# MathPractice

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## Implicit Differentiation

Q Differentiate y = arccos(2t/(1+t2))

A A trick to do this question is to convert the question to cos(y) = 2t/(1 + t2). Now do implicit differentiation to get -sin(y) y' = (-2t2 + 2)/(1 + t2)2. Since as know cos(y) = 2t/(1 + t2), then sin(y) = (t2 - 1)/(t2 + 1). (there are two ways to get this. One way is to use trig identity sin2 + cos2 = 1; the other way is to draw a right-angled triangle with angle y, adjacent side = 2t and hypotenuse = (1 + t2). Then the opposite side is (t2 - 1), and sin(y) = opp/hyp). So going back to the derivative and isolate y' = (-2t2 + 2)/(t4 - 1).

## Maxima/Minima

Q Find all values of ${\displaystyle a}$ such that ${\displaystyle f(x)=x^{3}-3x^{2}+ax+1}$ has a local maximum

A first derivative ${\displaystyle f'(x)=3x^{2}-6x+a}$ and second derivative ${\displaystyle f''(x)=6x-6}$. For a local maximum to exist there should be a critical point ${\displaystyle x}$ so that ${\displaystyle f'(x)=0}$ and ${\displaystyle f''(x)<0}$. Thus, there should exist an ${\displaystyle x}$ such that ${\displaystyle 3x^{2}-6x+a=0}$ and ${\displaystyle x<1}$. Solution to the quadratic equation is ${\displaystyle 1\pm {\frac {\sqrt {36-12a}}{6}}}$. Clearly one root is less than 1 if the ${\displaystyle 36-12a>0}$, i.e square root is defined. This gives ${\displaystyle a<3}$

## Graphing

Q Find this equation y = ax3 + bx2 + cx + d if the local max is 3 at -3 and local min is 0 at 1.

A First we differentiate the equation y' = 3ax2 + 2bx + c. We know that x = 1 and x = -3 are solutions to y' = 0, so we know 3a + 2b + c = 0 and 27a -6b + c = 0. Subtract these two equations to get 24a - 8b = 0, or 3a = b. Moreover, we know that at x = 1, y = a + b + c + d = 0, and at x = -3, y = -27a + 9b - 3c + d = 3. Subtract the last two equations to get -28a + 8b - 4c = 3. Now we add this equation to 4*(3a + 2b + c) = 0 to get -16a + 16b = 3. Solving gives a = 3/32, b = 9/32, c = -27/32, and d = -15/32.

## Related Rates

Q A candle is placed at a distance l1 from a thin block of wood of height H. The block is a distance l2 from a wall as shown in figure 7.1. The candle burns down so that the height of the flame, h1 decreases at the rate of 3 cm/hr. Find the rate at which the length of the shadow y cast by the block on the wall increases. ( note: your answer will be in terms of the constants l1 and l2).

A: note that a decrease in the candle height is equivalent to an increase in H. Thus we can use similar triangle to write H/l1 = y/(l1 + l2). Now differentiate the equation with respect to time t, to get H'/l1 = y'/(l1 + l2). Since H' = 3cm/hr (note that it is positive because it's increasing in length), y' = 3(1 + l2/l1) cm/hr.

Q An ostrich 1.5 m tall is walking toward a street light 4 m above the ground at a speed of 5 m/s. How fast is the length of the ostrich's shadow decreasing? At what speed is the tip of the shadow moving? If "a" is the angle between the light and the ground measured at the tip of the shadow, as shown in the picture, what is the rate of change of when the shadow is 1.5 m long?

