Calc Question

[AgaBoogaBoo]

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

I do know the distance formula between the origin and a point on a graph incase that helps somehow. I don't need someone to do it but rather guide me on what I should do...

Thanks 🙂

 

[Random Variable]

directional derivative?

 

[cirthix]

dy/dx=2x, dx/dt=2cm/s. the derivative of distance is velocity. that should be enough of a hint 🙂

 

[AgaBoogaBoo]

Originally posted by: cirthix
dy/dx=2x, dx/dt=2cm/s. the derivative of distance is velocity. that should be enough of a hint 🙂

Which equation for distance should I find the derivative of?

 

[RadioHead84]

Err dont you have to take the Derivative and each time you take a derivative of X or Y you need to write down Y prime or X prime next to it. Then basically plug and play untill you only have one peice to solve for?

 

[PurdueRy]

d = SQRT(y2^2-y1^2.....yadadad

Derivative of both sides and substitute

 

[cirthix]

doh! misunderstood your problem. you've got a point moving along a line at some speed, the problem is asking how quckly the point is moving away from the orgin. we have an equation for its position (and distance if you change it around a bit) and its speed, combine the two to get dy/dt

 

[RadioHead84]

Originally posted by: cirthix
doh! misunderstood your problem. you've got a point moving along a line at some speed, the problem is asking how quckly the point is moving away from the orgin. we have an equation for its position (and distance if you change it around a bit) and its speed, combine the two to get dy/dt

Yeah kinda what I was getting at

Draw a picture..it helps

 

[AgaBoogaBoo]

Originally posted by: cirthix
doh! misunderstood your problem. you've got a point moving along a line at some speed, the problem is asking how quckly the point is moving away from the orgin. we have an equation for its position (and distance if you change it around a bit) and its speed, combine the two to get dy/dt

Ok, that makes sense so far, my confusion comes in when I try to combine them to get dy/dt, I'm fairly sure I'm missing a step just not sure what

 

[Random Variable]

distance = (x^2+(x^2+1)^2)^(1/2) = (3x^2+x^4+1)^(1/2)

d(distance)/dt = (1/2)(3x^2+x^4+1)^(-1/2)*(6x+4x^3)(dx/dt)

d(istance)/dt = (6x+4x^3)/(3x^2+x^4+1)^(1/2) ??

 

[AgaBoogaBoo]

Originally posted by: Random Variable
distance = (x^2+(x^2+1)^2)^(1/2) = (3x^2+x^4+1)^(1/2)

d(distance)/dt = (1/2)(3x^2+x^4+1)^(-1/2)*(6x+4x^3)(dx/dt)

d(istance)/dt = (6x+4x^3)/(3x^2+x^4+1)^(1/2) ??

How did you get that?

 

[Random Variable]

Originally posted by: AgaBoogaBoo

Originally posted by: Random Variable
distance = (x^2+(x^2+1)^2)^(1/2) = (3x^2+x^4+1)^(1/2)

d(distance)/dt = (1/2)(3x^2+x^4+1)^(-1/2)*(6x+4x^3)(dx/dt)

d(istance)/dt = (6x+4x^3)/(3x^2+x^4+1)^(1/2) ??

How did you get that?

On any point on the curve, the x-coordinate is x cm from the origin and the y-coordinate is x^2+1 cm from the origin. Then I used the distance formula.

 

[AgaBoogaBoo]

Originally posted by: Random Variable

Originally posted by: AgaBoogaBoo

Originally posted by: Random Variable
distance = (x^2+(x^2+1)^2)^(1/2) = (3x^2+x^4+1)^(1/2)

d(distance)/dt = (1/2)(3x^2+x^4+1)^(-1/2)*(6x+4x^3)(dx/dt)

d(istance)/dt = (6x+4x^3)/(3x^2+x^4+1)^(1/2) ??

How did you get that?

On any point on the curve, the x-coordinate is x cm from the origin and the y-coordinate is x^2+1 cm from the origin. Then I used the distance formula.

Oh wow, that makes much more sense now, thank you very much 🙂

 

[IAteYourMother]

what calculus is this?

 

[AgaBoogaBoo]

Originally posted by: IAteYourMother
what calculus is this?

Calc AB AP - 1/2 a semester of college credit

I'm a senior in HS