- Name the two quantities that change over time. - A derivative's units come from the output quantity divided by the input's units.

1. The side-length rate is \(s'(t)\), in \(\text{cm/s}\). 2. The area rate is \(A'(t)\), in \(\text{cm}^2/\text{s}\). These are different rates even though the area depends on the side length.

Side length: \(s'(t)\) in \(\text{cm/s}\). Area: \(A'(t)\) in \(\text{cm}^2/\text{s}\).

- If the radius increases a little, what happens to the circumference? - Which part of the circumference formula changes as time passes?

1. Differentiate the relation with respect to time: \(C'(t)=2\pi r'(t)\). 2. Substitute the radial rate \(r'(t)=0.3\,\text{cm/s}\). Thus \(C'(t)=0.6\pi\,\text{cm/s}\).

a) \(C'(t)=2\pi r'(t)\). b) \(C'(t)=0.6\pi\,\text{cm/s}\), increasing.

- Is the side length fixed for all time or only known at one instant? - Consider what happens when a function inside a square depends on time.

1. The value \(s=5\) describes only one instant; the side length is not constant over time. 2. Differentiate the changing relation first: \(A'(t)=2s(t)s'(t)\). The value \(s=5\) can then be used, giving \(A'=10s'\) at that instant.

The student treated a changing side length as a constant. The correct relation is \(A'(t)=2s(t)s'(t)\), or \(A'=10s'\) at the stated instant.

- Treat the radius as a quantity that changes with time. - Differentiate the area relation with respect to time before substituting the instant's values. - Check that an area rate has square-units per unit time.

1. Differentiate \(A=\pi r^2\) with respect to time: \(\frac{dA}{dt}=2\pi r\frac{dr}{dt}\). 2. Substitute \(r=3\) and \(\frac{dr}{dt}=2\): \(\frac{dA}{dt}=2\pi(3)(2)=12\pi\). 3. The area is increasing at \(12\pi\,\text{cm}^2/\text{s}\).

\(12\pi\,\text{cm}^2/\text{s}\)

- Differentiate the area equation with respect to time. - The given rate is the area rate, while the requested rate is the side-length rate. - Substitute the side length only after forming the rate equation.

1. Differentiate \(A=s^2\) with respect to time: \(\frac{dA}{dt}=2s\frac{ds}{dt}\). 2. Substitute \(s=6\) and \(\frac{dA}{dt}=9\): \(9=12\frac{ds}{dt}\). 3. Therefore, \(\frac{ds}{dt}=\frac{3}{4}\,\text{ft/min}\).

\(\frac{3}{4}\,\text{ft/min}\)

- Radius and volume are both functions of time in this situation. - Differentiate the volume formula before inserting the instantaneous measurements. - Use cubic-units per unit time for a volume rate.

1. Differentiate the volume relation with respect to time: \(\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}\). 2. Substitute \(r=5\) and \(\frac{dr}{dt}=0.4\): \(\frac{dV}{dt}=4\pi(25)(0.4)=40\pi\). 3. The volume is increasing at \(40\pi\,\text{cm}^3/\text{s}\).

\(40\pi\,\text{cm}^3/\text{s}\)

- In the area formula, are both factors changing with time? - Differentiate the entire relation with respect to the common time variable before substituting snapshot values. - Your final rate equation should account separately for changing length and changing width.

1. Jordan's equation treats the derivative of a product as the product of the derivatives. Both \(L\) and \(W\) vary with time, so differentiating \(A=LW\) requires one term for the change in each factor. 2. The correct rate equation is \(\frac{dA}{dt}=W\frac{dL}{dt}+L\frac{dW}{dt}\). 3. At the stated instant, \(\frac{dA}{dt}=5(2)+8(-1)\,\text{m}^2/\text{min}\).

Jordan incorrectly multiplies the two derivative rates. The correct differentiated relation is \(\frac{dA}{dt}=W\frac{dL}{dt}+L\frac{dW}{dt}\), so at the stated instant \(\frac{dA}{dt}=5(2)+8(-1)\,\text{m}^2/\text{min}\).

- Express area in terms of radius. - Use a negative rate for a decreasing area. - Differentiate before substituting the given radius.

1. Use the area relation \(A=\pi r^2\), where both quantities change with time. 2. Differentiate with respect to time: \(A'=2\pi r r'\). 3. Substitute \(A'=-12\pi\) and \(r=3\): \(-12\pi=6\pi r'\), so \(r'=-2\,\text{cm/min}\).

