To solve this problem, let’s go through it step-by-step:

Problem Summary:

A particle moves along the curve $y = 4\sqrt{4x + 4}$. As it passes through the point $(3, 16)$, its x-coordinate increases at a rate of 2 units per second. We need to find the rate of change of the distance from the particle to the origin at this instant.

Solution:

1. Define the Distance from the Origin: Let $D$ represent the distance from the particle to the origin. Using the distance formula: $D = \sqrt{x^2 + y^2}$

2. Express $y$ in Terms of $x$: The curve is given by $y = 4\sqrt{4x + 4}$. To find $\frac{dy}{dx}$, let’s rewrite $y$ explicitly: $y = 4\sqrt{4(x + 1)} = 8\sqrt{x + 1}$

3. Differentiate $y$ with Respect to $x$: To use in our calculations, we find $\frac{dy}{dx}$: $\frac{dy}{dx} = 8 \cdot \frac{1}{2\sqrt{x + 1}} = \frac{4}{\sqrt{x + 1}}$

4. Differentiate $D$ with Respect to Time $t$: Since $D = \sqrt{x^2 + y^2}$, we differentiate both sides with respect to $t$: $\frac{dD}{dt} = \frac{1}{2\sqrt{x^2 + y^2}} \cdot (2x \frac{dx}{dt} + 2y \frac{dy}{dt})$ Simplify: $\frac{dD}{dt} = \frac{x \frac{dx}{dt} + y \frac{dy}{dt}}{\sqrt{x^2 + y^2}}$

5. Substitute Values at the Given Point:

   • At $(x, y) = (3, 16)$: $x = 3, \quad y = 16, \quad \frac{dx}{dt} = 2$
   • Calculate $\frac{dy}{dt}$ using the chain rule: $\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt} = \frac{4}{\sqrt{3 + 1}} \cdot 2 = \frac{4}{2} \cdot 2 = 4$

6. Calculate $D$ and $\frac{dD}{dt}$:

   • Compute $D$ at $(3, 16)$: $D = \sqrt{3^2 + 16^2} = \sqrt{9 + 256} = \sqrt{265}$
   • Substitute all known values into $\frac{dD}{dt}$: $\frac{dD}{dt} = \frac{3 \cdot 2 + 16 \cdot 4}{\sqrt{265}} = \frac{6 + 64}{\sqrt{265}} = \frac{70}{\sqrt{265}}$

7. Simplify the Result (Optional): $\frac{dD}{dt} = \frac{70}{\sqrt{265}} \approx 4.29 \text{ units per second}$

Answer:

The rate of change of the distance from the particle to the origin at this instant is approximately $\frac{70}{\sqrt{265}}$ or about 4.29 units per second.

Would you like more details on any specific step?

Related Questions

1. What would happen to $\frac{dD}{dt}$ if the $x$-coordinate’s rate of change was different?
2. How would we solve this problem if the curve equation was different?
3. Can we generalize this method for any point on a given curve?
4. How do implicit differentiation and related rates connect in this context?
5. What real-world applications are there for related rate problems involving distance?

Tip:

Always identify which quantities are constant and which are changing with respect to time when dealing with related rates.