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

  • Thread starter RyanMcStylin
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In summary, the problem is asking to find dy/dt when x=3, and the equation given is Y=2((x^2)-3x) with a given dx/dt=2. The attempt at a solution involved finding the derivative of the given equation with respect to x, but the problem is asking for the derivative with respect to t. The value of x=3 should be used when finding dy/dt, and the given value of dx/dt=2 should also be incorporated in the solution.
  • #1
RyanMcStylin
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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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  • #2
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].)
 
  • #3
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.
 
  • #4
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
 

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

What is the formula for finding dy/dt?

The formula for finding dy/dt (the derivative of y with respect to t) is d/dt(y), where y is the function and t is the independent variable.

How do you find the derivative of a function?

To find the derivative of a function, you can use the power rule, product rule, quotient rule, or chain rule, depending on the form of the function. In this case, we will use the power rule.

What is the power rule?

The power rule states that for a function y = x^n, the derivative is dy/dx = nx^(n-1). This can be extended to more complicated functions by using the chain rule.

How do you apply the power rule to find dy/dt for y = 2((x^2)-3x) when x=3?

First, we rewrite the function as y = 2(x^2 - 3x)^1. Using the power rule, we can find the derivative as dy/dt = 2 * 1 * (x^2 - 3x)^(1-1) * (2x - 3). Plugging in x=3, we get dy/dt = 2 * 1 * (3^2 - 3*3)^(1-1) * (2*3 - 3) = 2 * 1 * 0 * 3 = 0.

What does the resulting value of 0 for dy/dt represent?

The resulting value of 0 for dy/dt represents that at the specific point where x=3, the rate of change of y (dy/dt) is 0. This means that at that point, the tangent line to the function is horizontal and there is no change in the value of y with respect to t.

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