The radius is shrinking at \(2\,\text{cm/min}\): \(r'=-2\,\text{cm/min}\).

- Both side lengths change, so neither is a constant during differentiation. - Keep the width-change rate negative. - Interpret the final rate's sign rather than the sign of one side's rate alone.

1. Write \(A(t)=L(t)W(t)\) and differentiate with respect to time: \(A'=L'W+LW'\). 2. Substitute both rates: \(A'=(1)(5)+(8)(-0.4)=5-3.2=1.8\,\text{cm}^2/\text{s}\). 3. Since the result is positive, the sheet's area is increasing even though its width is decreasing.

\(A'=1.8\,\text{cm}^2/\text{s}\); the area is growing.

- The small water triangle and the full triangular end have the same shape. - Compare width to depth in the two similar triangles. - Both the water's width and depth change with time.

1. The triangles at the water line and at the rim are similar, so \(\frac{w}{h}=\frac{6}{4}\) and \(w=\frac{3}{2}h\). 2. Differentiate with respect to time: \(w'=\frac{3}{2}h'\). 3. With \(h'=0.5\,\text{ft/min}\), \(w'=\frac{3}{2}\cdot0.5=0.75\,\text{ft/min}\). The surface width is growing.

\(w'=0.75\,\text{ft/min}\), increasing.

- Differentiate the sphere's volume with respect to time. - In the resulting expression, identify which factor changes between the two instants. - Compare the two rates as a ratio rather than finding their actual values.

1. Sphere volume is \(V=\frac{4}{3}\pi r^3\), so \(V'=4\pi r^2r'\). 2. Since \(r'\) is unchanged, the volume-change rate is proportional to \(r^2\). 3. Doubling the radius multiplies its square by \(4\), so the second volume-increase rate is four times the first.

Four times as fast, because \(V'=4\pi r^2r'\) and the radius doubles while \(r'\) stays the same.

- Identify the right triangle whose hypotenuse stays constant. - Use the stated horizontal distance to determine the vertical height at the requested instant. - Differentiate the geometric relation with respect to time before substituting the rates. - The sign of the vertical rate determines whether the top is moving up or down.

1. Let \(x\) be the bottom's distance from the wall and \(y\) the top's height. The fixed ladder length gives \(x^2+y^2=169\). 2. At \(x=5\), \(y=\sqrt{169-25}=12\). 3. Differentiate with respect to time: \(2x\frac{dx}{dt}+2y\frac{dy}{dt}=0\). 4. Substitute \(x=5\), \(y=12\), and \(\frac{dx}{dt}=2\): \(10(2)+24\frac{dy}{dt}=0\). 5. Thus, \(\frac{dy}{dt}=-\frac{5}{6}\,\text{ft/s}\). The negative sign means the top is moving downward.

The top moves downward at \(\frac{5}{6}\,\text{ft/s}\).

- The water surface radius and water depth are linked by similar triangles. - Express the volume using only one changing length before differentiating. - A volume rate should have cubic-units per unit time.

1. Let \(h\) be the water depth and \(r\) the water-surface radius. Similar triangles give \(\frac{r}{h}=\frac{4}{12}=\frac{1}{3}\), so \(r=\frac{h}{3}\). 2. Substitute into \(V=\frac{1}{3}\pi r^2h\): \(V=\frac{\pi}{27}h^3\). 3. Differentiate with respect to time: \(\frac{dV}{dt}=\frac{\pi}{9}h^2\frac{dh}{dt}\). 4. At \(h=6\) and \(\frac{dh}{dt}=0.5\), \(\frac{dV}{dt}=\frac{\pi}{9}(36)(0.5)=2\pi\). 5. The volume is increasing at \(2\pi\,\text{ft}^3/\text{min}\).

\(2\pi\,\text{ft}^3/\text{min}\)

- Identify the two similar triangles formed by the light ray, the ground, and the vertical heights. - Distinguish the person's distance from the lamp from the shadow's length. - The shadow tip's distance is the sum of those two changing lengths.

1. Let \(x\) be the distance from the lamp to the person and \(y\) the shadow length. The distance from the lamp to the shadow tip is \(x+y\). 2. Similar triangles give \(\frac{15}{x+y}=\frac{6}{y}\). Thus, \(15y=6x+6y\), so \(y=\frac{2}{3}x\). 3. Differentiate with respect to time: \(\frac{dy}{dt}=\frac{2}{3}\frac{dx}{dt}\). 4. Since \(\frac{dx}{dt}=4\), \(\frac{dy}{dt}=\frac{8}{3}\,\text{ft/s}\). 5. Therefore, \(\frac{d}{dt}(x+y)=4+\frac{8}{3}=\frac{20}{3}\,\text{ft/s}\).

