Free math worksheets

- What quantity does \(Q(t)\) measure, and what does \(t\) measure? - A derivative compares the output quantity with the input quantity, so combine their units as “output units per input unit.” - Use the sign of the derivative to decide whether the amount is increasing or decreasing.

1. Because \(Q\) is measured in gallons and \(t\) in minutes, \(Q'(t)\) is measured in gallons per minute. 2. The positive value \(Q'(6)=4.2\) means that exactly \(6\) minutes after the pump starts, the amount of water in the tank is increasing at an instantaneous rate of \(4.2\) gallons per minute.

At \(t=6\) minutes, the amount of water in the tank is increasing at an instantaneous rate of \(4.2\,\text{gal/min}\).

- A negative derivative describes the direction of change, not whether the original quantity is negative. - What does a zero derivative say about change at one instant? - Keep the units of energy divided by the units of time.

1. \(E'(3)=-7\,\text{MJ/h}\) means that at \(t=3\) hours, the stored energy is decreasing at an instantaneous rate of \(7\) megajoules per hour. 2. \(E'(8)=0\,\text{MJ/h}\) means that at \(t=8\) hours, the stored energy is momentarily not changing. 3. Neither derivative value tells the amount of energy stored at those times; each describes only its instantaneous rate of change.

At \(t=3\) hours, the stored energy is decreasing at \(7\,\text{MJ/h}\). At \(t=8\) hours, the stored energy is momentarily not changing. These statements describe rates, not the amount of stored energy.

- What geometric quantity is represented by \(x^2\)? What about \(x^3\)? - Differentiate each function with respect to \(x\). - For each derivative, compare the units of the output quantity with the units of \(x\).

1. By the power rule, \(A'(x)=2x\) and \(V'(x)=3x^2\). 2. \(A(x)=x^2\) is the area of a square with side length \(x\). Therefore, \(A'(x)=2x\) is the instantaneous rate of change of the square's area with respect to side length, measured in square units per unit of length. 3. \(V(x)=x^3\) is the volume of a cube with edge length \(x\). Therefore, \(V'(x)=3x^2\) is the instantaneous rate of change of the cube's volume with respect to edge length, measured in cubic units per unit of length.

1. \(A'(x)=2x\) and \(V'(x)=3x^2\) 2. \(A\) gives the area of a square, and \(A'\) gives its instantaneous area-change rate with respect to side length, in square units per unit of length. \(V\) gives the volume of a cube, and \(V'\) gives its instantaneous volume-change rate with respect to edge length, in cubic units per unit of length.

- Identify the variable of differentiation and treat the other letters as constants. - Which circle measurement is represented by \(2\pi r\)? - Think about how a cylinder's lateral area changes when only its height changes. - For the sphere, interpret the derivative as a rate for an extremely small change in radius.

1. For part a), treat \(r\) as constant: \(M'(h)=2\pi r\). This derivative equals the circumference of the circular base and gives the exact increase in lateral surface area per unit increase in height. 2. For part b), \(S'(r)=8\pi r\). This is the instantaneous rate of change of surface area with respect to radius, in square units per unit of radius. It is also four times the circumference \(2\pi r\) of a great circle.

a) \(M'(h)=2\pi r\); it is the cylinder base's circumference and the rate at which lateral area changes with height. b) \(S'(r)=8\pi r\); it is the instantaneous rate at which a sphere's surface area changes with radius.

- A smooth connection requires the two graphs to share a point and a slope. - Find the point on the hill at \(x=2\). - Use the derivative to find the hill's slope at that point. - Write the line using the point and slope.

1. Find the connection point: \(f(2)=-0.2(2)^2+2(2)+1=4.2\), so the point is \((2,4.2)\). 2. Differentiate: \(f'(x)=-0.4x+2\). 3. Find the slope at the connection point: \(f'(2)=-0.4(2)+2=1.2\). 4. A smooth connection requires the line to have the same point and slope. Using point-slope form, \(y-4.2=1.2(x-2)\). 5. Simplify: \(y=1.2x+1.8\).

\(y=1.2x+1.8\)

- Decide whether each statement describes a value at one time or a change over an interval. - Recall the difference between an instantaneous rate of change and an average rate of change. - Which notation represents the instantaneous rate of change? - Pay attention to whether the statement gives a height, a total change, or a rate.