A The main concept is to use similar triangles. Let the length of the shadow be s, and the distance of the ostrich to the lamp post be d. Then using similar triangles, we have the relationship s/1.5 = (s + d)/4. Now differentiate both sides w.r.t. time t, we get s'/1.5 = s'/4 + d'/4, or 5s'/3 = d'. Given that d' = -5, we know that s' = -3. The tip of the shadow is moving towards the lamppost at the speed of s' + d' (because both the shadow is decreasing in length and the ostrich is moving), so answer = -8. If we look at tangent of the angle a, we have tan(a) = 4/(s + d). It is easier to deal with 1/tan(a) = (s + d)/4. Differentiate w.r.t. time again, we get -sec2(a)/tan2(a) a' = -1/sin2(a) a' = s'/4 + d'/4. When s = 1.5, we know d = 2.5, so a = pi/4. Plug in a = pi/4, s' and d' to solve for a'.

## Optimization

Q Your room has a window whose height is 1.5 meters.. The bottom edge of the window is 10 cm above your eye level. How far away from the window should you stand to get the best view? ("Best view" means the largest visual angle, ie. angle between the lines of sight to the bottom and to the top of the window.)

A First we formulate a couple of relationships: tan(alpha) = 0.1/x, and tan(alpha + theta) = 1.6/x. Then we use the multiple angle formula for tangent to make it tan(alpha + theta) = tan(alpha) + tan(theta)/ (1 - tan(alpha)tan(theta) ) = 1.6/x. Simplifying a bit, and substitute in tan(alpha) = 0.1/x to get tan(theta) = 1.5x/(x2 + 0.16). Because tan(theta) is a function of x, we define f(x) = tan(theta) = 1.5x/(x2 + 0.16). As we know that tan(theta) is an increasing function of theta, maximizing theta is the same as maximizing tan(theta) (THIS IS THE KEY INSIGHT TO SOLVING THIS QUESTION). Therefore all we need to do is to maximize tan(theta), or equivalently, maximize f(x). To maximize, we differentiate f'(x) = (-1.5x2 + 0.24)/(x2 + 0.16)2 = 0, and solving yields x = 0.4.

Note: another method is to write theta = arctan(1.6/x) - arctan(0.1/x), and then use derivative formulas for arctan to differentiate.

Q Jack and Jill have an on-again off-again love affair. The sum of their love for one another is given by the function y( t) = sin(2t) + cos(2t). ( a) Find the times when their total love is at a maximum. (b ) Find the times when they dislike each other the most.

A We need to find the max and the min, so we differentiate. y' = 2cos(2t) - 2sin(2t) = 0. To solve this, we have cos(2t) = sin(2t). We need to find the angles (2t) where sine and cosine are the same. From high school (if you don't remember this, you have to review), sine and cosine are equal at angles pi/4 and 5pi/4. However, because the function oscillates, we have to find the general representation of the solution, which is pi/4 + 2n*pi, and 5pi/4 + 2n*pi, for any integer n. The reason we add this 2n*pi term is because the period of trig functions are 2pi. So we have 2t = pi/4 + 2n*pi and 5pi/4 + 2n*pi. Then, t = pi/8 + n*pi and 5pi/8 + n*pi. Checking second derivative to see that t = pi/8 + n*pi is the maximum and t = 5pi/8 + n*pi is the minimum.

## Differential Equations

Q A barrel initially contains 2 kg of salt dissolved in 20 L of water. If water flows in the rate of 0.4 L per minute and the well-mixed salt water solution flows out at the same rate. What is the concentration of salt after 8 minutes?

A We know that the amount of water in the barrel is constant, so we only need to worry about the amount of salt inside the barrel. If we take out 0.4L of salt water from a well-mixed salt solution of 20L, then we are taking out 0.4/20 = 0.02 portion of salt per minute. In other words, the rate of change of the amount of salt, dS/dt = -0.02S, with initial condition S(0) = 2.

Solving the differential equation yields S = S0 e-0.02t, with S0 = 2. In other words, S = 2e-0.02t. Plug in t = 8 to get the amount of salt = 2e-0.16, and the concentration is therefore 0.1e-0.16.

Q The population k(t) of a certain microorganism grows continuously and follows an exponential behavior over time. Its doubling time is found to be 0.27 hours. What differential equation would you use to describe its growth? (Note: you will have to find the value of the rate, k, using the doubling time).

