2)x is taken and used it simply results in an answer of -2.2 instead of -2.6.
 
  • #9
Emethyst said:
I used 1 because no information is given for dy/dt and the question is not asking for it
So whenever you have any equation with a variable and you aren't told what that variable is equal to; you automatically just assume it is equal to one?!:confused:

unless I am required to take the derivative of the equation given and plug it in for dy/dt? I think the light buld may have clicked :tongue:

Of course, how else are you supposed to find dy/dt?
 
  • #10
Emethyst said:
Nope, even if the derivative of y=(1/2)x is taken and used it simply results in an answer of -2.2 instead of -2.6.

I get -0.6 when I do it, so you must be making a mistake...show me the calculations
 
  • #11
gabbagabbahey said:
So whenever you have any equation with a variable and you aren't told what that variable is equal to; you automatically just assume it is equal to one?!:confused:

No, I just overlooked it in this question by accident, spent too much time on the other questions, it drained my mind :tongue:

And I just got -0.6 for my answer. I forgot to include dx/dt in the derivative, for I simply took it to be dy/dt=1/2 instead of dy/dt=(1/2)(dx/dt); I fixed that little error and came up with the right answer at last. Thanks for all that help gabbagabbahey, I don't know if I have spotted my mistakes if it was not for your assistance :smile:
 

1. What is the rate of change of distance?

The rate of change of distance, also known as velocity, is the measure of how quickly an object changes its displacement with respect to time. It is a vector quantity that includes both magnitude (speed) and direction.

2. How is rate of change of distance calculated?

The rate of change of distance is calculated by dividing the change in distance by the change in time. This can be represented by the equation v = Δd/Δt, where v is velocity, Δd is the change in distance, and Δt is the change in time.

3. What is the difference between average and instantaneous rate of change of distance?

The average rate of change of distance is calculated by taking the total change in distance and dividing it by the total change in time. It gives an overall measure of how quickly an object is moving. In contrast, instantaneous rate of change of distance is calculated at a specific moment in time and gives the exact velocity of the object at that instant.

4. How does acceleration relate to rate of change of distance?

Acceleration is the rate of change of velocity, which is the rate of change of distance. In other words, acceleration is the rate at which an object's velocity changes over time, and velocity is the rate at which an object's distance changes over time.

5. What factors can affect the rate of change of distance?

The rate of change of distance can be affected by various factors, such as the object's mass, the force acting on it, and any external forces (such as friction or air resistance). Additionally, changes in the object's direction or speed can also affect its rate of change of distance.

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