1. The height at the planting time is represented by \(h(0)\), so \(h(0)=1.2\). 2. An instantaneous growth rate is represented by the derivative, so \(h'(10)=0.8\). 3. The total growth from year \(5\) to year \(10\) is the difference in heights, so \(h(10)-h(5)=3.5\). 4. Stopping growth after year \(150\) means the instantaneous growth rate remains zero, so \(h'(t)=0\) for all \(t\geq150\). 5. The average growth rate over \([0,20]\) is the difference quotient, so \(\frac{h(20)-h(0)}{20}=0.6\).

a) \(h(0)=1.2\) b) \(h'(10)=0.8\) c) \(h(10)-h(5)=3.5\) d) \(h'(t)=0\) for all \(t\geq150\) e) \(\frac{h(20)-h(0)}{20}=0.6\)

- Differentiate using the power rule. - Compare each derivative directly with the corresponding perimeter or surface-area formula. - Find the constant factor that multiplies the boundary measure to produce the derivative. - Check whether the factor is the same in both cases.

1. For the square, \(A'(x)=2x\). Since \(P(x)=4x\), \(A'(x)=\frac{1}{2}P(x)\). 2. For the cube, \(V'(x)=3x^2\). Since \(S(x)=6x^2\), \(V'(x)=\frac{1}{2}S(x)\). 3. In both cases, the ratio of the derivative to the boundary measure is \(\frac{1}{2}\).

a) \(A'(x)=2x=\frac{1}{2}P(x)\) b) \(V'(x)=3x^2=\frac{1}{2}S(x)\) c) The ratio is \(\frac{1}{2}\), or \(1:2\).

- Which derivative represents growth rate? - What must the derivative of a differentiable function equal at an interior maximum? - How does an increasing or decreasing slope relate to concavity? - Picture a graph that first becomes steeper and then begins to flatten.

1. The growth rate is \(V'\). Since \(V'\) changes from increasing to decreasing at \(t_w\), it has a local maximum there. Thus, the graph of \(V\) changes concavity at \(t_w\), making \((t_w, V(t_w))\) an inflection point. 2. At an interior extremum of \(V'\), Fermat's theorem gives \(V''(t_w)=0\). 3. Before \(t_w\), the growth rate increases, so \(V''(t)>0\) and the graph is concave up. After \(t_w\), the growth rate decreases, so \(V''(t)<0\) and the graph is concave down.

1. An inflection point 2. \(V''(t_w)=0\) 3. Concave up before \(t_w\); concave down after \(t_w\)

- Use the derivative column, not the amount column, to compare instantaneous rates. - For decreasing quantities, compare the negative derivative values by how negative they are. - Interpret \(D'(5)\) using milligrams per minute.

1. A negative derivative means dissolved oxygen is decreasing. Among the listed rates, \(-4\,\text{mg/min}\) has the greatest decrease magnitude, so the fastest listed decrease occurs at \(t=9\) minutes. 2. Since \(D'(5)=-1\,\text{mg/min}\), at \(t=5\) minutes the amount of dissolved oxygen is decreasing at an instantaneous rate of \(1\) milligram per minute.

a) \(t=9\) minutes b) At \(t=5\) minutes, the dissolved oxygen is decreasing at \(1\,\text{mg/min}\).

- The derivative at the marked time is represented by the slope of the tangent line, not the slope of the entire curve. - Choose two readable points on the tangent line and compute rise over run. - Translate the slope units into gallons per minute and use its sign in the interpretation.

1. Read two convenient grid points on the drawn tangent line. For example, the tangent passes through approximately \((1,25.5)\) and \((5,29.5)\). 2. Its slope is \(\frac{29.5-25.5}{5-1}=1\,\text{gal/min}\). 3. Therefore, \(Q'(3)\approx1\,\text{gal/min}\). At \(t=3\) minutes, the amount of liquid in the tank is increasing at about \(1\) gallon per minute.

\(Q'(3)\approx1\,\text{gal/min}\). At \(t=3\) minutes, the amount of liquid is increasing at about \(1\) gallon per minute.

- First identify the units of \(P'(t)\), then differentiate the units once more with respect to time. - Interpret the first and second derivatives separately before combining them. - A negative second derivative does not by itself mean that the original power output is decreasing.

1. Since \(P'(t)\) is measured in kilowatts per minute, \(P''(t)\) is measured in kilowatts per minute squared. 2. \(P''(5)=-3\,\text{kW/min}^2\) means that at \(t=5\) minutes, the rate at which power output is changing is itself decreasing by \(3\,\text{kW/min}\) per minute. 3. Because \(P'(5)=12\,\text{kW/min}>0\), the power output is increasing at that instant. Because \(P''(5)<0\), that positive increase is becoming slower.

\(P''(5)=-3\,\text{kW/min}^2\) means the power-change rate is decreasing by \(3\,\text{kW/min}\) per minute at \(t=5\). Since \(P'(5)=12\,\text{kW/min}>0\), the power output is increasing, but its rate of increase is decreasing.