\(\frac{20}{3}\,\text{ft/s}\)

- Connect the steepness of the height graph with the speed at which the water rises. - Think about how much volume a thin horizontal layer of water occupies. - Convert milliliters to cubic centimeters and use the height rate at the requested time.

1. The height graph is steepest at \(t=6\,\text{s}\), when the level rises fastest. 2. During a short time interval, the added volume is approximately horizontal cross-sectional area times height gained. Thus \(V'(t)=A(h(t))h'(t)\). Since \(V'(t)=10\,\text{cm}^3/\text{s}\), a greater slope \(h'(t)\) means a smaller horizontal area. That area decreases toward the middle height and increases again higher up. The graph does not determine the container's exact shape or width. 3. Differentiate the given height model: \(h'(t)=\frac{12e^{-0.8(t-6)}}{(1+e^{-0.8(t-6)})^2}\). At \(t=6\), \(h'(6)=3\,\text{cm/s}\), so \(A(h(6))=\frac{10}{3}\approx3.33\,\text{cm}^2\).

a) At \(t=6\,\text{s}\). b) Horizontal cross-sectional area decreases toward the middle height and then increases. The relation is \(V'(t)=A(h(t))h'(t)\). c) Approximately \(3.33\,\text{cm}^2\).

- Use a right triangle whose legs are the signed distances from the intersection. - Pay attention to the sign of the north-south rate. - Differentiate the distance relation before substituting the instantaneous values.

1. Let \(x\) and \(y\) be the cyclists' signed distances east and north of the intersection, and let \(d\) be the distance between them. Then \(d^2=x^2+y^2\). 2. At the instant, \(x=3\), \(y=4\), so \(d=5\). The rates are \(\frac{dx}{dt}=20\) and \(\frac{dy}{dt}=-15\), because cyclist \(B\) is moving south. 3. Differentiate: \(2d\frac{dd}{dt}=2x\frac{dx}{dt}+2y\frac{dy}{dt}\). 4. Substitute the instantaneous values: \(5\frac{dd}{dt}=3(20)+4(-15)=0\). 5. Therefore, \(\frac{dd}{dt}=0\,\text{mph}\). At this instant, the cyclists' separation is momentarily unchanged.

\(0\,\text{mph}\); their separation is momentarily unchanged.

- Relate the angle of elevation to the fixed horizontal distance and changing height. - Differentiate the trigonometric relation with respect to time. - Express any trigonometric factor using the triangle's instantaneous side lengths.

1. Let \(y\) be the drone's height. The right triangle gives \(\tan\theta=\frac{y}{50}\). 2. Differentiate with respect to time: \(\sec^2\theta\frac{d\theta}{dt}=\frac{1}{50}\frac{dy}{dt}\). 3. Since \(\sec^2\theta=1+\tan^2\theta=1+\left(\frac{120}{50}\right)^2=\frac{169}{25}\), substitute \(\frac{dy}{dt}=12\). 4. Thus, \(\frac{d\theta}{dt}=\frac{6}{169}\,\text{rad/s}\approx0.0355\,\text{rad/s}\).

\(\frac{6}{169}\,\text{rad/s}\approx0.0355\,\text{rad/s}\)

- Write separate equations for area and perimeter before differentiating. - Each differentiated relation gives one equation involving the two unknown rates. - Interpret a negative rate as a decreasing dimension.

1. From \(A=LW\), differentiation gives \(\frac{dA}{dt}=W\frac{dL}{dt}+L\frac{dW}{dt}\). At the instant, \(6\frac{dL}{dt}+10\frac{dW}{dt}=14\). 2. From \(P=2L+2W\), differentiation gives \(\frac{dP}{dt}=2\frac{dL}{dt}+2\frac{dW}{dt}\). Thus, \(\frac{dL}{dt}+\frac{dW}{dt}=\frac{1}{2}\). 3. Solving the two rate equations gives \(\frac{dL}{dt}=-\frac{9}{4}\,\text{m/min}\) and \(\frac{dW}{dt}=\frac{11}{4}\,\text{m/min}\). 4. The length is decreasing while the width is increasing; together these changes produce the stated area and perimeter rates.

\(\frac{dL}{dt}=-\frac{9}{4}\,\text{m/min}\) and \(\frac{dW}{dt}=\frac{11}{4}\,\text{m/min}\). The length is shrinking and the width is growing.