An object is moving along the parabola y = 3x^2. (a) When it passes through the point (2,12), its horizontal velocity is dx/dt = 3. what is its vertical velocity at that instant? (b) If it travels in such a way that dx/dt = 3 for all t, then what happens to dy/dt as t --> +infinity? (c) If, however, it travels in such a way that dy/dt remains constant, then what happens to dy/dt as t --> +infinity?
y=3x^2
dy/dt=6xdx/dt
I don't understand c). IF dy/dt is constant, it is constant.
If vertical velocity is the derivative of the displacement function, then shouldn't the slope be undefined?
So once you get the vertical velocity to dy/dt=6xdx/dt, then you can plug in dx/dt=3 and for x you can plug in 2 from the given point (2,12)?
Then you would get a vertical velocity of (6)(2)(3)= 36? And how would you draw a visual representations or pictures of all these parts?
What are they asking for on b and c?
To solve these questions, we need to understand the relationship between velocity and the derivatives of the given equation. Let's start by analyzing the equation y = 3x^2.
(a) To find the vertical velocity (dy/dt) at the point (2,12), we need to differentiate the equation y = 3x^2 with respect to time. The derivative of y with respect to t will give us dy/dt.
Differentiating y = 3x^2 with respect to t:
dy/dt = d/dt(3x^2)
To find dy/dt, we must use the chain rule, which states that if y = f(u) and u = g(t), then dy/dt = dy/du * du/dt.
In this case, u = x and f(u) = 3u^2.
Using the chain rule, we have:
dy/dt = d/dt(3x^2)
= dy/du * du/dt
= d/dx(3x^2) * dx/dt
Since dx/dt is given as 3, we substitute it into the equation above:
dy/dt = d/dx(3x^2) * 3
= (6x) * 3
= 18x
Substituting x = 2 into the equation, we can find the vertical velocity (dy/dt) at that instant:
dy/dt = 18 * 2
= 36
Therefore, the vertical velocity at the instant the object passes through the point (2, 12) is 36.
(b) If dx/dt = 3 for all t (constant horizontal velocity), we need to examine the behavior of dy/dt as t approaches infinity.
We know that dx/dt = 3, which means that the horizontal velocity remains constant. Since the equation y = 3x^2 represents a parabola, it implies that as t proceeds to infinity, the object moves along the parabola indefinitely.
As the object moves along the parabola, dy/dt corresponds to the rate of change of y with respect to time. Since the object is moving along the parabola, and there are no external forces acting upon it, dy/dt stays constant. Therefore, as t approaches infinity, dy/dt remains the same.
(c) If dy/dt remains constant, we need to assess its behavior as t approaches infinity.
If dy/dt remains constant, it means that y changes at a constant rate over time, regardless of x. This indicates that the object is moving vertically at a constant speed, independent of the horizontal position.
As t proceeds towards infinity, the object moves along the parabola indefinitely, while its vertical velocity (dy/dt) remains constant. Thus, as t approaches infinity, dy/dt also remains constant.
In conclusion:
(a) The vertical velocity at the instant the object passes through the point (2,12) is 36.
(b) If dx/dt = 3 for all t, then dy/dt remains the same as t approaches infinity.
(c) If dy/dt remains constant, then dy/dt remains constant as t approaches infinity.