- What equation identifies the point where the object reaches the ground? - The derivative at the impact point gives the tangent slope. - Use the magnitude of the slope when finding the acute angle with the horizontal. - Round the angle as requested.

1. Find where the shot put reaches the ground by solving \(h(x)=0\). The solutions are \(x=-2\) and \(x=10\), so the physically relevant impact point is \(x=10\). 2. Differentiate: \(h'(x)=-0.2x+0.8\). 3. Find the slope at impact: \(h'(10)=-0.2(10)+0.8=-1.2\). 4. The acute angle \(\alpha\) with the horizontal satisfies \(\tan(\alpha)=|-1.2|=1.2\). Thus, \(\alpha=\arctan(1.2)\approx 50.2^\circ\).

\(50.2^\circ\)

- What does the derivative tell you about the graph at a point? - For a smooth transition to the ground, which two quantities must match? - Use the relationship between slope and angle.

1. Differentiate: \(f'(x)=\frac{1}{4}x-1\). 2. At \(x=0\), the slope is \(f'(0)=-1\). The acute angle \(\alpha\) satisfies \(\tan(\alpha)=|-1|=1\), so \(\alpha=45^\circ\). 3. At \(x=4\), \(f(4)=\frac{1}{8}\cdot 4^2-4+2=0\), so the ramp reaches the ground. 4. Also, \(f'(4)=\frac{1}{4}\cdot 4-1=0\), which matches the slope of the horizontal line \(y=0\). Since both the height and slope agree, the ramp joins the ground smoothly at \((4,0)\).

a) \(45^\circ\) b) \(f(4)=0\) and \(f'(4)=0\), so the ramp and the ground line \(y=0\) have the same point and slope at \((4,0)\).

- A tangent continuation has the same value and slope as the curve at the transition point. - Differentiate the given function. - Use point-slope form for the tangent line. - Verify that the line passes through the point on the curve.

1. Find the transition point: \(g(2)=2^3-4\cdot 2^2+5\cdot 2=2\), so the point is \((2,2)\). 2. Differentiate: \(g'(x)=3x^2-8x+5\). 3. Find the slope at \(x=2\): \(g'(2)=3\cdot 2^2-8\cdot 2+5=1\). 4. Use point-slope form: \(y-2=1(x-2)\). 5. Simplify: \(y=x\).

\(y=x\)

- Find the point where the straight continuation begins. - A smooth connection requires the line to have the same slope as the curve. - Write the tangent line at \(x=10\). - Set the line's output equal to the target height.

1. Find the transition point: \(f(10)=0.01(10)^3-0.2(10)^2+1.5(10)+2=7\), so the point is \((10,7)\). 2. Differentiate: \(f'(x)=0.03x^2-0.4x+1.5\). 3. Find the slope at the transition: \(f'(10)=0.03(100)-0.4(10)+1.5=0.5\). 4. A smooth straight continuation is the tangent line: \(y-7=0.5(x-10)\), so \(y=0.5x+2\). 5. Set the height equal to \(12\): \(12=0.5x+2\). Solving gives \(x=20\).

\(x=20\,\text{m}\)

- A smooth transition means the linear model has the same value and slope at the joining time. - Find the temperature and rate of change at \(t=10\). - Write the linear model for times after \(10\) minutes. - Convert the final time to the requested unit.

1. Find the temperature at the transition: \(T(10)=0.1(10)^2-4(10)+80=50\). 2. Differentiate: \(T'(t)=0.2t-4\). The rate at \(t=10\) is \(T'(10)=-2\) degrees Celsius per minute. 3. A smooth linear continuation has the same value and rate at \(t=10\): \(L(t)-50=-2(t-10)\), so \(L(t)=-2t+70\). 4. Solve \(25=-2t+70\). This gives \(t=22.5\) minutes. 5. Convert to seconds: \(22.5\cdot 60=1350\) seconds.

\(1350\,\text{s}\)

- The derivative gives the local slope of the trail. - How are slope, percent grade, and angle of inclination related? - A tangent line requires both the point and the slope. - Use point-slope form for the tangent line.

1. Differentiate: \(h'(x)=-0.006x^2+0.12x\). 2. At \(x=5\), \(h'(5)=-0.006(25)+0.12(5)=0.45\). The percent grade is \(0.45\cdot 100\%=45\%\). 3. At \(x=15\), \(h'(15)=0.45\). Therefore, \(\alpha=\arctan(0.45)\approx 24.23^\circ\). 4. At \(x=25\), the slope is \(h'(25)=-0.75\), and \(h(25)=16.25\). The tangent line is \(y-16.25=-0.75(x-25)\), so \(y=-0.75x+35\).

a) \(45\%\) b) \(24.23^\circ\), approximately c) \(y=-0.75x+35\)

- A line perpendicular to the curve at a point is the normal line. - Use the derivative to find the curve's tangent slope. - Tangent and normal slopes are negative reciprocals. - Intersect the normal line with \(y=7\).

