Solving for y', we have , and, at the point (3,4), y' = 3/4. – In the previous example, it would have been easy to explicitly solve for y, and then we could differentiate y to get y '. Because we could explicitly solve for y , we had a choice of methods for calculating y '. Sometimes, however, we can not explicitly solve for y , and the only way of determining y ' is implicit differentiation. Related Rates If several variables or quantities are related to each and some of the variables are changing at a known rate, then we can use derivatives to determine how rapidly the other variables must be changing. Example 3 Suppose the border of a town is roughly circular, and the radius of that circle has been increasing at a rate of 0.1 miles each year. Find how fast the area of the town has been increasing when the radius is 5 miles. We could get an approximate answer by calculating the area of the circle when the radius is 5 miles ( A = = miles)2 78.6 miles2 ) and 1 year later when the radius is 0.1 feet larger πr2 π(5 ≈ than before ( A = = miles)2 81.7 miles2 ) and then finding = (81.7 πr2 π(5.1 ≈ ∆Area/∆time mi2 78.6 mi2)/(1 year) = 3.1 mi2/yr. This approximate answer represents the average change – in area during the 1 year period when the radius increased from 5 miles to 5.1 miles, and would correspond to the secant slope on the area graph. To find the exact answer, though, we need derivatives. In this case both radius and area are functions of time: = radius at time = area at time r(t) t A(t) t We know how fast the radius is changing, is a statement about the derivative: . We also know that = 5 at our moment of interest. r We are looking for how fast the area is increasing, which is . Now we need an equation relating our variables, which is the area equation: . Taking the derivative of both sides of that equation with respect to we can use implicit t, differentiation: Plugging in the values we know for and r dr/dt,
Solving for y', we have , and, at the point (3,4), y' = – 3/4. In the previous example, it would have been easy to explicitly solve for y, and then we could differentiate y to get y '. Because we could explicitly solve for y , we had a choice of methods for calculating y '. Sometimes, however, we can not explicitly solve for y , and the only way of determining y ' is implicit differentiation. Related Rates If several variables or quantities are related to each and some of the variables are changing at a known rate, then we can use derivatives to determine how rapidly the other variables must be changing. Example 3 Suppose the border of a town is roughly circular, and the radius of that circle has been increasing at a rate of 0.1 miles each year. Find how fast the area of the town has been increasing when the radius is 5 miles. We could get an approximate answer by calculating the area of the circle when the radius is 5 miles ( A = πr2 = π(5 miles)2 ≈ 78.6 miles2 ) and 1 year later when the radius is 0.1 feet larger than before ( A = πr2 = π(5.1 miles)2 ≈ 81.7 miles2 ) and then finding ∆Area/∆time = (81.7 mi2 – 78.6 mi2)/(1 year) = 3.1 mi2/yr. This approximate answer represents the average change in area during the 1 year period when the radius increased from 5 miles to 5.1 miles, and would correspond to the secant slope on the area graph. To find the exact answer, though, we need derivatives. In this case both radius and area are functions of time: r(t) = radius at time t A(t) = area at time t We know how fast the radius is changing, is a statement about the derivative: . We also know that r = 5 at our moment of interest. We are looking for how fast the area is increasing, which is . Now we need an equation relating our variables, which is the area equation: . Taking the derivative of both sides of that equation with respect to t, we can use implicit differentiation: Plugging in the values we know for r and dr/dt,
BusCalc_Chp2_wExercises.docx