George L.

engineering graduate, tutored friends for years

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Question:

$$y$$ varies inversely with the square root of $$x$$. When $$x = 9, y = 2$$. what is $$y$$ when $$ x = 4$$?

George L.

Answer:

"y varies inversely with the square root of x" denotes the relation $$y = \frac{1}{\sqrt{x}}$$ Let variable $$k$$ be a number such that $$y = \frac{k}{\sqrt{x}}$$, then solve for k to complete the function. Solve for $$k$$ where $$x = 9, y = 2$$: $(y = \frac{k}{\sqrt{x}}$) $(k = y\sqrt{x}=(2)\sqrt{(9)}=2 \mathrm{x} 3=6 $) Thus the complete relation is: $(y = \frac{6}{\sqrt{x}}$) Now to answer the question, solve for $$y$$ when $$x = 4$$. $(y = \frac{6}{\sqrt{(4)}}=\frac{6}{2}=3$) $(y = 3$)

Calculus

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Question:

Find $$ \int x\mathrm{^2} \mathrm{sin} x \mathrm{d}x $$

George L.

Answer:

Use integration by parts. Let $$ u = x^2, dv = \mathrm{sin}xdx$$. Thus, $$ du = 2xdx, v = -\mathrm{cos}x$$. Then $(\int x^2 sinx dx = -x^2 \mathrm{cos}x - \int -2x \mathrm{cos}xdx$) $( = -x^2 \mathrm{cos}x +2 \int x \mathrm{cos}xdx$) Then we apply integration by parts to $$\int x \mathrm{cos}x dx, \mathrm{with} u = x, dv = \mathrm{cos}x dx$$ $(\int x\mathrm{cos} x = x\mathrm{sin}x - \int \mathrm{sin}x dx = x\mathrm{sin}x + \mathrm{cos}x $) Thus $(\int x^2 \mathrm{sin}xdx = -x^2\mathrm{cos}x + 2(x\mathrm{sin}x+\mathrm{cos}x) + C$)

Physics

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Question:

A 3 gram bullet is fired from a 5 kilogram gun with a speed of 600 m/sec. What is the speed of recoil of the gun?

George L.

Answer:

The initial momentum of the system is zero. The "collision" between the gun and the bullet is small enough that after the bullet has been fired from the gun, the total momentum of the system remains zero. Thus, let $$m_1$$ be the mass of the gun, $$m_2$$ be the mass of the bullet, and $$v_1$$and $$v_2$$ be their respective velocities. Then: $$m_1 v_1 + m_2 v_2$$ = 0 $$v_1 = - \frac{m_2}{m_1} v_2$$ $$v_1 = - \frac{0.003 kg}{5.0 kg} (600 m/s) $$ $$v_1$$ = - 0.48 m/sec Minus sign means the gun moves in the opposite direction of the bullet.

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