1. Differentiate: \(f'(x)=0.4x\). At \(x=5\), the tangent slope is \(f'(5)=2\). 2. The support follows the normal line, whose slope is \(-\frac{1}{2}\). 3. Through \(P(5,3)\), the normal line is \(y-3=-0.5(x-5)\), or \(y=-0.5x+5.5\). 4. Set \(y=7\): \(7=-0.5x+5.5\), so \(x=-3\). 5. Therefore, \(Q=(-3,7)\).

\(Q=(-3,7)\)

- If the first derivative is a rate, what does the derivative of that rate represent? - What does the sign of the second derivative tell you about how the first derivative changes? - What must be true of \(W'(t)\) while the water volume is increasing? - Examine the value of the first derivative at a time beyond the model's interval.

1. The first derivative \(W'(t)\) is the net rate of change of the water volume. Therefore, \(W''(t)\) is the rate at which that net inflow rate is changing, measured in \(\frac{\text{m}^3}{\text{min}^2}\). 2. Differentiate: \(W'(t) = -0.6t^2 + 6t + 20\) and \(W''(t) = -1.2t + 6\). Then \(W''(2)=3.6\,\frac{\text{m}^3}{\text{min}^2}\), so the net inflow rate is increasing at \(t=2\). Also, \(W''(8)=-3.6\,\frac{\text{m}^3}{\text{min}^2}\), so the net inflow rate is decreasing at \(t=8\). 3. Extending the model to \(t=15\) gives \(W'(15)=-0.6 \cdot 15^2 + 6 \cdot 15 + 20 = -25\,\frac{\text{m}^3}{\text{min}}\). This negative rate would mean the water volume is decreasing, so the extended model would no longer describe a continuously filling basin.

a) \(W''(t)\) is the rate of change of the basin's net inflow rate. b) \(W''(2)=3.6\,\frac{\text{m}^3}{\text{min}^2}\), so the net inflow rate is increasing. \(W''(8)=-3.6\,\frac{\text{m}^3}{\text{min}^2}\), so the net inflow rate is decreasing. c) \(W'(15)=-25\,\frac{\text{m}^3}{\text{min}}\), which would make the modeled volume decrease.

- Distinguish conditions that must be true from conditions that are only sufficient. - The growth rate is the first derivative. - At an interior local extremum of a differentiable function, its derivative is \(0\). - A negative second derivative can confirm a local maximum, but it is not required.

1. Situation (1) directly states the function value \(h(12)=80\), so card A must be true. 2. The growth rate is \(h'\). If \(h'\) has a local maximum at \(t=6\), Fermat's theorem applied to \(h'\) gives \(h''(6)=0\). The value \(h'(6)\) need not be positive, and \(h'''(6)<0\) is a sufficient but not necessary test for the maximum. Thus only card B must be true. 3. A local maximum height of \(160\,\text{cm}\) at \(t=25\) gives \(h(25)=160\). Since \(h\) is differentiable, Fermat's theorem gives \(h'(25)=0\). The condition \(h''(25)<0\) is sufficient but not necessary for a local maximum. Thus cards C and D must be true.

(1) A (2) B (3) C, D

- Distinguish conditions that must be true from conditions that are only sufficient. - The first derivative represents vertical velocity. - “Descending” determines the sign of the velocity. - At an interior local extremum of a differentiable function, its derivative is \(0\).

1. Descending at the greatest local rate means that the vertical velocity \(H'\) has a local minimum at \(t=10\) and is negative there. Fermat's theorem applied to \(H'\) gives \(H''(10)=0\), and descending gives \(H'(10)<0\). The condition \(H'''(10)>0\) is sufficient but not necessary for the local minimum. Thus cards D and E must be true. 2. A local maximum altitude of \(120\,\text{m}\) at \(t=30\) gives \(H(30)=120\). Since \(H\) is differentiable, Fermat's theorem gives \(H'(30)=0\). The condition \(H''(30)<0\) is sufficient but not necessary for a local maximum. Thus cards A and B must be true.

(1) D, E (2) A, B

- What quantity does the derivative compare with time in this model? - After differentiating, evaluate the same rate formula at each requested time and use its sign in your interpretation. - Average rate uses two function values over an interval, while instantaneous rate describes one time.

