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Tutor profile: Daniel D.

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Daniel D.
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Questions

Subject: Calculus

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Question:

Using the concept of limit, show that the derivative of $$f(x)=x^2$$ is $$f'(x)=2x$$

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Daniel D.
Answer:

By definition $$ f'(x) = \lim_{\Delta x\to0} \frac{f(x +\Delta x)-f(x)} {\Delta x} $$ Thus, $$ f'(x) = \lim_{\Delta x\to0} \frac{(x+ \Delta x)^2 - x^2 } {\Delta x} $$ That is: $$ f'(x) = \lim_{\Delta x\to0} \frac{(x^2 + 2x\Delta x+ \Delta x^2) - x^2 } {\Delta x} $$ Simplifying the $$x^2$$ we get: $$ f'(x) = \lim_{\Delta x\to0} \frac{ 2x\Delta x+ \Delta x^2 } {\Delta x} $$, Then using the fact that $$\Delta x \neq 0 $$, we can simplify too by $$\Delta x$$ $$ f'(x) = \lim_{\Delta x\to0} 2x + \Delta x $$ Therefore the result: $$ f'(x) = 2x $$

Subject: Physics

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Question:

1. Derive the equations of motion of falling object of mass (m) in a constant gravitational field g 2. Given that the initial altitude is $$ h_0$$, the initial speed is $$v_0$$, solve the equation of motion. 3. What amount of time is spent by the object before touching the ground?

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Daniel D.
Answer:

1. Using Newton law of motion: $$m \vec{a} = m \vec{F}_{total} $$ $$m \frac{\partial^2 h}{\partial t^2} = -m g $$ Thus, $$ \frac{\partial^2 h}{\partial t^2} = - g $$, note that the EOM does NOT depend on the mass m of the object. 2. By integrating one time the EOM, we get: $$ \frac{\partial h}{\partial t} = - g t +v_0 $$ We integrate one more time and get $$ h(t) = - \frac{1}{2} g t^2 +v_0t + h_0 $$ 3. We solve previous equation for h(t) =0. We immediately recognize a second degree equation in variable t, solution is of the form : $$t = \frac { -v_0 +/- \sqrt{ v_0^2-4* (-1/2g)h_0} } { 2 (-1/2g) } $$ We keep only the positive solution $$ t = \frac { v_0 + \sqrt{ v_0^2+2gh_0} } { g } $$ $$ t = \frac { v_0} {g} \sqrt{ 1+\frac{2gh_0}{v_0^2} } $$

Subject: Algebra

TutorMe
Question:

Solve this system of equation: $$ \begin{cases} 2r + s -t= 0 \\ r - s + t = 6 \\ r + 2s +t= 9 \end{cases} $$

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Daniel D.
Answer:

At first look, this is a system of 3 linear equation with 3 variables. This is a good sign as we have as many equation as variables. Recognize that by adding the first 2 equations side by side we obtain: $$ (2r+s-t) +(r-s+t) = 0+ 6 $$ That is : $$ 3r = 6 $$, thus $$ r=2 $$ We can use the value of r in equation (2) and (3) an rewrite it as: $$ \begin{cases} 2 - s + t = 6 \\ 2 + 2s +t= 9 \end{cases} $$ Or as well, simplifying $$ \begin{cases} - s + t = 4 \\ 2s +t= 7 \end{cases} $$ By substracting the first equation to the second one, we get: $$ 2s - (-s) + t -t = 7 -4 $$ That is, $$3s = 3 $$, or $$s = 1 $$ Bu re-injecting s=1 in the first equation, we get: $$-1 +t = 4 $$ , thus t=5 The solution of this system are: $$ ( r, s, t) = ({2, 1, 5 }) $$

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