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# Tutor profile: Daniel B.

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Daniel B.
Civil Engineer
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## Questions

### Subject:Physics (Fluid Mechanics)

TutorMe
Question:

Why I can't use the Bernoulli equation to analyze the flow in a pipe system with a pump?

Inactive
Daniel B.
Answer:

The Bernoulli equation is derived considering the forces acting along an differential element of a streamline; the forces considered in that balance are only the pressure and weight of that element, so no friction (viscous effects) is considered. In one hand, in a pipe, due to viscosity and the velocity profile, there is friction, so, if you want to take into account the effect of that friction force, the Bernoulli equation won't help you. On the other hand, the pump exerts some kind of work over the flow through a tangential force that isn't taken into account by the Bernoulli equation either. So, even though is a common practice to say that the energy and Bernoulli equations are the same, strictly speaking, you need to define a control volume to use the energy equation.

### Subject:Linear Algebra

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Question:

If $$A \in \mathbb{R}^{n\times n}$$ and $$A$$ satisfies the equation $$A^2-2A+I=0$$ where $$I$$ is the identity, probe that A is invertible

Inactive
Daniel B.
Answer:

$$A$$ is invertible if there is a matrix $$B$$ (also in $$\mathbb{R}^{n\times n}$$) such that $$AB = I$$. We want to put the equation $$A^2-2A+I=0$$ in those terms. You can start isolating the matrix $$I$$, so we add the matrix $$2A-A^2$$ to both sides of the equation and get $$I=2A-A^2$$. Now notice that if could get a common factor on the r.h.s. of the equation we would get something like $$A B=I$$, unfortunately that doesn't exist in matrix algebra. However, we do know that $$A=AI$$, $$A^2=AA$$ and $$C(D+F)=CD+CF$$ (if $$C, D$$ and $$F$$ are in compatible spaces); therefore, $$2A-A^2=2AI-AA=A(2I)-AA=A(2I-A)$$ So, if we make $$B=2I-A$$, we see that every matrix $$A\in\mathbb{R}^{n\times n}$$ who satisfies the equation $$A^2-2A+I=0$$ is invertible with $$B=2I-A$$ as his inverse.

### Subject:Calculus

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Question:

Why does the gradient vector maximize the directional derivative?

Inactive
Daniel B.
Answer:

If $$f$$ is such a function that you can calculate its gradient, the directional derivative of $$f$$ in direction of the $$\vec{u}$$ vector (where $$\vec{u}$$ is unitary) is defined from the gradient vector as $$\vec{\nabla f} \cdot \vec{u}$$. In general, the scalar product between vectors is defined as the product of their magnitudes, multiplied by the cosine of the angle between them ($$\theta$$). Therefore $$\vec{\nabla f} \cdot \vec{u}=|\vec{\nabla f}| |\vec{u}| \cos{\theta}=|\vec{\nabla f}| |\vec{u}| \cos{\theta}$$ which is maximum when $$\cos{\theta}=1$$, so $$\theta=0$$. Thus, the maximum absolute value of the directional derivative is obtained when direction in which it is calculated (given by the direction of $$\vec{u}$$) coincides with the direction of the gradient vector

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