A (note you can by-pass the steps here if you find the well-established formula from some books) When something grows exponentially, it is described by a differential equation dk/dt = rk, and the solution to the differential equation is k(t) = k0 ert. When we plug in t = 0, we find that k(0) = k0, so we know k_0 is the initial population. Then, to find the constant r we plug in when t = 0.27, k(t) = 2k0 (it doubles): k(0.27) = 2k0 = k0 e0.27r. Solving yields r = ln(2)/0.27. So we know the equation is k(t) = k0 eln(2)t/0.27.

Q (a similar question to previous one) If 70% of a radioactive substance remains after one year, find its half-life.

Let the amount of the radioactive substance by y(t), then we know the differential equation is dy/dt = ry, or y(t) = y0 ert, which is the solution to the differential equation. We know that at t = 0, y(0) = y0 e0 = y0, so y0 is the initial value. Now we know that at t = 1, y(t) = 0.7y0 (because 70% remains), so plug this in to get y(1) = 0.7y0 = y0 e1r). Solving for r to get r = ln(0.7). To find half life, we want to solve for t such that y(t) = 0.5y0. Plug it into the equation to get 0.5y0 = y0eln(0.7)t. So t = ln(0.5)/ln(0.7).

Q A bacterial population grows at a rate proportional to the population size at time t. let y(t) be the population size at time t. By experiment it is determined that the population at t=10 min is 15000 and t=30 min is 20000 . What was the initial population?

A We know solutions to differential equations dy/dt = ry is y(t) = y0 ert). We are given two sets of data: (t = 10, y = 15000), and (t = 30, y = 20000). We plug it into the solution to get y(10) = 15000 = y0 e10r and y(30) = 20000 = y0 e30r). To save some typing we will divide the two equations to get 15000/20000 = e10r(10r)/e30r = e-20r. We can solve for r, which is -ln(3/4)/20. Plug it back into one of the two equations and we can find y0.

Q Alcohol enters the blood stream at a constant rate of "k" gm per unit time during a drinking session. The liver gradually converts the alcohol to other, nontoxic by-products. The rate of conversion per unit time is proportional to the current blood alcohol level, so that the differential equation satisfied by the blood alcohol level is dc/dt=k-sc where k, s are positive constants. Suppose initially there is no alcohol in the blood. Find the blood alcohol level c as a function of time from t=0, when the drinking started.

A To solve the differential equation dc/dt = k - sc. We can let u = k - sc, thus du/dt = -s dc/dt. Substitute this into the equation to get du/dt = -s dc/dt = -s(k - sc) = -su. We know how to solve a simple differential equation du/dt = -su, which is u = Ae-st. So we can solve the equation from there.

Initially there is no alcohol in the blood, meaning c(0) = 0. How about u(t)? u(t) = k - sc(t), so u(0) = k - sc(0) = k. Now we know A = k. The solution is therefore u(t) = ke-st). Because u = k - sc, or c = 1/s(k - u), we know that c(t) = 1/s (k - ke-st). Done!

## Euler's Method

Q For each of the following differential equations, find value of y at the specific point by Euler’s method using the given initial condition and step size (display three decimal places for your answer).

dy/dx = √(x2 + y2) , y(0.2) = 0.5, find y(0.4) in 4 steps.


A Note that the formula for Euler's method is y(i+1) = yi + f(yi, xi) dx. The initial value is given by y(0.2) = 0.5, ie, x0 = 0.2, and y0 = 0.5. Since we need to do it in 4 steps, up to the final value x4 = 0.4, the step size is (0.4 - 0.2)/4 = 0.05. This is the value of dx. Then we know the rest of the x values are x0 = 0.2, x1 = 0.25, x2 = 0.3, x3 = 0.35 and x4 = 0.4. We can start the approximation with the formula y1 = y(0.25) = 0.5 + sqrt(0.22 + 0.52) (0.05). Next y2 = y(0.3) = y1 + sqrt(0.252 + y12) (0.05)... until y4.