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Aaron D.

Master's Student in Biomedical Engineering at Pitt

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Calculus

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Question:

Evaluate the following Integral where C is the closed boundary of a triangle with vertices (0,0); (2,0); (2,4) $$ \oint_C xy^{2}dx + 3x^{2}y^4dy$$

Aaron D.

Answer:

The easiest way to solve this line integral around the circle is to use Green's Theorem which says $$\oint_C Pdx+Qdy = \iint_D (\frac {\partial Q}{\partial x} - \frac{\partial P}{\partial y}) dx dy$$ where D is the area inside the closed boundary C. $$P=xy^2$$ , $$Q=3x^2y^4$$ $$\frac{\partial P}{\partial y}=2xy$$ , $$\frac{\partial Q}{\partial x}=6xy^4$$ D can be expressed as $$0\leq x \leq 2$$ $$0\leq y \leq 2x$$ Therefore $$\int_{0}^{2} \int_{0}^{2x} (6xy^4-2xy)dydx$$ $$\int_{0}^{2} (\frac{6}{5}x(2x)^5 - x(2x)^2 - (\frac{6}{5}x(0)^5-x(0)^2)dx$$ $$\int_{0}^{2} (\frac{192}{5}x^6 - 4x^3)dx$$ $$\frac{192}{5}(\frac{2^7}{7}-\frac{0^7}{7})-(2^4-0^4)$$ $$\boldsymbol{\frac{24016}{35}=686.2}$$

Biological Engineering

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Question:

If we can model systemic arterial circulation based on an input of Aortic pressure $$P_{a}(t)$$ and an output of Aortic blood flow of $$Q_{a}(t)$$ with a resistor $$r$$ and capacitor $$ C$$ in series, parallel to a resistor $$R$$, identify the state variables and derive a system of equations.

Aaron D.

Answer:

The best way to setup this problem is to draw a picture. With $$P_{a}(t)$$ and $$Q_{a}(t)$$ accounting for the pressure and flow of the whole circuit respectively and the resistors and capacitor positioned as described in the prompt. For this model, we will identify pressure as voltage and flow as current. From here, we can identify that one of the components, the capacitor, has a differential equation associated with it since the flow through the capacitor varies with respect to pressure which varies with time $$Q_{C}(t)=C \frac {dP_{C}}{dt}$$ where C is the capacitance. Because of this, we can identify $$\textbf{1 state variable}$$, and to make our system of equations as simple as possible, we can name $$ \boldsymbol{P_{C}}$$ our state variable. Now we can setup our system of differential equations using $$P_{a}$$ as our input and $$Q_{a}$$ as our output. The total flow is equal to the flow through the first resistor and the parallel resistor. The total flow is also equal to the flow through the capacitor in series and the parallel resistor. $$Q_{a}(t) = Q_{r}(t) + Q_{R}(t) = Q_{C}(t) + Q_{R}(t)$$ The total aortic pressure across the circuit can be expressed as the pressure across the parallel resistor or the sum of pressures across the first resistor and capacitor. $$P_{a}(t)=P_{R}(t)=P_{r}(t)+P_{C}(t)$$ Knowing that the flow through the first resistor is expressed as $$Q_{r}(t)= \frac{P_{r}(t)}{r}$$ and we know $$Q_{r}(t)=Q_{C}(t)$$ therefore $$Q_{C}(t)= \frac{P_{r}(t)}{r}$$. Now we can say $$C \frac {dP_{C}}{dt}= \frac{P_{r}(t)}{r} \rightarrow \frac {dP_{C}}{dt}= \frac{P_{r}(t)}{rC}$$ This is one of the equations we need, but we need it in terms of our given input, state variables, and output . To do this we use our total pressure equation to solve for $$P_{r}(t)$$ and plug it into our equation. $$P_{a}(t)=P_{r}(t)+P_{C}(t) \rightarrow P_{r}(t)=P_{a}(t)-P_{C}(t)$$ $$ \frac {dP_{C}}{dt}= \frac {P_{a}(t)-P_{C}(t)}{rC}$$ Our other equation should be easier now using $$Q_{a}(t) = Q_{C}(t) + Q_{R}(t)$$ $$Q_{C}(t)=C \frac {dP_{C}}{dt}$$ and $$Q_{R}(t)=\frac{P_{R}(t)}{R}=\frac{P_{a}(t)}{R}$$ Therefore: $$Q_{a}(t)=C \frac {dP_{C}}{dt} + \frac{P_{a}(t)}{R}$$ Our System of Equations: $$\boldsymbol{Q_{a}(t)=C \frac {dP_{C}}{dt} + \frac{P_{a}(t)}{R}}$$ $$\boldsymbol {\frac {dP_{C}}{dt}= \frac {P_{a}(t)-P_{C}(t)}{rC}}$$

Physics

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Question:

A cannon positioned 30 degrees from the ground ($$\theta = 30 ^{\circ}$$) shoots a cannonball with a mass of 5kg. If the initial velocity of the cannonball is 100 m/s ($$v_{0}=100 m/s$$)and air resistance is negligible, at what distance will the cannonball land?

Aaron D.

Answer:

The primary equations we are going to use for this equation are the following: $$x(t) = x_{0} + v_{0x}t + 0.5a_{x}t^{2} $$ $$ y(t) = y_{0} + v_{0y}t + 0.5a_{y}t^{2} $$ We can set our coordinate system where the horizontal direction is our x coordinate and our vertical direction is the y coordinate. Let's say the initial position of the cannonball is the origin of our coordinate system, i.e. $$ x_{0}=0 , y_{0}=0 $$. So we need to find the time at which the cannonball lands, to determine the displacement. First, we will look at $$ y(t) = y_{0} + v_{0y}t + 0.5a_{y}t^{2} $$. The acceleration in the y direction is solely gravity, the only force acting on the cannonball so $$a_{y} = -g = -9.8m/s^{2}$$ From the prompt we can find $$v_{0y} $$. $$v_{0y} = v_{0} * sin(\theta)= 50 m/s$$. Our y equation simplifies to the following: $$y(t) = 0 + 50t+0.5(-9.8)t^{2}=50t-4.9t^{2}$$ We know the vertical velocity of the cannonball at the top of its height, halfway through its trajectory, is equal to 0 as it begins its descent. Using an equation for acceleration $$v_{1/2}-v_{0}=at$$ we can solve for the time at the height now. $$0-50m/s=-9.8m/s^{2}*t_{1/2} \\t_{1/2}=\frac {-50m/s}{-9.8m/s^{2}}=5.1 $$ The force of gravity is the only force acting on the cannonball as it goes up as it comes down and the initial velocity is equal to the final velocity so we know the time to get the cannonball from 50m/s to 0m/s is the same as from 0m/s to 50m/s so the total time of flight is equal to twice the time necessary to full height. $$t_{1}=10.2s$$ Knowing the x velocity is $$v_{0x}=v_{0}*cos(\theta)=86.6m/s$$ and since no force is acting on the cannonball in the x direction $$a_{x}=0$$ our equation becomes $$ x(10.2)=0+86.6*10.2+0=\textbf{883.32m}$$. The mass of the cannonball is extra irrelevant information.

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