Find dy/dt for Y=2((x^2)-3x) when x=3

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Homework Statement


Suppose that Y=2((x^2)-3x) and dx/dt = 2
Find dy/dt when x=3

Homework Equations


the only calc is taking the derivative of the equation, i am wondering if i am doing the whole problem right.

The Attempt at a Solution


dy/dx = 4x-6
find the equation for x=3 and multiply in 2 for the rate of change of time
Y=4(2*3)-6
Y=18?

I think my answer goes back to the clac. I don't know what number i am supposed to plug in for X, i know it has to do when what X equals at dy/dt and the rate of chage of X over time.
 
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You found
[tex]\frac{dy}{dx}[/tex]
(The derivative with respect to [itex]x[/itex] when [itex]x=3[/itex].)

The problem is asking for
[tex]\frac{dy}{dt}[/tex]
(The derivative with respect to [itex]t[/itex].)
 
RyanMcStylin said:

Homework Statement


Suppose that Y=2((x^2)-3x) and dx/dt = 2
Find dy/dt when x=3

Homework Equations


the only calc is taking the derivative of the equation, i am wondering if i am doing the whole problem right.

The Attempt at a Solution


dy/dx = 4x-6
find the equation for x=3 and multiply in 2 for the rate of change of time
Y=4(2*3)-6
Y=18?

I think my answer goes back to the clac. I don't know what number i am supposed to plug in for X, i know it has to do when what X equals at dy/dt and the rate of chage of X over time.

The problem SAYS "Find dy/dt when x= 3"! What value of x do you think you should put in? The phrase "what x equals at dy/dt" is meaningless.
 
i understand that 3 must be replaced for x, but where does the dx/dt = 2 fit into the equation? I am guessing around somewhere around the radius portion of the equation