Tutor profile: Simon L.
How do you identify critical points and how the function behaves outside of these points.
Critical points are defined as locations on a function where the first derivative, also known as slope, is zero or undefined. To determine the slope, you must differentiate the function. Then solve for the point of the function where the derivative goes to zero. There are cases where the derivative does not exist. These could occur at the junction of a piece-wise function, or when the denominator of the first derivative expression goes to zero. Critical points, when the first derivative exist, implies the function is horizontal instantaneously at that point. Between critical points, the function changes depending on the sign of the first derivative. Positive first derivation implies the function increases, and vice versa. The concavity is another way to represent the behavior of functions. This is determined by finding the second derivative (differentiate the function twice). A positive second derivative means concave up, where the function appears to curve upwards, and vice versa. The location where a function changes from concave up to concave down (where second derivative is 0), is called the inflection point.
How do you work with absolute values in an equation?
Having absolute values in an algebraic equation adds more workload than complexity to the solution. The following outlines the necessary steps required when solving an equation with absolute value functions. The example $$|x-3| = |3x+2|-1$$ will be used. 1. Determine the breakpoints of each absolute value. What value of x will cause any of the absolute values to be 0? The two absolute functions in the sample equation have breakpoints x = -1.5 and x = 3. 2. Create cases for x that are separated by the breakpoints. These cases must span the entirety of the domain of x, from negative to positive infinity. In the sample equation, we would have 3 cases: $$x<-1.5$$ , $$-1.5\leq x < 3 $$ , $$x\geq3 $$. Remember, when expressing these cases, the inequalities must not overlap. This means I cannot write one case as $$x\leq-1.5$$ and the other $$-1.5\leq x < 3 $$ 3. Manipulate the equation based on the first case. Remove the absolute value signs and replace them with open and end brackets. If the value inside the brackets is less than 0 based on the current case, add a negative sign in front of the bracket. This is to take into account of the absolute sign flipping the value from negative to positive. Then solve the equation. In the sample equation, the case $$x<-1.5$$ would result in the equation: $$-(x-3) = -(3x+2)-1$$. Notice how the two absolute values are replaced with brackets, and a negative sign appears in front of each bracket since $$x<-1.5$$ causes the value inside both brackets to be negative. 4. Solve the equation and check the answer. Solving case 1 for the sample equation gives the solution x = -3. We now simply check whether x = -3 satisfy this case, which is $$x<-1.5$$. Indeed it does, so x = -3 is a solution. At times, a solution will not satisfy the case. As an example, in case 3, $$x\leq3 $$, we can simplify the equation to $$(x-3) = (3x+2)-1$$, which gives the solution x = -2. This solution does not satisfy the case, so we must omit it. 5. Repeat steps 3 and 4 for all cases, and record all solutions. In this sample equation, our final solutions are x = -3 and x = 0.5.
Subject: Aerospace Engineering
Explain each term in the governing equation of the thrust of a jet engine. What would happen if the nozzle exit pressure decreases?
The thrust equation of a jet engine is as follows: $$T = m_aV_e(1+f)-m_aV_a+(P_e-P_a)A_e $$ The first two terms result from the conservation of momentum between the inlet and nozzle of mass from the engine. These two terms are the product of the mass and velocity at the nozzle and inlet, respectively. The third term arises due to the pressure difference between the nozzle exit and the atmosphere. This produces an additional force on the jet engine. The sum of all three forces is the net thrust of the engine. Keeping all other variables constant, if we were to decrease nozzle pressure, the thrust would decrease as well. However, it is important to know that one cannot do this in practice, as changing the nozzle pressure would result in the change in other variables as well in real world engine design. It also happens that, the greatest thrust is produced when Pe = Pa. This means, in practice, it is best to keep Pe close to Pa, and it will generally result in greater thrust and better propulsive efficiency. You can achieve Pe = Pa by having subsonic exhaust flow, or at very specific conditions in supersonic exhaust flow.
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