Math 131 01/02 Midterm 2 Reviews

The midterm questions are similar to that of past exams. Please make sure you understand all of them. All examples in this page are selected from past exams.

Part 0. Fundamental Formulas

Power Rule:

\[(x^n)' = n x^{n-1}\]

Exponential Rule:

\[(a^x)' = a^x \ln(a)\]

Product Rule:

\[(fg)' = f' g + g' f\]

Quotient Rule:

\[(f/g)' = (f' g - g' f)/g^2\]

Trig functions:

\[\sin'(x) = \cos(x), \cos'(x) = -\sin(x)\] \[\tan'(x) = \sec^2(x), \cot'(x) = -\csc^2(x)\] \[\sec'(x) = \sec(x) \tan(x), \csc'(x) = - \csc(x) \cot(x)\]

Chain Rule:

\[ (f(g(x)))' = f'(g(x)) g'(x)\]

Log functions:

\[(\log_a(x))' = \frac{1}{\ln(a) x}\]

Inverse trig functions:

\[\arctan'(x) = \frac{1}{1 + x^2}, arccot'(x) = -\frac{1}{1 + x^2}\] \[\arcsin'(x) = \frac{1}{\sqrt{1-x^2}}, \arccos'(x) = -\frac{1}{\sqrt{1 - x^2}}\]

Well-known limits:

\[ \lim_{x \to 0} \frac{\sin(x)}{x} = 1, \lim_{x \to 0} \frac{\tan(x)}{x} = 1\]

\[ \lim_{n \to \infty} (1 + \frac{1}{n})^n = e\]

Examples:

1. \[(e^x \cos(x) - x^3 \sin(x))' = e^x(\cos(x) - \sin(x)) - 3x^2 \sin(x) - x^3 \cos(x)\]

2. \[\left(\frac{4 \tan(x)}{x^2 +1}\right)' = \frac{4 \sec^2(x)(x^2 +1) - 8x\tan(x)}{(x^2+1)^2}\]

3. \[(\arctan(kx))' = \frac{k}{1 + k^2x^2}\]

4. \[(\sqrt{\ln(5x + 9)})' = \frac{1}{2\sqrt{\ln(5x + 9)}} \frac{5}{5x + 9}\]

5. \[(\sqrt{2-e^x})' = \frac{1}{2\sqrt{2-e^x}}(-e^x)\]

6. \[\cos((1-2x)^3)' = -\sin((1-2x)^3) (3 (1-2x)^2) (-2)\]

Part 1. Implicit differentials and log differentials.

Implicit differentials:

Step 1. Take derivative with respect to \(x\) on whole equation.

Step 2. Collect terms involving \(y'\).

Step 3. Solve for \(y'\)

Examples:

1. \[2 x^3 + 10 y^4 = 8ye^x +2\]

Step 1. \[6x^2 + 40 y^3 y' = 8e^x y' + 8ye^x\]

Step 2. \[(40 y^3 - 8e^x )y' = 8y e^x - 6x^2\]

Step 3. \[y' = \frac{8y e^x - 6x^2}{40 y^3 - 8e^x}\]

The tangent line at (0, 1) has the slope \(m = y' = \frac{8e}{32} = \frac{e}{4}\). So the line equation is \(y = 1 + \frac{e}{4}x\).

2. \[\sin(x + y) = 2x - 2y\]

Step 1. \[\cos(x + y) (1 + y') =2 - 2y'\]

Step 2. \[(\cos(x + y) + 2) y' = 2 - \cos(x+y)\]

Step 3. \[y' = \frac{2 - \cos(x+y)}{\cos(x + y) + 2}\]

The tangent line at \((\pi, \pi)\) has the slope \(m = y' = \frac{2-(-1)}{-1+2} = 3\). So the line equation is \(y = \pi + 3(x-\pi)\).

log differentials

Step 1. Apply natural log and simplify with the help of properties:

\[\ln(ab) = \ln(a) + \ln(b)\]

\[\ln(a/b) = \ln(a) - \ln(b)\]

\[\ln(a^b) = b \ln(a)\]

Step 2. Treat with implicit differentials

Step 3. Solve for \(y'\).

