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The path of a particle is described by y^2 + 3x^2 = 10. If dx/dt = 2 and x = 1, what is...

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The path of a particle is described by y^2 + 3x^2 = 10. If dx/dt = 2 and x = 1, what is dy/dt

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Posted (Answer #1)

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The path of the particle is described by y^2 + 3x^2 = 10. If x = 1,

y^2 + 3*1 = 10

=> y^2 = 7

=> `y = +-sqrt 7`

Take the derivative of y^2 + 3x^2 = 10 with respect to t,

`2y*(dy/dt) + 6x*(dx/dt) = 0`

=> `dy/dt = (-3x)/y*(dx/dt)`

At x = 1, `dx/dt = 2` and y can take on the values `-sqrt 7` and `+sqrt 7`

This gives `dy/dt = -6/sqrt 7` and `dy/dt = 6/sqrt 7` .

The position of the particle satisfies the given relation at two different points. The value of `dy/dt` is different for each of them.

At `y = sqrt 7` , `dy/dt = -6/sqrt 7` and at `y = -sqrt 7` , `dy/dt = 6/sqrt 7`

oldnick's profile pic

Posted (Answer #2)

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first have to set y = y(x)

`y=sqrt(10-3x^2)`

so that:

`dy/dx=` `-3x/sqrt(10-3x^2)`

for x= 1:

`dy/dx=` `-3sqrt(7)/7`

since:

`dy/dx dx/dt = dy/dt`

then:   `dy/dt= -6 sqrt(7)/7 `

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