The volume \(V\) of liquid in a container is measured in gallons, and time \(t\) is measured in minutes. Average rates \(\frac{\Delta V}{\Delta t}\) over intervals near \(t=6\) approach \(1.8\). Which unit is appropriate for the instantaneous rate of change of volume at \(t=6\): gallons, minutes per gallon, gallons per minute, or gallons per square minute? Explain from the average-rate expression.

A cyclist's average speeds on intervals of width \(0.1\) and \(0.01\) seconds ending at \(t=5\) seconds are \(5.9\) and \(5.99\) feet per second. On intervals of the same widths starting at \(t=5\), they are \(6.1\) and \(6.01\) feet per second. Estimate the cyclist's speed at \(t=5\). Do these four averages prove the exact speed?

A camera lift's height is represented by the curve. At \(t=1\), is its instantaneous vertical rate positive, negative, or zero? Explain how nearby heights help you decide.

For \(f(x)=|x|\), consider the average rate of change from \(x=0\) to \(x=h\), where \(h\ne0\). For positive \(h\), the average rate is \(1\). For negative \(h\), the average rate is \(-1\). Does a single instantaneous rate of change at \(x=0\) follow from these shrinking intervals? Explain.

Average rates from intervals immediately to the left of \(t=7\) are \(4.7,\ 4.97,\ 4.997,\ 4.9997\). No information is given from the right of \(t=7\). Anouk concludes that the instantaneous rate at \(t=7\) must be \(5\). Is that guaranteed? Explain what the data suggest and what they cannot prove.

A cart's average velocities over intervals shrinking toward \(t=6\) seconds approach \(-4\,\text{ft/s}\). a) What is the cart's instantaneous velocity at \(t=6\)? b) What is its instantaneous speed at \(t=6\)? c) Explain why the sign appears in one answer but not the other.

A particle has position \(s(t)=t^2\) feet, where \(t\) is measured in seconds. For each \(h=1\), \(0.5\), and \(0.1\), compute the average rate of change on \([2-h,2]\) and on \([2,2+h]\). Use the two patterns to estimate the instantaneous rate of change at \(t=2\).

The graph shows a curve \(f\), the point \(A=(0,0)\), six labeled points on the curve, and secant segments from \(A\) to each labeled point. Read the coordinates of \(B\) through \(G\) from the graph. Compute the slope of each secant from \(A\). Compare the slopes as the second point approaches \(A\) from the left and from the right. What instantaneous rate of change at \(x=0\) do the secants suggest?

A square's side length is measured in centimeters, and time \(t\) is measured in seconds. The table shows average rates of change of the square's area from \(t=1\) to \(t=1+h\). Negative \(h\) values use times before \(t=1\), and positive \(h\) values use times after \(t=1\). <table><tr><th>\(h\) (seconds)</th><td>\(-0.5\)</td><td>\(-0.1\)</td><td>\(-0.01\)</td><td>\(0.01\)</td><td>\(0.1\)</td><td>\(0.5\)</td></tr><tr><th>Average area-change rate</th><td>\(5.5\)</td><td>\(5.9\)</td><td>\(5.99\)</td><td>\(6.01\)</td><td>\(6.1\)</td><td>\(6.5\)</td></tr></table> Estimate the instantaneous rate of change of area at \(t=1\), state its units, and explain how the data from both sides support your estimate.

Three sets of average-rate evidence are proposed for a function at \(t=4\). In each set, the interval widths shrink toward \(0\). I. From the right only, the average rates are \(8.1\), \(8.01\), and \(8.001\). II. From the left, the average rates are \(7.9\), \(7.99\), and \(7.999\); from the right, they are \(8.1\), \(8.01\), and \(8.001\). III. From the left, the average rates are \(7.1\), \(7.01\), and \(7.001\); from the right, they are \(8.9\), \(8.99\), and \(8.999\). Which set supports a two-sided instantaneous rate of change at \(t=4\)? Give the estimated rate and explain why the other two sets are insufficient.

For a position function near \(t=4\), average velocities over shrinking intervals ending at \(4\) approach \(k\,\text{ft/s}\). Average velocities over shrinking intervals starting at \(4\) approach \(3\,\text{ft/s}\). What must \(k\) equal for the particle to have a two-sided instantaneous velocity at \(t=4\)? What would that velocity be?

For a cart, \(s\) is position in feet and \(t\) is time in seconds. Line \(m\) is tangent to the position curve at \(t=1\). Use two grid points on line \(m\) to find its slope. What does that slope tell you about the cart's instantaneous velocity at \(t=1\)?

A roller coaster car's height is modeled by \(H(t)=40-5(t-3)^2\) feet, where \(t\) is measured in seconds. For each symmetric interval \([2,4]\), \([2.5,3.5]\), and \([2.9,3.1]\), its average vertical rate is \(0\,\text{ft/s}\). Sibusiso says, “Because all three average rates are zero, the car is not moving up or down near \(t=3\), and its instantaneous vertical rate at \(t=3\) must be zero.” a) Explain why the claim about no vertical motion near \(t=3\) is false. b) Explain why the symmetric average rates alone do not prove the instantaneous rate at \(t=3\). c) For \(h>0\), find the average vertical rate on \([3-h,3]\) and on \([3,3+h]\). Use both to find the instantaneous vertical rate at \(t=3\).

For a function \(F\), the average rate of change from \(t=2\) to \(t=2+h\) simplifies to \(Q(h)=8+2h\) for every \(h\ne0\) near \(0\). Use this information to find the instantaneous rate of change at \(t=2\). Explain why substituting \(h=0\) into the original average-rate quotient would still be invalid even though the simplified expression has a value there.

A chemical concentration is modeled only for \(t\ge0\) by \(C(t)=3+2t+0.5t^2\), where \(C\) is measured in milligrams per liter and \(t\) in minutes. The average rates from \(t=0\) to \(t=1\), \(0.1\), and \(0.01\) are \(2.5\), \(2.05\), and \(2.005\) milligrams per liter per minute. Estimate the initial instantaneous rate at \(t=0\). Why is right-side information appropriate here?