1. The derivative \(c'(t)\) gives the instantaneous rate of change of dye concentration with respect to time. Its units are \(\frac{\text{g}}{\text{L}\cdot\text{h}}\). 2. Differentiate using the quotient rule: \(c'(t)=\frac{10(t+4)^2-20t(t+4)}{(t+4)^4}=\frac{40-10t}{(t+4)^3}\). 3. Evaluate the derivative: \(c'(0)=\frac{5}{8}=0.625\), \(c'(4)=0\), and \(c'(8)=-\frac{5}{216}\approx -0.0231\), all in \(\frac{\text{g}}{\text{L}\cdot\text{h}}\). Thus the concentration is increasing initially, is momentarily unchanged at \(t=4\), and is decreasing at \(t=8\). 4. The average rate of change over \([0,8]\) is \(\frac{c(8)-c(0)}{8}=\frac{5}{72}\approx 0.0694\,\frac{\text{g}}{\text{L}\cdot\text{h}}\). 5. The average rate is positive because the concentration at \(t=8\) is greater than at \(t=0\), while \(c'(4)=0\) describes only the instantaneous change at \(t=4\).

a) \(c'(t)\) is the instantaneous rate of change of dye concentration with respect to time, measured in \(\frac{\text{g}}{\text{L}\cdot\text{h}}\). b) \(c'(0)=0.625\,\frac{\text{g}}{\text{L}\cdot\text{h}}\), \(c'(4)=0\,\frac{\text{g}}{\text{L}\cdot\text{h}}\), and \(c'(8)=-\frac{5}{216}\,\frac{\text{g}}{\text{L}\cdot\text{h}}\approx -0.0231\,\frac{\text{g}}{\text{L}\cdot\text{h}}\). These mean the concentration is increasing initially, momentarily unchanged at \(t=4\), and decreasing at \(t=8\). c) The average rate is \(\frac{5}{72}\,\frac{\text{g}}{\text{L}\cdot\text{h}}\approx 0.0694\,\frac{\text{g}}{\text{L}\cdot\text{h}}\). It is positive even though \(c'(4)=0\) because the two rates describe different kinds of change.

- Use symmetry to locate the right edge from the total width. - Convert a percent grade to a decimal slope. - The derivative at the edge gives a condition on \(a\). - Evaluate the model at the edge to find the depth.

1. Because the bowl is \(12\,\text{m}\) wide and symmetric, the right edge is at \(x=6\). 2. A grade of \(120\%\) corresponds to slope \(1.2\), so \(f'(6)=1.2\). 3. Since \(f'(x)=2ax\), the condition gives \(12a=1.2\), so \(a=0.1\). Thus \(f(x)=0.1x^2\). 4. The depth is the height of the edge above the origin: \(f(6)=0.1(6)^2=3.6\,\text{m}\).

a) \(f(x)=0.1x^2\) b) \(3.6\,\text{m}\)

- Differentiate a constant multiple of \(r^2\). - Substitute the fixed height into the lateral-area formula and compare expressions. - Expand the exact volume difference for radius \(r+\Delta r\) and identify its first-order term.

1. Differentiate: \(V'(r)=20\pi r\). 2. With \(h=10\), the lateral area is \(M(r)=2\pi r(10)=20\pi r\). Thus \(V'(r)=M(r)\). 3. If the radius increases by a small amount \(\Delta r\), then \(\Delta V=10\pi[(r+\Delta r)^2-r^2]=20\pi r\Delta r+10\pi(\Delta r)^2\). The first-order change is \(20\pi r\Delta r=M(r)\Delta r\); the quadratic term becomes negligible relative to \(\Delta r\) in the derivative limit.

a) \(V'(r)=20\pi r\) b) \(V'(r)=M(r)\) c) For a very small radial increase, the first-order volume change is the lateral surface area times the radial change.

- Is the function increasing or decreasing? - If a decline slows, does the slope become more negative or less negative? - Relate the sign of the second derivative to concavity. - What type of point separates intervals of different concavity?

1. A decline means the stock price is decreasing, so \(f'(t)<0\). 2. A slowing decline means the negative slope is becoming less negative. Therefore, the slope is increasing and \(f''(t)>0\). 3. Since \(f''(t)>0\), the graph is concave up during the slowing decline. A transition from \(f''<0\) to \(f''>0\) is an inflection point and corresponds to the most negative slope.

1. \(f'(t)<0\) 2. \(f''(t)>0\) 3. The graph is concave up. The transition point is an inflection point.

- Write the tangent-line equation at a general input \(x_0\). - Use the derivative for the tangent slope. - Substitute the sensor coordinates into the tangent-line equation. - Solve the resulting quadratic equation. - Check each solution against the stated domain.

