Model and interpret a contextual rate of change

Given

\(\displaystyle Y(t)=100-3t+\tfrac12t^2,\quad t=2\)

Find the output mass at two minutes

Compare the two columns at every time. Notice that positive mass can have a negative or zero rate

When is the instantaneous rate zero?

Differentiate the quantity, then interpret the units

For mass Y grams and time t minutes, the derivative has units grams per minute. Differentiate before substituting the time

1. 1Name both quantities A rate describes change in one quantity with respect to another. Its units are dependent units per independent unit. A rate with respect to length is not automatically a rate with respect to time
2. 2Differentiate before substitution Form the derivative rule first, then substitute the stated input. The original quantity value is not its rate

Given

\(\displaystyle Y^{\prime}(t)=-3+t,\quad t=2\)

Find the signed instantaneous mass rate at two minutes

At two minutes, the output mass is 96 grams and its instantaneous rate is −1 gram per minute. What does the negative rate mean?

Area can change with respect to a length

A rectangle has length x centimetres and the width shown above. Its area and derivative are shown above. The derivative measures square centimetres of area per centimetre of length, not per second

1. 1Name both quantities A rate describes change in one quantity with respect to another. Its units are dependent units per independent unit. A rate with respect to length is not automatically a rate with respect to time
2. 2Model before differentiating Use each geometric or proportional relation to express the quantity in the named independent variable. Keep conversion factors such as 1000 cm³ = 1 litre in the model

Given

\(\displaystyle A^{\prime}(x)=3x,\quad x=4\)

Find the area rate at x = 4 cm

A fixed-base tank gives a linear depth model

A rectangular tank has base dimensions 6 cm by 4 cm and water depth h cm. Its volume is 24h cubic centimetres. Changing depth by a small amount changes volume at a constant rate

1. 1Model before differentiating Use each geometric or proportional relation to express the quantity in the named independent variable. Keep conversion factors such as 1000 cm³ = 1 litre in the model
2. 2Differentiate before substitution Form the derivative rule first, then substitute the stated input. The original quantity value is not its rate

Given

\(\displaystyle V(h)=24h\)

Find dV/dh at h = 3 cm

Similarity removes the cone radius

A cone has full radius 3 cm and height 6 cm. At depth h, similar triangles give the radius shown above. Substitute it into the volume formula before differentiating

1. 1Model before differentiating Use each geometric or proportional relation to express the quantity in the named independent variable. Keep conversion factors such as 1000 cm³ = 1 litre in the model
2. 2Keep exact geometric coefficients For pi-bearing models, enter the rational coefficient of pi only when that is explicitly requested. Do not substitute a decimal approximation for pi

Given

\(\displaystyle V^{\prime}(4)=\frac{\pi\cdot4^2}{4}\)

At h = 4, enter only the rational coefficient of π in dV/dh

A proportional cylinder is a different model

If a family of cylinders has height 2r, its volume is 2πr³. Both radius and height vary under this relation; treating height as constant would give the wrong derivative

1. 1Model before differentiating Use each geometric or proportional relation to express the quantity in the named independent variable. Keep conversion factors such as 1000 cm³ = 1 litre in the model
2. 2Keep exact geometric coefficients For pi-bearing models, enter the rational coefficient of pi only when that is explicitly requested. Do not substitute a decimal approximation for pi

Given

\(\displaystyle V^{\prime}(r)=6\pi r^2,\quad r=3\)

At r = 3, enter only the coefficient of π in the volume rate

Convert capacity inside the model

For a cube with internal side s centimetres, volume is s³ cubic centimetres. Divide by 1000 to obtain capacity C in litres, then differentiate with respect to s.

1. 1Name both quantities A rate describes change in one quantity with respect to another. Its units are dependent units per independent unit. A rate with respect to length is not automatically a rate with respect to time
2. 2Model before differentiating Use each geometric or proportional relation to express the quantity in the named independent variable. Keep conversion factors such as 1000 cm³ = 1 litre in the model

Given

\(\displaystyle C^{\prime}(20)=\frac{3\cdot20^2}{1000}\)

Find the capacity rate at side length 20 cm

Shadow similarity specifies the independent distance

For a 6 m lamp and a 1.5 m person, x is metres walked from the lamp and s is shadow length. Similar triangles give the first equation above; rearrange it to express shadow length in terms of distance walked

1. 1Name both quantities A rate describes change in one quantity with respect to another. Its units are dependent units per independent unit. A rate with respect to length is not automatically a rate with respect to time
2. 2Model before differentiating Use each geometric or proportional relation to express the quantity in the named independent variable. Keep conversion factors such as 1000 cm³ = 1 litre in the model

Given

\(\displaystyle s=x/3\)

Find the change in shadow length per metre walked

Shadow length s is one third of distance x walked from the lamp, both in metres. Can its rate with respect to time in metres per second be concluded from this relation alone?

A zero rate need not mean zero mass

The model Y(t) = 90 − 2(t−3)² has derivative −4(t−3). At t = 3, its mass is 90 grams while its instantaneous rate is zero. The function still changes at nearby times

1. 1Zero rate is local information Zero instantaneous rate does not mean zero quantity or a constant quantity throughout the interval

Given

\(\displaystyle Y(t)=90-2(t-3)^2,\quad t=3\)

Y is mass in grams and t is time in minutes. Find the instantaneous rate at the turning input

Which conclusion is supported by Y′(3) = 0?

Recover a stated quadratic from a table

Suppose Y is quadratic and its values at inputs zero, one and two are 80, 83 and 88. Substitute these values into the quadratic form above to find its coefficients, then differentiate

1. 1Use enough model information Three exact observations at distinct inputs determine a quadratic only when the model is specified to be quadratic. They do not identify an arbitrary function
2. 2Differentiate before substitution Form the derivative rule first, then substitute the stated input. The original quantity value is not its rate

Given

\(\displaystyle Y^{\prime}(t)=2t+2,\quad t=3\)

Y is mass in grams and t is time in minutes. Evaluate the recovered model’s rate at t = 3.