Free math worksheets

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

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

- Identify the right triangle whose hypotenuse stays constant. - Differentiate the geometric relation with respect to time. - 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, \(\frac{d}{dt}(x+y)=4+\frac{8}{3}=\frac{20}{3}\,\text{ft/s}\).

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

- 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.