1. Differentiate: \(f'(x)=-0.5x\). 2. The tangent line at \(x=x_0\) is \(t(x)=f'(x_0)(x-x_0)+f(x_0)\). Substitution and simplification give \(t(x)=-0.5x_0x+0.25x_0^2+4\). 3. Since the line must pass through \(S(4,1)\), set \(t(4)=1\): \(1=-2x_0+0.25x_0^2+4\). 4. Rearranging gives \(x_0^2-8x_0+12=0\), so \((x_0-2)(x_0-6)=0\). Thus, \(x_0=2\) or \(x_0=6\). 5. Only \(x_0=2\) lies in the rail's domain \([0,4]\).

The laser must be at \(x_0=2\).

- Find the vertex of the parabola. - Translate “stay at or above the roadway” into an inequality for the minimum value. - Differentiate to find the slope at the tower. - Use the tangent relationship between slope and angle.

1. The parabola's lowest point occurs halfway between the towers, at \(x=50\). Its height is \(f_a(50)=30-2500a\). 2. To stay at or above the roadway, the minimum height must satisfy \(30-2500a\ge0\). Therefore, \(a\le0.012\). Together with \(a>0\), this gives \(0<a\le0.012\). 3. If the minimum height is \(5\,\text{m}\), then \(30-2500a=5\), so \(a=0.01\). 4. For \(a=0.01\), \(f_a'(x)=0.02x-1\). At \(x=0\), the slope is \(-1\). The acute angle relative to the horizontal is \(\arctan(1)=45^\circ\).

a) \(0<a\le0.012\) b) \(a=0.01\) c) \(45^\circ\)

- Perpendicular lines have slopes that are negative reciprocals. - Use the derivative to find the walking path's tangent slope at \(A\). - Write the connector line through \(A\). - Set the connector and creek equations equal to find their intersection.

1. Differentiate: \(f'(x)=-x+2\). At \(x=3\), the tangent slope is \(f'(3)=-1\). 2. The perpendicular connector has slope \(1\). 3. Through \(A(3,3)\), its equation is \(y-3=x-3\), or \(y=x\). 4. Intersect this line with the creek: \(x=-2x+6\). Thus, \(3x=6\), so \(x=2\) and \(y=2\). 5. Therefore, the bridge location is \(B=(2,2)\).

The connector path is \(y=x\), and the bridge is at \(B=(2,2)\).

- First find the exact takeoff point on the ramp. - Use the derivative to find the ramp's slope at takeoff. - Model the initial path with the tangent line. - Substitute the wall's x-coordinate into the tangent-line equation.

1. Find the takeoff point: \(f(2)=-0.1\cdot2^3+0.6\cdot2^2=1.6\), so the point is \((2,1.6)\). 2. Differentiate: \(f'(x)=-0.3x^2+1.2x\). 3. Find the slope at takeoff: \(f'(2)=-0.3\cdot2^2+1.2\cdot2=1.2\). 4. The tangent line is \(y-1.6=1.2(x-2)\), or \(y=1.2x-0.8\). 5. At the wall, \(y=1.2\cdot6-0.8=6.4\). Therefore, \(h=6.4\,\text{m}\).

\(h=6.4\,\text{m}\)

- Use the midpoint coordinates in the cable model. - The derivative gives the slope of the cable at the tower. - Relate slope to angle using the tangent function. - For an upward-opening parabola, the minimum occurs at the vertex.

1. At the midpoint, \(x=0\). Set the height equal to \(30\): \(50-2500k=30\). Thus, \(k=\frac{20}{2500}=\frac1{125}\). 2. The right tower is at \(x=50\). Differentiate: \(f_k'(x)=2kx\). For \(k=\frac1{125}\), the slope at the tower is \(f_k'(50)=\frac45\). 3. If \(\theta\) is the angle of inclination, then \(\tan(\theta)=\frac45\), so \(\theta=\arctan\left(\frac45\right)\approx38.66^\circ\). 4. Because \(k>0\), the lowest point is at \(x=0\). The cable touches or passes below the roadway when \(50-2500k\le0\). Therefore, \(k\ge\frac1{50}\). It first touches the roadway when \(k=\frac1{50}\).

a) \(k=\frac1{125}\) b) \(\theta\approx38.66^\circ\) c) \(k\ge\frac1{50}\); it first touches the roadway at \(k=\frac1{50}\).

- Landing at the same height means the function value is zero at the basin. - Use the vertex of the downward-opening parabola. - Convert slope to angle with the inverse tangent.

