A particle moves along the curve y = sqr(1+x^3). As it reaches the point (2,3), the y-coordinate is increasing at a rate of 4 cm/s. How fast is the x-coordinate of the point changing at that instant.

Question

A particle moves along the curve y = sqr(1+x^3). As it reaches the point (2,3), the y-coordinate is increasing at a rate of 4 cm/s. How fast is the x-coordinate of the point changing at that instant.

Answer 1

y = √(1+x^3)

dy/dt = 3x^2 / 2√(1+x^3) dx/dt
4 = 3*4 / 2*3 dx/dt
dx/dt = 2

Answer 2

Well, let's start by finding the derivative of the equation y = √(1+x^3) with respect to x:

dy/dx = (3x²)/(2√(1+x^3))

Now, we know that at the point (2,3), the y-coordinate is increasing at a rate of 4 cm/s. So, we can plug this information into our equation:

4 = (3(2)²)/(2√(1+(2)³))

Simplifying this equation, we get:

4 = (12)/(2√9)
4 = 6/6
4 = 1

Now, I'm not sure if you caught that, but it seems like we have stumbled upon a mathematical paradox! According to the information given, the y-coordinate is increasing at a rate of 4 cm/s, which means the x-coordinate should be changing, but somehow it's not! It's remaining constant at 2!

So, to answer your question, the x-coordinate of the point is NOT changing at that instant. It's frozen in time, like a statue! It seems this particle has decided to take a little break from moving.

I hope this whimsical response brings a smile to your face! If you have any more questions, feel free to ask!

Answer 3

To find the rate of change of the x-coordinate, we need to find dy/dx, and then use the relationship dy/dx = (dy/dt)/(dx/dt).

Given the equation of the curve: y = √(1 + x^3)

Differentiating both sides with respect to x using the chain rule:
dy/dx = (1/2)(1 + x^3)^(-1/2) * 3x^2

Now we can find dx/dt by finding the derivative of x with respect to t, using the equation x = 2:
dx/dt = 0, since x is a constant.

Substituting the values into the relationship dy/dx = (dy/dt)/(dx/dt), we get:
dy/dx = (dy/dt)/(dx/dt)
dy/dx = 4/0

Since the denominator is zero, we cannot divide by zero and the rate of change of the x-coordinate cannot be determined at this instant.

Answer 4

To find how fast the x-coordinate of the point is changing, we can use the chain rule of differentiation.

Let's differentiate both sides of the equation with respect to time (t) by applying the chain rule:

d/dt(y) = d/dt(√(1+x^3))

To find dy/dt (how fast the y-coordinate is changing), we need to differentiate both sides of the equation with respect to time (t).

Using the chain rule, we have:

dy/dt = d/dt(√(1+x^3))

Now, let's differentiate the right side of the equation:

dy/dt = d/dx(√(1+x^3)) * dx/dt

We know that dy/dt is given as 4 cm/s.

4 = d/dx(√(1+x^3)) * dx/dt

We are given that (2,3) is a point on the curve, which means x = 2 and y = 3.

Substitute these values into the equation:

4 = d/dx(√(1+2^3)) * dx/dt

Simplify the equation:

4 = d/dx(√(1+8)) * dx/dt

4 = d/dx(√9) * dx/dt

4 = d/dx(3) * dx/dt

Since d/dx(3) is zero (the derivative of a constant), the equation simplifies to:

4 = 0 * dx/dt

Since anything multiplied by zero is zero, we have no information about dx/dt at this point. Therefore, we cannot determine the rate at which the x-coordinate of the point is changing without additional information.