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# Tutor profile: Daniel D.

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Daniel D.
Engineer, Parent, Tutor
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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.

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?

Inactive
Daniel D.

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

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Question:

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

Inactive
Daniel D.

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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