1. Problem Description

1. Problem Description

The problem consists of three sub-problems.

4. 1: Two cars move from the same point. One goes south at 30 km/h and the other west at 72 km/h. We need to find the rate at which the distance between them increases after 2 hours.

5. 2: We need to find the dimensions of a rectangle with an area of 1000 m$^2$ that has the smallest perimeter.

6. 3: We need to use the given substitution to evaluate two integrals:

(a) $\int 2^x dx$, $u = 2^x$

(b) $\int_0^{\sqrt{b}} \frac{1}{x^2+b} dx$, $u = \frac{x}{\sqrt{b}}$

7. Solution Steps

4. 1:

After 2 hours, the car traveling south will have traveled $30 \text{ km/h} \times 2 \text{ h} = 60 \text{ km}$.

After 2 hours, the car traveling west will have traveled $72 \text{ km/h} \times 2 \text{ h} = 144 \text{ km}$.

The distance between the two cars after 2 hours, $d$, can be found using the Pythagorean theorem:

$d = \sqrt{(60)^2 + (144)^2} = \sqrt{3600 + 20736} = \sqrt{24336} = 156 \text{ km}$.

Let $s(t)$ be the distance traveled by the car going south at time $t$, and $w(t)$ be the distance traveled by the car going west at time $t$.

$s(t) = 30t$

$w(t) = 72t$

The distance between the cars at time $t$ is $D(t) = \sqrt{s(t)^2 + w(t)^2} = \sqrt{(30t)^2 + (72t)^2} = \sqrt{900t^2 + 5184t^2} = \sqrt{6084t^2} = t\sqrt{6084} = 78t$.

The rate at which the distance is increasing is the derivative of $D(t)$ with respect to $t$.

$\frac{dD}{dt} = 78$ km/h.

5. 2:

Let the dimensions of the rectangle be $l$ and $w$. The area is given by $A = lw = 1000$.

The perimeter is given by $P = 2l + 2w$. We want to minimize $P$.

From $lw = 1000$, we have $w = \frac{1000}{l}$.

Substituting into the perimeter equation, we get $P = 2l + 2(\frac{1000}{l}) = 2l + \frac{2000}{l}$.

To minimize $P$, we take the derivative with respect to $l$ and set it equal to

0. $\frac{dP}{dl} = 2 - \frac{2000}{l^2} = 0$.

$2 = \frac{2000}{l^2}$

$l^2 = 1000$

$l = \sqrt{1000} = 10\sqrt{10}$

Then $w = \frac{1000}{10\sqrt{10}} = \frac{100}{\sqrt{10}} = \frac{100\sqrt{10}}{10} = 10\sqrt{10}$.

Thus $l = w = 10\sqrt{10}$, which means the rectangle is a square.

6. 3:

(a) $\int 2^x dx, u = 2^x$.

This substitution doesn't seem very helpful. A better approach is to remember that $\frac{d}{dx} a^x = a^x \ln a$.

Thus, $\int 2^x dx = \frac{2^x}{\ln 2} + C$.

Alternatively, if we try $u = 2^x$, then $du = 2^x \ln 2 dx$.

Then $dx = \frac{du}{2^x \ln 2} = \frac{du}{u \ln 2}$.

$\int 2^x dx = \int u \frac{du}{u \ln 2} = \int \frac{1}{\ln 2} du = \frac{u}{\ln 2} + C = \frac{2^x}{\ln 2} + C$.

(b) $\int_0^{\sqrt{b}} \frac{1}{x^2+b} dx; u = \frac{x}{\sqrt{b}}$

$x = u\sqrt{b}$

$dx = \sqrt{b} du$

When $x = 0$, $u = \frac{0}{\sqrt{b}} = 0$.

When $x = \sqrt{b}$, $u = \frac{\sqrt{b}}{\sqrt{b}} = 1$.

Then $\int_0^{\sqrt{b}} \frac{1}{x^2+b} dx = \int_0^1 \frac{1}{(u\sqrt{b})^2 + b} \sqrt{b} du = \int_0^1 \frac{\sqrt{b}}{bu^2 + b} du = \int_0^1 \frac{\sqrt{b}}{b(u^2+1)} du = \frac{\sqrt{b}}{b} \int_0^1 \frac{1}{u^2+1} du = \frac{1}{\sqrt{b}} \int_0^1 \frac{1}{u^2+1} du$.

$\int_0^1 \frac{1}{u^2+1} du = [\arctan(u)]_0^1 = \arctan(1) - \arctan(0) = \frac{\pi}{4} - 0 = \frac{\pi}{4}$.

So the integral is $\frac{1}{\sqrt{b}} \cdot \frac{\pi}{4} = \frac{\pi}{4\sqrt{b}}$.

8. Final Answer

9. 1: 78 km/h

1

0. 2: $10\sqrt{10}$ m by $10\sqrt{10}$ m

1

1. 3:

(a) $\frac{2^x}{\ln 2} + C$

(b) $\frac{\pi}{4\sqrt{b}}$