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

1. Define Variables and Relationships:

   • Let $P = (x, y)$ be a point on the curve.
   • The equation of the curve is $y = 4x$, so $y$ depends on $x$.
   • We need to find the rate at which the distance between $P$ and the fixed point $(2, 0)$ is changing when $P$ is at $(2, 8)$ and $x$ is decreasing at $3$ units/second.

2. Distance Formula: The distance $D$ between $P = (x, y)$ and $(2, 0)$ is given by: $D = \sqrt{(x - 2)^2 + (y - 0)^2} = \sqrt{(x - 2)^2 + y^2}$ Since $y = 4x$, substitute $y = 4x$ into the distance formula: $D = \sqrt{(x - 2)^2 + (4x)^2} = \sqrt{(x - 2)^2 + 16x^2}$

3. Differentiate with Respect to Time $t$: We want to find $\frac{dD}{dt}$, the rate at which $D$ is changing. Differentiate both sides of the distance formula with respect to $t$: $\frac{dD}{dt} = \frac{1}{2} ( (x - 2)^2 + 16x^2 )^{-\frac{1}{2}} \cdot \left( 2(x - 2) \frac{dx}{dt} + 32x \frac{dx}{dt} \right)$ Simplifying, we get: $\frac{dD}{dt} = \frac{(x - 2) \frac{dx}{dt} + 16x \frac{dx}{dt}}{\sqrt{(x - 2)^2 + 16x^2}}$

4. Substitute Known Values: At the instant we’re considering:

   • $x = 2$
   • $y = 8$ (since $y = 4x$)
   • $\frac{dx}{dt} = -3$ units/s

   Substitute these values into the formula: $\frac{dD}{dt} = \frac{(2 - 2)(-3) + 16 \cdot 2 \cdot (-3)}{\sqrt{(2 - 2)^2 + 16 \cdot (2)^2}}$ Simplify each part:

   • The numerator: $(0)(-3) + 16 \cdot 2 \cdot (-3) = -96$
   • The denominator: $\sqrt{0 + 16 \cdot 4} = \sqrt{64} = 8$

   So, $\frac{dD}{dt} = \frac{-96}{8} = -12$

5. Conclusion: The rate at which the distance between $P$ and $(2, 0)$ is changing at that instant is $-12$ units/second. The negative sign indicates that the distance is decreasing.

Would you like further details on any step?

Related Questions

1. What is the importance of implicit differentiation in related rates problems?
2. How would the solution change if $x$ were increasing instead of decreasing?
3. Why does the negative rate signify a decreasing distance?
4. What other scenarios might involve a point moving along a curve with changing distances?
5. How can we use related rates in physics problems involving speed and distance?

Tip

When solving related rates problems, always start by identifying the relationships between the quantities involved before differentiating. This helps prevent errors in differentiation steps.