1. The basin condition is \(h_c(4)=0\): \(-0.5(4)^2+4c=0\). Thus \(-8+4c=0\), so \(c=2\). 2. For \(c=2.5\), the vertex occurs at \(x=2.5\). The maximum height is \(h_{2.5}(2.5)=3.125\,\text{m}\). 3. The initial slope is \(h_c'(0)=c\). For \(c=1.5\), \(\tan(\alpha)=1.5\), so \(\alpha=\arctan(1.5)\approx56.31^\circ\).

a) \(c=2\) b) \(3.125\,\text{m}\) c) \(56.31^\circ\)

- The rate of change is zero at local maxima and minima of the temperature graph. - Determine when the temperature is increasing or decreasing. - Match the candidate zeros and signs to the temperature graph. - Use the steepest parts of the temperature graph as an additional check.

1. The temperature graph has a local maximum at \(t=3\) and a local minimum at \(t=7\). 2. The rate-of-change graph must have zeros at those times because the temperature graph has horizontal tangent lines there. 3. Graphs 1 and 2 both have zeros at \(t=3\) and \(t=7\). Graph 3 has zeros at \(t=1\) and \(t=5\), and graph 4 has no zeros on the displayed interval. Thus, only graphs 1 and 2 remain possible. 4. The temperature decreases between \(t=3\) and \(t=7\), so its rate of change must be negative there. Graph 1 is negative on this interval, while graph 2 is positive. Therefore, graph 1 represents the rate of change.

Graph 1 represents the instantaneous rate of change of temperature.

- The greatest rate of increase occurs where the graph is steepest. - Interpret a horizontal asymptote as a limiting value in context. - The derivative records the slope of the original graph. - Compare slopes before and after the inflection point.

1. The user count increases most rapidly where the graph of \(N\) has its greatest slope, at the inflection point of the S-shaped curve. This occurs at \(t=5\) months. 2. Line \(g\) is the horizontal asymptote \(N=12\). In context, it represents a long-term saturation level of \(12\) million users. 3. The derivative \(N^{\prime}\) gives the instantaneous growth rate in millions of users per month. Its graph is bell-shaped, stays positive, and has its maximum at \(t=5\). Thus, the growth rate increases until month \(5\) and then decreases as the user count approaches saturation.

a) At \(t=5\) months b) Line \(g\) represents a long-term saturation level of \(12\) million users. c) The graph of \(N^{\prime}\) is a positive bell-shaped curve with a maximum at \(t=5\). It represents the instantaneous user-growth rate.

- A fraction with a fixed positive numerator is largest when its denominator is smallest. - Substitute \(t=12\) into the model. - At an inflection point, the rate of change reaches a local extremum.

1. The numerator is constant, so \(k(t)\) is greatest when the denominator is smallest. The denominator \((t-10)^2+12\) is minimized at \(t=10\). 2. The maximum concentration is \(k(10)=\frac{80}{12}=\frac{20}{3}\approx6.67\,\text{mg/L}\). 3. At the inflection point, \(k(12)=\frac{80}{(12-10)^2+12}=\frac{80}{16}=5\,\text{mg/L}\). 4. On the decreasing side of the curve, the inflection point is where \(k'(t)\) is most negative. Thus, at \(t=12\), the medication concentration is decreasing at its fastest rate.

a) Maximum at \(t=10\) hours: \(\frac{20}{3}\approx6.67\,\text{mg/L}\) b) \(k(12)=5\,\text{mg/L}\). At that time, the concentration is decreasing at its fastest rate.

- Interpret a function value as height and a derivative value as slope. - Use vertex form because the highest point is given. - The water reaches the surface at a zero of the function. - The tangent of the launch angle equals the slope at the nozzle.

1. The condition \(f(0)=2\) means the water leaves the nozzle \(2\,\text{ft}\) above the surface. The condition \(f'(4)=0\) means the path has a horizontal tangent at a horizontal distance of \(4\,\text{ft}\), where it reaches its maximum height. 2. Using vertex form with vertex \((4, 10)\), write \(f(x)=a(x-4)^2+10\). Since \(f(0)=2\), \(16a+10=2\), so \(a=-\frac{1}{2}\). Thus, \(f(x)=-\frac{1}{2}(x-4)^2+10=-\frac{1}{2}x^2+4x+2\). 3. Set \(f(x)=0\): \(-\frac{1}{2}x^2+4x+2=0\), or \(x^2-8x-4=0\). The positive solution is \(x=4+2\sqrt{5}\approx8.47\). The stream hits the water about \(8.47\,\text{ft}\) from the nozzle. 4. The initial slope is \(f'(0)\). Since \(f'(x)=-x+4\), \(f'(0)=4\). Therefore, \(\tan(\theta)=4\), so \(\theta=\arctan(4)\approx76.0^\circ\).

a) \(f(0)=2\) gives the nozzle's height, and \(f'(4)=0\) identifies the horizontal tangent at the maximum height. b) \(f(x)=-\frac{1}{2}x^2+4x+2\) c) \(4+2\sqrt{5}\,\text{ft}\approx8.47\,\text{ft}\) d) \(\theta\approx76.0^\circ\)

- Decide first whether \(R'(4)\) is a quantity value or a rate value. - Check which statement uses units of “thousand gallons per hour.” - Distinguish a derivative at one instant from a difference quotient over an interval.

