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

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SUMMARY

The discussion focuses on finding the derivative dy/dt for the function Y=2((x^2)-3x) when x=3, given that dx/dt=2. The correct derivative dy/dx is calculated as 4x-6, which evaluates to 6 when x=3. To find dy/dt, the chain rule is applied, resulting in dy/dt = dy/dx * dx/dt. Substituting the known values yields dy/dt = 6 * 2 = 12. Therefore, the final answer is that dy/dt equals 12 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
\frac{dy}{dx}
(The derivative with respect to x when x=3.)

The problem is asking for
\frac{dy}{dt}
(The derivative with respect to t.)
 
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
 

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