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### Problem 1
Given:
\[
y = x^2 + 5 \quad \text{and} \quad \frac{dx}{dt} = 3 \, \text{when} \, x = 2
\]
Find \(\frac{dy}{dt}\) when \(x = 2\).

**Solution:**

1. Differentiate \(y = x^2 + 5\) with respect to \(t\) using implicit differentiation:
\[
\frac{dy}{dt} = \frac{d}{dt}(x^2 + 5) = 2x \frac{dx}{dt}.
\]
2. Substitute \(x = 2\) and \(\frac{dx}{dt} = 3\):
\[
\frac{dy}{dt} = 2(2)(3) = 12.
\]
Thus, \(\frac{dy}{dt} = 12\).

---

### Problem 2
A circular plate's radius is increasing at a rate of 0.03 cm/min. Find the rate at which the plate's area is increasing when the radius is 45 cm.

**Solution:**

1. The area \(A\) of a circle is given by:
\[
A = \pi r^2.
\]
2. Differentiate both sides with respect to \(t\):
\[
\frac{dA}{dt} = 2\pi r \frac{dr}{dt}.
\]
3. Substitute \(r = 45\) cm and \(\frac{dr}{dt} = 0.03\) cm/min:
\[
\frac{dA}{dt} = 2\pi (45)(0.03) = 2.7\pi \, \text{cm}^2/\text{min}.
\]
Thus, the rate of area change is \(2.7\pi \, \text{cm}^2/\text{min}\).

---

### Problem 3
Given:
\[
x^2 + 4y^2 + 2y = 16, \quad \frac{dx}{dt} = 7 \, \text{when} \, x = 2 \, \text{and} \, y = -2
\]
Find \(\frac{dy}{dt}\).

**Solution:**

1. Differentiate both sides of the equation implicitly with respect to \(t\):
\[
\frac{d}{dt}(x^2) + \frac{d}{dt}(4y^2 + 2y) = \frac{d}{dt}(16).
\]
This gives:
\[
2x \frac{dx}{dt} + (8y + 2)\frac{dy}{dt} = 0.
\]
2. Substitute \(x = 2\), \(y = -2\), and \(\frac{dx}{dt} = 7\):
\[
2(2)(7) + (8(-2) + 2)\frac{dy}{dt} = 0,
\]
\[
28 - 14 \frac{dy}{dt} = 0,
\]
\[
\frac{dy}{dt} = \frac{28}{14} = 2.
\]
Thus, \(\frac{dy}{dt} = 2\).

---

### Problem 4
The edge length \(x\) of a cube decreases at the rate of 4 m/min. 

**(a)** At what rate does the surface area change when \(x = 6\) m?

The surface area \(S\) of a cube is:
\[
S = 6x^2.
\]
1. Differentiate with respect to \(t\):
\[
\frac{dS}{dt} = 12x \frac{dx}{dt}.
\]
2. Substitute \(x = 6\) m and \(\frac{dx}{dt} = -4\) m/min (since the edge length is decreasing):
\[
\frac{dS}{dt} = 12(6)(-4) = -288 \, \text{m}^2/\text{min}.
\]
Thus, the surface area is decreasing at a rate of \(288 \, \text{m}^2/\text{min}\).

**(b)** At what rate does the volume change when \(x = 6\) m?

The volume \(V\) of a cube is:
\[
V = x^3.
\]
1. Differentiate with respect to \(t\):
\[
\frac{dV}{dt} = 3x^2 \frac{dx}{dt}.
\]
2. Substitute \(x = 6\) m and \(\frac{dx}{dt} = -4\) m/min:
\[
\frac{dV}{dt} = 3(6^2)(-4) = 3(36)(-4) = -432 \, \text{m}^3/\text{min}.
\]
Thus, the volume is decreasing at a rate of \(432 \, \text{m}^3/\text{min}\).

---

Would you like further details or have any questions on these solutions?

### Relative Questions:
1. How do you apply the chain rule for implicit differentiation?
2. What happens to the surface area of a cube if its edge length increases?
3. How would the solution change in Problem 2 if the radius increased faster?
4. What if \(\frac{dx}{dt}\) in Problem 3 were negative? How would it affect the result?
5. Can you think of a real-world scenario where understanding rates of change for volume and area is critical?

**Tip:** Always check whether a quantity is increasing or decreasing when determining the sign of your rates of change.

The user submitted several calculus problems involving rates of change and implicit differentiation. These include finding dy/dt for different functions and real-world applications involving surface area and volume.

Learn how to solve calculus problems involving related rates using implicit differentiation. Problems include finding dy/dt for functions and calculating rates of change for the surface area and volume of a cube, and area of a circle. Suitable for college-level calculus students.

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