A particle moves along the path x^3 - 3x^2 + 2. If the particle's horizontal rate of change when x = 4 is -3 ft/sec, what is its vertical rate of change at that moment.

The given problem describes a particle moving along a path defined by the equation:

\[
y = x^3 - 3x^2 + 2
\]

To find the particle's vertical rate of change at the moment \( x = 4 \), we need to apply related rates. The horizontal rate of change (rate of change of \( x \) with respect to time, \( \frac{dx}{dt} \)) is given as:

\[
\frac{dx}{dt} = -3 \, \text{ft/sec}
\]

The vertical rate of change refers to \( \frac{dy}{dt} \), which we can find using the chain rule:

\[
\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}
\]

### Step 1: Differentiate \( y = x^3 - 3x^2 + 2 \) with respect to \( x \)
The derivative of \( y \) with respect to \( x \) is:

\[
\frac{dy}{dx} = 3x^2 - 6x
\]

### Step 2: Evaluate \( \frac{dy}{dx} \) at \( x = 4 \)

\[
\frac{dy}{dx} = 3(4)^2 - 6(4) = 3(16) - 24 = 48 - 24 = 24
\]

### Step 3: Use the chain rule

Now, using \( \frac{dy}{dx} = 24 \) and \( \frac{dx}{dt} = -3 \, \text{ft/sec} \):

\[
\frac{dy}{dt} = 24 \cdot (-3) = -72 \, \text{ft/sec}
\]

Thus, the particle's vertical rate of change at \( x = 4 \) is:

\[
\frac{dy}{dt} = -72 \, \text{ft/sec}
\]

### Summary:
The vertical rate of change of the particle at the moment when \( x = 4 \) is \(-72 \, \text{ft/sec}\).

Would you like a deeper explanation of any part, or have any questions?

### Related Questions:
1. How would the solution change if \( \frac{dx}{dt} \) was positive instead of negative?
2. What is the significance of the negative sign in the vertical rate of change?
3. How do we interpret the second derivative \( \frac{d^2y}{dx^2} \) in this context?
4. Can we use the same approach if the path equation was more complex, involving trigonometric functions?
5. How would the problem change if the particle’s path was in three dimensions?

### Tip:
When solving related rates problems, always identify all variables and rates first, and use the chain rule to connect them properly.

A particle moves along the path x^3 - 3x^2 + 2. If the particle's horizontal rate of change when x = 4 is -3 ft/sec, what is its vertical rate of change at that moment?

Learn how to calculate the vertical rate of change of a particle moving along the path y = x^3 - 3x^2 + 2 when its horizontal rate of change is -3 ft/sec at x = 4. This calculus problem uses related rates and differentiation, suitable for students in Grades 10-12.

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