Examples:

1. \[y = (x^4 +1)^{1/3}(5x -1)^{-1/2}\]

Step 1. \[\ln(y) = \frac{1}{3} \ln(x^4 + 1) -\frac{1}{2} \ln(5x-1)\]

Step 2. \[\frac{y'}{y} = \frac{1}{3} \frac{4x^3}{x^4+1} - \frac{1}{2}\frac{5}{5x-1}\]

Step 3. \[y' = (x^4 +1)^{1/3}(5x -1)^{-1/2}\cdot \left(\frac{1}{3} \frac{4x^3}{x^4+1} - \frac{1}{2}\frac{5}{5x-1} \right)\]

2. \[y = \frac{x^5\sqrt{3x+1}}{e^x +4}\]

Step 1. \[\ln(y) = 5 \ln(x) + \frac{1}{2} \ln(3x+ 1) - \ln(e^x + 4)\]

Step 2. \[\frac{y'}{y} = \frac{5}{x} + \frac{1}{2}\frac{3}{3x+1} - \frac{e^x}{e^x+4}\]

Step 3. \[y' = \frac{x^5\sqrt{3x+1}}{e^x +4}\cdot \left(\frac{5}{x} + \frac{1}{2}\frac{3}{3x+1} - \frac{e^x}{e^x+4} \right)\]

Part 2. Particle Motion and Exponential Growth Model

Particle Motion

1. velocity \(v(t) = s'(t)\), acceleration \(a(t) = v'(t) = s''(t)\).

2. When particle is at rest? those t values \(v(t) = 0\), or critical point of \(s(t)\).

3. Positive direction \(\iff\) \(s(t)\) increasing \(\iff\) \(v(t) > 0\).

4. Negative direction \(\iff\) \(s(t)\) decreasing \(\iff\) \(v(t) < 0\).

5. Speeding up \(\iff\) absolute value \(|v(t)|\) increasing \(\iff\) \(a(t), v(t)\) same sign.

6. Slowing down \(\iff\) absolute value \(|v(t)|\) decreasing \(\iff\) \(a(t), v(t)\) different sign.

7. Total distance: (a) find critical points of \(s(t)\) or t values such hat \(v(t) =0\); (b) separate interval by those critical points; (c) add absolute distance traveled on each sub-interval.

Examples:

1. \[s(t) = -2t^3 + 15 t^2 - 36 t + 28\]

\(v(t) = -6 t^2 + 30 t - 36 = -6(t-2)(t-3)\).

\(a(t) = -12 t + 30 = -12(t-2.5)\).

The critical points are t values \(v(t) = 0\), \(t = 2, 3\).

Total distance on the first 4 seconds: separate \([0, 4]\) into \([0, 2] \cup [2, 3] \cup [3, 4]\), total distance = \(|s(2) - s(0)| + |s(3)-s(2)| + |s(4)- s(3)| = 28 + 1 + 5 = 34\).

\[a(t) > 0 \iff (0, 2.5), a(t) < 0 \iff (2.5, \infty)\]

\[v(t) > 0 \iff (2,3), v(t) < 0 \iff (0, 2) \cup (3 , \infty)\]

When particle is moving left? It is when \(v(t) < 0\) \(\iff\) \(t \in (0, 2) \cup (3 , \infty)\).

When particle is speeding up? It is when \(v(t), a(t)\) take the same sign. So both positive when \(t \in (2, 2.5)\), both negative when \(t \in (3, \infty)\). The answer is \((2, 2.5) \cup (3, \infty)\).

When particle is slowing down? It is when \(v(t), a(t)\) take the opposite sign. So \(v(t) > 0, a(t) < 0\) when \(t \in (2.5, 3)\), \(v(t) < 0, a(t) > 0\) when \(t \in (0, 2)\). The answer is \((0, 2) \cup (2.5, 3)\).

2. A height function \(h(t) = 30 t - 5t^2\)

velocity \(v(t) = 30 - 10t\).

acceleration \(a(t) = -10 m/s^2\).

After 2 seconds, \(v(2) = 10 m/s\). It is positive, means rock is moving up.

When rock reaches the peak, velocity becomes \(0\). So let \(v(t) = 30 - 10t = 0\), \(t = 3\).

What is velocity hitting the ground? The moment when \(h(t) = 0\). Let \(30 t - 5 t^2 = 0\), \(t = 0, 6\). Plug in t = 6 for obvious reason, \(v(6) = -30 m/s\).

Exponential growth/decay

\[P(t) = P(0) e^{kt}\] \(P(0)\) the initial size, \(k > 0\) is growing, while \(k < 0\) is decaying. Remember that \[\frac{dP}{dt} = k P\]

Examples:

1. A bacterial starts with 10 bacterial and grows at a rate proportional to its size. After 2 hours there are 30 bacteria.

\(P(0) = 10\). As \(P(2) = 10 e^{2k} = 30\), \(k = \frac{1}{2} \ln(3)\).

So \(P(t) = 10 e^{\frac{1}{2} \ln(3) t} = 10 \cdot 3^{t/2}\).

When will the population reach 270? Let \(270 = 10 3^{t/2}\) or \(270 = 10 e^{\frac{1}{2} \ln(3) t}\). So \(\ln(27) = \frac{1}{2} \ln(3) t\), \(t = 6\) hours.