1. Interpretation B is correct. The derivative value \(R'(4)=-6\) describes the instantaneous rate of change at \(t=4\), with units of thousand gallons per hour. 2. Interpretation A is incorrect because \(R'(4)\) is a rate, not the reservoir volume \(R(4)\). 3. Interpretation C is incorrect because \(R'(4)\) is an instantaneous rate at one time, not an average rate over the interval \([0,4]\).

B is correct. \(R'(4)=-6\) means that at \(t=4\) hours, the reservoir's volume is decreasing at \(6\) thousand gallons per hour. A confuses \(R'\) with \(R\), and C confuses an instantaneous rate with an average rate.

- The upper edges lie on the x-axis, so use their coordinates in the model. - A \(50\%\) grade corresponds to derivative value \(0.5\). - The parabola's lowest point occurs at its vertex. - Report depth as a positive distance.

1. The top edges are at \(x=\pm10\), and \(f(10)=0\). 2. Since \(f'(x)=2ax\) and \(f'(10)=0.8\), \(20a=0.8\), so \(a=0.04\). 3. Use \(f(10)=0\): \(0.04(10)^2+c=0\), giving \(c=-4\). Thus \(f(x)=0.04x^2-4\). 4. The lowest point is \(f(0)=-4\), so the maximum depth is \(4\,\text{m}\). 5. A \(50\%\) slope is \(0.5\). Solve \(f'(x)=0.08x=0.5\), giving \(x=6.25\).

a) \(f(x)=0.04x^2-4\) b) \(4\,\text{m}\) c) \(6.25\,\text{m}\) from the center

- A tangential pulse follows the tangent line at the release point. - Use the derivative to find the tangent slope at each candidate input. - Write each tangent line using its release point and slope. - Evaluate each tangent line at the sensor's x-coordinate.

1. Differentiate: \(g'(x)=0.5\cos(0.5x)\). 2. At \(x_1=\frac{3\pi}{2}\approx4.712\), \(g(x_1)\approx2.707\) and \(g'(x_1)\approx-0.354\). The tangent gives \(y\approx-0.354(10-4.712)+2.707\approx0.835\), which does not match the sensor height. 3. At \(x_2=2\pi\approx6.283\), \(g(x_2)\approx2.000\) and \(g'(x_2)\approx-0.500\). The tangent gives \(y\approx-0.500(10-6.283)+2.000\approx0.142\), which matches the sensor height. 4. At \(x_3=\frac{5\pi}{2}\approx7.854\), \(g(x_3)\approx1.293\) and \(g'(x_3)\approx-0.354\). The tangent gives \(y\approx-0.354(10-7.854)+1.293\approx0.533\), which does not match the sensor height. 5. Therefore, the pulse should be released at \(x_2=2\pi\).

\(x_2=2\pi\)

- A point on the path gives a condition on the function value. - At the highest point, the tangent is horizontal. - Start with a general quadratic and solve for its coefficients. - The tangent slope and the angle with the horizontal are related by tangent.

1. The takeoff condition is \(g(0)=0\). A horizontal tangent at the highest point gives \(g'(6)=0\). The remaining given point gives \(g(10)=5\). 2. Let \(g(x)=ax^2+bx+c\). From \(g(0)=0\), \(c=0\). Since \(g'(x)=2ax+b\), the condition \(g'(6)=0\) gives \(12a+b=0\), so \(b=-12a\). Using \(g(10)=5\), \(100a+10b=5\). Substitution gives \(-20a=5\), so \(a=-\frac{1}{4}\) and \(b=3\). Therefore, \(g(x)=-\frac{1}{4}x^2+3x\). 3. The maximum occurs at \(x=6\): \(g(6)=-\frac{1}{4}(36)+18=9\). The maximum height is \(9\,\text{m}\). 4. The slope at landing is \(g'(12)=-\frac{1}{2}(12)+3=-3\). If \(\theta\) is the magnitude of the angle with the horizontal ground, then \(\tan(\theta)=3\). Thus \(\theta=\arctan(3)\approx71.6^\circ\).

a) \(g(0)=0\) and \(g'(6)=0\). b) \(g(x)=-\frac{1}{4}x^2+3x\). c) \(9\,\text{m}\). d) \(\arctan(3)\approx71.6^\circ\).