A bug crawls along the graph of
2
y x x 4 1
, where x and y are positive and measured in
centimeters. If the x-coordinate of the bug’s position
x y, changes at a constant rate of 3
cm/min, how fast is the y-coordinate changing when the bug is 1 cm above the x-axis?
your equation came out looking rather jibberish on my computer, but I see it contans x's and y's
I can give you the general method of doing this question ....
Find the derivative with respect to t, then your result should contain dx/dt and dy/dt , as well as x and y
you are given :
y = 1, sub into the original equation and solve for x
you are also given: dx/t = 3
which leaves dy/dt as the only unknown in your derivative equation.
Sub in all known values and solve for it.
To find the rate at which the y-coordinate is changing, we need to use the concept of derivatives. In this case, since the x-coordinate is changing at a constant rate, we can assume it is a function of time, denoted by x(t). Similarly, the y-coordinate is also a function of time, denoted by y(t).
We are given that the rate of change of x with respect to time, dx/dt, is a constant 3 cm/min. Let's find an expression for dy/dt, the rate of change of y with respect to time.
First, let's rewrite the equation of the graph in differential form:
y = 2x^2 + x + 4
Next, let's differentiate both sides of the equation with respect to time t. Using the chain rule, we get:
dy/dt = d(2x^2 + x + 4)/dt
To differentiate 2x^2 with respect to t, we apply the chain rule which states that d(uv)/dt = u(dv/dt) + v(du/dt). In this case, u = 2x^2 and v = 1.
So, differentiating 2x^2 with respect to t gives:
d(2x^2)/dt = 2(x^2)(dx/dt)
The term dx/dt represents the rate of change of x with respect to t, which is given as 3 cm/min. Substituting this value, we have:
d(2x^2)/dt = 2(x^2)(3)
Simplifying, we have:
d(2x^2)/dt = 6x^2
We can differentiate the other terms in a similar manner. The derivative of x with respect to t is dx/dt, which is given as 3 cm/min. So:
d(x)/dt = dx/dt = 3
Finally, to find dy/dt, we substitute the values we've calculated into the differential form of the equation:
dy/dt = 6x^2 + dx/dt + 0
dy/dt = 6x^2 + 3
Now we can find the rate of change of the y-coordinate when the bug is 1 cm above the x-axis. This means that y = 1, and solving for x in the equation y = 2x^2 + x + 4:
1 = 2x^2 + x + 4
2x^2 + x + 3 = 0
Solving this quadratic equation, we find that x = -1 or x = -1.5. Since x is measured in centimeters and it is given that x and y are positive, we take x = -1.5.
Substituting this value of x into the expression for dy/dt, we have:
dy/dt = 6(-1.5)^2 + 3
dy/dt = 6(2.25) + 3
dy/dt = 13.5 + 3
dy/dt = 16.5 cm/min
Therefore, the y-coordinate is changing at a rate of 16.5 cm/min when the bug is 1 cm above the x-axis.