2. A population of bacteria triples in size every hour. The initial population is 100 bacterial.

As \(P(t) = P(0) e^{kt} = 100 e^{kt}\). The first sentence means \(3\cdot100 = 100 e^k\). So \(k = \ln(3)\).

After 2 hours, \(P(2) = 100 e^{\ln(3) \cdot 2} = 100 \cdot 3^2 = 900\).

So the rate of change at \(t = 2\) is \(k \cdot P(2) = \ln(3) \cdot 900 = 219.72\) bacteria/hour.

Part 3. Related rates and linear approximation.

Related rates

Step 1. Draw a diagram to find what are given what is unknown.

Step 2. Often in the right triangle to build a relation for the given quantity and unknown quantity.

Step 3. Take derivative with respect to time t on the relation found in step 2.

Step 4. Solve for the rate of change of the unknown.

Examples:

1. Two hikers. Hiker A is 1 mile east of B. Hiker A walks to east at a constant speed 2mi/h. Hiker B walks north at 4mi/h.

Step 1. Let x be the distance of A to starting point, y the distance of B to initial point. Draw a diagram.

Step 2. So the distance between A and B is \(z = \sqrt{y^2 + (x +1)^2}\).

Step 3. \(z' = \frac{1}{2\sqrt{y^2 + (x +1)^2}}( 2y y' + 2(x+1)x')\).

Step 4. After 1 hour, x = 2, y = 4, then z = 5. Therefore, \(z' = 4.4\) mi/h.

2. A plane flying horizontally at altitude 10 miles above a radar station, with constant speed at 400 mi/h. Find the rate of change of angle of the plane to the ground when it is 20 miles from the station.

Step 1. Let x be the horizontal distance traveled by the plane, \(\theta\) be angle of plane to the ground. Draw a diagram.

Step 2. So the relation between \(x\) and \(\theta\) is \(\tan(\theta) = \frac{3}{x}\).

Step 3. \(\sec^2(\theta) \theta' = -\frac{3}{x^2}x'\).

Step 4. When plane is 20 away from the station, \(x^2 = 20^2 - 10^2 = 300\), \(\sec(\theta) = \frac{1}{\cos(\theta)} = 2/\sqrt{3}\). Therefore, \(\theta' = -3\) radiance/hour.

Linear approximation

Linear approximation to \(f(x)\) near point a, \[L(x) = f(a) + f'(a) (x - a)\] that is using values at tangent line through \(a\) to approximate values of \(f(x)\).

Examples:

1. Function \(f(x) = (x+5)^{1/3}\).

\(f'(x) = \frac{1}{3}(x+5)^{-2/3}\). The linear approximation at a = 3, is \(L(x) = 2 + \frac{1}{12} (x - 3)\).

Use it to estimate \((8.06)^{1/3}\). When \(x = 3.06\), \(f(3.06) = (8.06)^{1/3}\). So \(L(3.06) = 2 + \frac{1}{12} (3.06 - 3) = 2.005\) as the estimate.

2. Function \(f(x) = \sqrt{2x^2 - 1}\). \(f'(x) = \frac{1}{2\sqrt{2x^2 - 1}} 4x\). The linear approximation at a = 1, is \(L(x) = 1 + 2 (x - 1)\).

Use it to estimate \(f(1.1)\). When \(x = 1.1\), \(L(1.1) = 1 + 2(1.1-1) = 1.2\) as the estimate.

Differentials

It is just a formal equation: \[ dy = f'(x) dx\]

Application to error estimates:

1. The absolute error is \(\Delta y = f(x + \Delta x) - f(x)\), roughly is equal to \(f'(x) \Delta x\) due to linear approximation.

2. The relative error is \(\frac{\Delta y}{y}\), the percentage error is \(\frac{\Delta y}{y} \cdot 100 \%\).

Examples:

1. A sphere was measured and its radius was found to be 45 inches with a possible error of no more that 0.01 inches. What is the maximum possible error in the volume if we use this value of the radius?

As \(V = \frac{4}{3} \pi r^3\), the differential is \(dV = 4 \pi r^2 dr\).

Now the error estimate is \(\Delta V \sim dV = 4 \pi (45)^2 (0.01) = 254.47\) in^3.

Part 4. Global Extrema and mean value theorem

Global Extrema on closed interval [a, b]

1. Local/relative extrema can’t occur at the endpoints.

2. So global/absolute extrema occurs either at endpoints or local extrema.

3. A list of possible local extrema is a list of possible critical points on \((a, b)\) by Fermat’s theorem.

Therefore, to find global extrema,

Step 1. Find all possible critical points falls inside \((a, b)\).

Step 2. Compare value at endpoints and the critical point found in step 1.

Step 3. Identify which is global maximum and which is global minimum.

Note that a critical point c by definition can be either \(f'(c) = 0\) or \(f'(c)\) DNE.

