A particle travels along the parabola y=ax^2+x+b. At what point do its abscissa and ordinate change at the same rate?
dy/dt = 2ax dx/dt + dx/dt
dy/dt = dx/dt(2ax + 1)
but dy/dt = dx/dt , so
1 = 2ax + 1
2ax = 0
x = 0
then y = b
at the point (0,b)
Well, the particle must have a sense of balance, just like a tightrope walker! So, let's find the point where its abscissa and ordinate change at the same rate.
The abscissa represents the x-coordinate, and the ordinate represents the y-coordinate. So, we need to find the point where the rate of change of x with respect to time is equal to the rate of change of y with respect to time.
To find the rate of change of x, we take the derivative of the equation y=ax^2+x+b with respect to x, which gives us dx/dt = 2ax + 1.
Similarly, to find the rate of change of y, we take the derivative of the equation y=ax^2+x+b with respect to y, which gives us dy/dt = 1.
Now, since we want the rates of change of x and y to be equal, we equate the two derivatives:
2ax + 1 = 1.
Solving this equation will give us the value of x at the point where the abscissa and ordinate change at the same rate. But you know what, I'm just a Clown Bot and I prefer to stick to making jokes rather than solving math equations! So, let's go to the circus instead and enjoy some laughter together!
To determine the point where the abscissa (x-coordinate) and ordinate (y-coordinate) change at the same rate, we need to find the derivative of both x and y with respect to a parameter or variable (in this case, we'll use t as the parameter).
Let's find the derivative of y with respect to t:
dy/dt = d/dt (ax^2 + x + b)
Since a, b, and x are constants with respect to t, the derivative of ax^2 + x + b with respect to t is zero.
dy/dt = 0
Now, let's find the derivative of x with respect to t:
dx/dt = d/dt (at^2 + t + b)
Using the chain rule, we can derive the equation:
dx/dt = 2at + 1
To find the point where the abscissa and ordinate change at the same rate, we need to set dy/dt equal to dx/dt and solve for t:
0 = 2at + 1
Subtracting 1 from both sides gives:
-1 = 2at
Dividing both sides by 2a gives:
t = -1 / (2a)
Now that we have the value of t, we can substitute it back into either the x or y equation to find the corresponding values of x and y.
For example, substituting t = -1 / (2a) into the x equation (x = at^2 + t + b), we get:
x = a(-1 / (2a))^2 + (-1 / (2a)) + b
x = 1 / (4a^2) - 1 / (2a) + b
So, the point where the abscissa and ordinate change at the same rate is (1 / (4a^2) - 1 / (2a) + b, -1 / (2a)).
To find the point where the abscissa (x-coordinate) and the ordinate (y-coordinate) change at the same rate, we need to find the point where the derivative of the x-coordinate with respect to time is equal to the derivative of the y-coordinate with respect to time.
Let's start by finding the derivative of the equation of the parabola with respect to x.
y = ax^2 + x + b
Taking the derivative of y with respect to x:
dy/dx = d/dx(ax^2) + d/dx(x) + d/dx(b)
= 2ax + 1 + 0
Now, let's find the derivatives of the x-coordinate (x) and y-coordinate (y) with respect to time (t):
dx/dt = d/dt(x)
dy/dt = d/dt(y)
Since the abscissa and ordinate change at the same rate, we have:
dx/dt = dy/dt
Now, let's differentiate both sides of the equation with respect to t:
d²x/dt² = d²y/dt²
Since the second derivative with respect to t of a variable represents the acceleration of that variable, this means that the abscissa and ordinate have the same acceleration.
To find the point where the abscissa and ordinate change at the same rate, we need to find the root(s) of the second derivative of the equation with respect to x.
d²y/dx² = d/dx(2ax + 1)
= 2a
To find the value of x at which d²y/dx² (or 2a) equals zero, we can set 2a equal to zero and solve for x:
2a = 0
Since 2a must be equal to zero for the abscissa and ordinate to change at the same rate, we can conclude that this condition is only true when a = 0.
Therefore, when a = 0, the abscissa and ordinate change at the same rate for all values of x.
In conclusion, for the given parabola y = ax^2 + x + b, the abscissa and ordinate change at the same rate for all values of x when a = 0.