Problems with complex numbers.

Q1. Convert the complex number z = 6 + 8j to polar form.

A1. The magnitude $|z|= \sqrt{6^2 + 8^2}= \sqrt{36+ 64} = \sqrt{100} = 10.$

Since both x and y components are positive, the complex number is in the first quadrant.
The argument $\theta = atan(\frac{8}{6}) = 0.9273$, in radians. Multiply by $180/\pi$ to obtain the argument degrees. The value would be $53.3^\circ$.

Q2, Convert to rectangular form the complex number in polar form $12 \angle 210^\circ$.

A2. The complex number is in the third quadrant. This means that both components are negative.
$x = 12 \cos(210^\circ) = -10.392, y = 12 \sin(210^\circ) = -6.0$,from which $z = -10.392 - 6j.$

Fomula mass of chemical compound

Q. Calculate the formula mass of the compounds KBr and $NH_4CN$

A. The atomic masses of K and Br are respectively 39.102 and 79.904 respectively. Thus the formula mass is 39.102+79.904= 119.006 amu.

On the other hand, NH4CN has a formula mass calculated in the following table:

Component	Amu	Total
N	14.00671	14.0067
H4	1.008(4)	4.032
C	12	12
N	14.0067	14.0067
Total		44.107

The formula mass is then 44.107.

Integrals of the form ∫ xn ex dx by integration by parts (IBP)

These problems originally appeared in my former alternate blog at http://www.optimal-learning-systems.org/eps.


Q1. What is the value of the integral $\int xe^xdx$?

A1. Let $u=x$. Then $du=dx$. Also let $dv=e^xdx$. Then $v= \int e^xdx=e^x$.

Now apply the integration by parts formula, $\int u dv=uv−\int v du$.

We have, $\int x e^xdx=x e^x−\int e^xdx=x e^x−e^x=e^x[x−1]$.

Therefore, $\int xe^xdx=e^x[x−1]$.

One may wonder why the constant of integration is not added in the formula for v. The IBP formula becomes in this case $u(v+C)−∫(v+C)du$ and one can see that the terms containing the constant of integration cancels out.

Q2. Determine the integral $\int x^2e^x dx$.

A2. Let $u=x^2$, then $du=2^xdx$. Also let $dv=e^xdx$, then $v=\int e^xdx$. Then $dv=e^x dx$. Substituting in the IBP formula,

$x^2e^x−\int e^x(2x)dx=x^2e^x−2∫xe^xdx$. But the last term was the problem we treated before! Hence, $\int x^2e^xdx=x2e^x−2[e^x(x−1)]$ and therefore

$$\int x^2e^xdx=e^x [x2−2(x−1)]$$

You may add the constant of integration in the final answer.

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Q3. Determine the integral $\int x^3e^x dx$.

A3. As usual, let $u=x^3$, then $du=3x^2dx$. Also let $dv=e^xdx$, then $v=e^x$. Substituting in the IBP formula,

$x^3e^x−\int e^x(3x^2)dx=x^3e^x−3∫x2e^xdx$. But the last term involved an integral we treated before! Therefore $$\int x^3e^x dx=x^3e^x−3e^x[x^2−2(x−1)]=e^x[x^3−3(x^2−2(x−1))]$$
.
Notice that $$\int x^ne^x dx=x^ne^x−n\int x^{n−1}e^xdx$$ . We leave the fun to the student to discover on his/her own a general formula for the answer.

Calculus: derivative of products of functions of x.

Q. What is the derivative of $x^2 e^x \sin(x)$ ?


A. The derivative of a product of three functions in $x$, say, $u, v$ and $w$,$\frac{d(uvw)}{d x}$ is given by

$$\frac{d(uvw)}{dx} =\frac{uvw}{u} \frac{dw}{dx} + \frac{uvw}{v} \frac{dw}{dv} + \frac{uvw}{u} \frac{dw}{dv}$$ .

It is easy to generalize this 'product' rule. Now let $u = x^2, v = e^x, w = \sin{(x)}$. Then we obtain the following result:

$$e^x \sin(x) \frac{d(x^2)}{dx} + x^2 \sin(x) \frac{d(e^x)}{dx}+ {x^2 e^x}\frac{d(\sin(x) )}{dx} = 2x e^x \sin(x) + x^2 \sin(x) e^x + x^2 e^x \cos(x)$$

Center and radius of a circle from the general equation.

Q. Determine the center and radius of a circle whose equation given by $x^2 - 4x + y^2 +10y -71 = 0$.


A. The standard form is given by $(x-h)^2 + (y-k)^2 = r^2$. Hence we should complete the squares for the terms involving $x$ and $y$. For x, the completion of $x^2 - 4x + \_\_$ has a constant term given by $[(-4x/ (2\sqrt{x^2})]^2 = 4$. On the other hand for y, we have $[10y/(2\sqrt{y^2})]^2= 25$. Thus we obtain $$(x-\sqrt{4})^2 +(y- -\sqrt{25})^2 -71 - 29 = 0$$ which we finalize to $(x -2)^2 + (y-(-5))^2 = 100$. Thus we find the center to be (2, -5) and the radius 10.

Welcome to my new blog. I will post some engineering problems with their solutions here. Stay tuned!