Examples:

1. Function \(f(x) = x^3 + x^2 - x + 7\). Find global maximum and minimum values of \(f(x)\) on \([-2, 0]\).

Step 1. Let \(f'(x) = 3x^2 + 2x - 1 = (3x - 1)(x+1) = 0\). \(x = -1, 1/3\). \(x = 1/3\) is not in \([-2, 0]\). So only one critical point \(x = -1\).

Step 2. Evaluate \(f(-2) = 5, f(0) = 7, f(-1) = 8\).

Step 3. The global maximum value is \(8\), global minimum value is 5.

2. Function \(f(x) = x \sqrt{25 - x^2}\). Find Find global maximum and minimum values of \(f(x)\) on \([-1, 5]\).

Step 1. Let \(f'(x) = x \frac{1}{2\sqrt{25 - x^2}}(-2x) + \sqrt{25 - x^2} = \frac{1}{\sqrt{25 - x^2}} (-x^2 + 25 - x^2)\). When \(x^2 = \frac{25}{2}\), \(x = \sqrt{\frac{5}{2}}, - \sqrt{\frac{5}{2}}\) are critical points such that \(f'(x) = 0\). When \(x^2 = 25\), \(x = 5, -5\) are critical points such that \(f'(x)\) DNE. As \(-5, 5, - \sqrt{\frac{5}{2}}\) are not in \((-1, 5)\). So only one critical point \(x = - \sqrt{\frac{5}{2}}\).

Step 2. Evaluate \(f(-1) = -2\sqrt{6}, f(5) = 0, f(\sqrt{\frac{5}{2}}) = \frac{25}{2}\).

Step 3. The global maximum value is \(\frac{25}{2}\), global minimum value is \(-2\sqrt{6}\).

Mean value theorem

If

1. \(f(x)\) is continuous on closed interval \([a,b]\),

2. \(f'(x)\) is differentiable on open interval \((a, b)\),

then there exists at least one such point \(c \in (a, b)\),

\[f'(c) = \frac{f(b) - f(a)}{b-a}\]

Note that Rolle’s theorem is just a special case when \(f(b) = f(a)\). The geometric interpretation is there exists at least one point such that its tangent line is paralell to the secant line connecting to the two endpoints.

Examples:

1. \(f(x) = (x-3)^{-2}\), closed interval \([2, 5]\). Can you find values of \(c\) in the theorem? If not, explain why?

As \(f'(x) = -2(x-3)^{-3}\), let \(f'(c) = -2(c-3)^{-3} = \frac{f(5) - f(2)}{5-2} = -1/4\), then \(c = 5\). Note that \(5\) is not in the open interval \((2, 5)\). So \(c\) doesn’t exist at all. The reason is \(f(x)\) is even not continuous at \(x = 3\), the condition 1 of the theorem fails.

2. \(f(x) = |x-1|\)

On the interval of \([0,1]\), \(f(x) = 1-x\). \(f'(x) = -1\) for any x. Also \(\frac{f(1)-f(0)}{1-0} = -1\). That means for any \(x \in (0, 1)\), it satisfies the theorem.

But \(\frac{f(2)-f(0)}{1-0} = 0\). There is no such point \(c \in (0, 2)\) such that \(f'(x) = 0\).

The reason is that the function \(f(x)\) is not differentiable at \(x = 1\). So the condition 2 of the theorem fails.

Applications:

1. If \(f(3) = 5\) and \(f'(x) \geq 2\) for \(3 \leq x \leq 5\), how small can \(f(5)\) possibly be?

Apply mean value theorem to \([3, 5]\): \(f'(c) = \frac{f(b) - f(a)}{b-a}\) or \(f(b) = f(a) + f'(c) (b-a)\). So \(f(5) = f(3) + f'(c) (5-3) = 5 + f'(c)\cdot 2 \geq 5 + 2\cdot 2 = 9\).

2. Show that an equation \(f(x) = L\) has exactly one root.

Step 1. Apply intermeidate value theorem on an interval \([a, b]\) to show there is at least one solution.

Step 2. Apply mean value theorem to prove by contradiction there can’t be two roots.

For example: show that \(8 x + \cos(x) = 0\) has exactly one root.

Step 1. Look at the interval \([-1, 1]\). As \(f(-1) < 0 < f(1)\), by intermediate value theorem, \(f(x) = 0\) has at least one solution on \([-1, 1]\).

Step 2. Suppose there are two roots, \(x_1 < x_2\). Apply mean value theorem to \([x_1, x_2]\). Then there exists such \(c \in (x_1, x_2)\) such that \(f'(c) =\frac{f(x_2) - f(x_1)}{x_2-x_1} = 0\). However, \(f'(x) = 8 - \sin(x)\) can’t be \(0\). The contradiction implies there is actually at most one solution.

Combining the above two arguments, the equation has exactly one solution.