Here’s a detailed solution to each of the problems provided in your uploaded file:

1. Find the point on the curve $y = x^2$ closest to the point $(3, 4)$.

To find the point closest to $(3, 4)$, minimize the distance:

$D = \sqrt{(x - 3)^2 + (x^2 - 4)^2}$

Minimizing $D^2$ (to simplify), we have:

$D^2 = (x - 3)^2 + (x^2 - 4)^2$

Simplify:

$D^2 = (x - 3)^2 + (x^4 - 8x^2 + 16)$

Take the derivative:

$\frac{d(D^2)}{dx} = 2(x - 3) + 4x^3 - 16x$

Set to zero:

$2(x - 3) + 4x^3 - 16x = 0$

$2x - 6 + 4x^3 - 16x = 0 \implies 4x^3 - 14x - 6 = 0$

Solve $4x^3 - 14x - 6 = 0$ numerically or using factoring techniques. Once $x$ is found, substitute into $y = x^2$ to get the point.

2. Packing Crate Problem

Given:

• Volume = $9 \, \text{ft}^3$
• Cost per square inch for the base: $0.003$
• Cost per square inch for sides and top: $0.002$

Let:

• $x$ = side length of square base (in feet)
• $h$ = height of the box (in feet)

Volume constraint:

$x^2 h = 9$

Surface area:

$\text{Area of base} = x^2, \quad \text{Area of sides and top} = 4xh + x^2$

Cost:

$\text{Total cost} = 0.003x^2 + 0.002(4xh + x^2)$

Substitute $h = \frac{9}{x^2}$:

$\text{Total cost} = 0.003x^2 + 0.002\left(4x \cdot \frac{9}{x^2} + x^2\right)$

Simplify:

$\text{Total cost} = 0.003x^2 + 0.002\left(\frac{36}{x} + x^2\right)$

$\text{Total cost} = 0.003x^2 + 0.002x^2 + \frac{0.072}{x}$

$\text{Total cost} = 0.005x^2 + \frac{0.072}{x}$

Minimize this cost function using calculus.

3. Graph Analysis

a. $g'(3)$

$g'(t)$ represents the slope of the tangent line to $g(t)$, which indicates the rate of change of $g$ at $t = 3$.

b. If $g(t)$ is velocity:

$g'(3)$ represents the acceleration at $t = 3$.

c. Estimate $g'(3)$:

Find the slope of the tangent line at $t = 3$ using the graph.

d. Displacement from $t = 1$ to $t = 3$:

Displacement is the definite integral of $g(t)$ from $t = 1$ to $t = 3$. Approximate the area under the curve between these bounds.

4. Evaluate Integrals

a. $\int \left(7x^3 - \sqrt{x} + \frac{4}{x^3}\right) dx$

Break into separate terms and integrate:

$\int 7x^3 dx = \frac{7x^4}{4}, \quad \int -\sqrt{x} dx = \int -x^{1/2} dx = -\frac{2}{3}x^{3/2}$

$\int \frac{4}{x^3} dx = \int 4x^{-3} dx = -\frac{4}{2}x^{-2} = -\frac{2}{x^2}$

Final result:

$\frac{7x^4}{4} - \frac{2}{3}x^{3/2} - \frac{2}{x^2} + C$

b. $\int (\sec x \tan x + \frac{2}{x}) dx$

Split the terms:

$\int \sec x \tan x dx + \int \frac{2}{x} dx$

$\int \sec x \tan x dx = \sec x, \quad \int \frac{2}{x} dx = 2\ln|x|$

Final result:

$\sec x + 2\ln|x| + C$

Let me know if you'd like further clarification or a deeper dive into any of these problems.

Here are some related questions to consider:

1. How do we determine local minima/maxima when minimizing cost functions?
2. What methods can be used to solve cubic equations like in Problem 1?
3. How does the graph of $g(t)$ relate to displacement and velocity?
4. Why is it more efficient to work with $D^2$ instead of $D$ in distance problems?
5. What are some common strategies for solving integrals involving trigonometric functions?

Tip: Always check units when solving applied math problems to ensure consistency and correctness.