#1

so yea does anybody know how to do this:

Find the rate of change of the distance between the origin and a moving point on the graph of y = x² + 1 if dx/dt = 2 centimeters per second.

Find the rate of change of the distance between the origin and a moving point on the graph of y = x² + 1 if dx/dt = 2 centimeters per second.

#2

I do not know.

#3

Is that the entire question?

Because the rate of change of distance is velocity, or

OHHHH... they're asking you the value of x when dx/dt=2. So what you have to do is differentiate (find the derivative of) x2+1, and set it equal to 2. Then solve for x.

EDIT: In other words, if f(x)=x2+1, find f'(x) and then find where f'(x)=2.

SECOND EDIT: The question is worded very oddly, gimme a sec to think this one over.

Because the rate of change of distance is velocity, or

**dx/dt**. So the rate of change of velocity should be 2, right?OHHHH... they're asking you the value of x when dx/dt=2. So what you have to do is differentiate (find the derivative of) x2+1, and set it equal to 2. Then solve for x.

EDIT: In other words, if f(x)=x2+1, find f'(x) and then find where f'(x)=2.

SECOND EDIT: The question is worded very oddly, gimme a sec to think this one over.

*Last edited by Desh627 at Oct 12, 2008,*

#4

do they give u a point to work with?

#5

Maths Thread. Be there, or be

*SQUARED!*
#6

dont set the derivative to 2, that would mean you are subbing in dy/dt = 2. From what im gathering, you take the derivative and with respect to time: dy/dt = 2x(dx/dt). But you dont know enough info from what I see, shouldn't they give you a point at which the dx/dt is changing.

#7

The

**rate of change of distance**is the derivative of distance, which is velocity, is it not?
#8

isn't it (f(x)-f(c))/(x-c) ??? where C = constant = 2

isn't that it? i'm not completely sure, i might be mixing that up with something else...

yeah nvm... I'm wrong. =( math doesn't come easy for me

isn't that it? i'm not completely sure, i might be mixing that up with something else...

yeah nvm... I'm wrong. =( math doesn't come easy for me

#9

Unless this is applying differentials, which means to find where dx/dt along the graph f(x)=x2+1, and then find the slope of the line between x=0 and f'(x)=2

#10

the derivative of x² + 1 is 2x. You set that equal to 2cm/sec and get x=1. So y=2. You can use the Pythagorean theorem to find the distance between the origin and (1,2) which is sqrt(3)

Basically, You have to find the velocity of the point in the x,y direction given the velocity in the x-direction. At least I think that's what it's going for.

Basically, You have to find the velocity of the point in the x,y direction given the velocity in the x-direction. At least I think that's what it's going for.

#11

the derivative of x² + 1 is 2x. You set that equal to 2cm/sec and get x=1. So y=2. You can use the Pythagorean theorem to find the distance between the origin and (1,2) which is sqrt(3)

Basically, You have to find the velocity of the point in the x,y direction given the velocity in the x-direction. At least I think that's what it's going for.

Which is what I thought/said in my first post

Although the fact that they're asking for the

**rate of change of distance**is what's throwing me off...

#12

ok so it doesn't actually wanna know the answer but instead this: [2(2x^3 + 3x)] / √x^4 + 3x^2 + 1)

which explains not giving any points

so now my next question is how they got that as an answer?

EDIT: and just in case it isn't clear its the square root of entire part (x^4 + 3x^2 + 1) not just x^4

which explains not giving any points

so now my next question is how they got that as an answer?

EDIT: and just in case it isn't clear its the square root of entire part (x^4 + 3x^2 + 1) not just x^4

*Last edited by oscarticas at Oct 12, 2008,*