AP Calculus BC Unit 4 Contextual Applications of Differentiation

Unit 4: Contextual Applications of Differentiation ● The derivative of a function determines the instantaneous rate of change with respect to its variable ○ Since the derivative of a function is basically the slope of that function at a certain point, the unit for the derivative is the unit of f(x) / unit of x ● Straight line motion: ○ The derivative of a position-time graph is its velocity, and the second derivative of a position-time graph is its acceleration. *note that speed is the absolute value of velocity ■ How did we derive this? ● Well, the unit of the derivative is unit of f(x) / unit of x, and thus, when taking the derivative of position, the unit of the derivative would be m/s (or any unit of distance/unit of time), which gives you velocity. The second derivative of position is acceleration, which is the same thing as the first derivative of velocity, and by dividing the unit of fx) by the unit of x, the unit of the second derivative of position would be m/ 𝑠 2 , which is the unit for acceleration. ● Rates of change in other contexts: ○ Similarly to motion, the rate of change is measured simply by taking the derivative of a function. Units are derived the same way. ● Related Rates: a problem that asks you to find the rate of change of a variable in relation to the rates of change of other variables ○ Ex. (taken from Princeton AP Calc BC Review book) : a circular pool of water is expanding at the rate of 16π 𝑖𝑛2 / sec. At what rate is the radius expanding when the radius is 4 inches? ■ Steps to approach a related rates problem: ● Set up an equation that connects the variables together. a. In this example, a circle’s radius and the rate of change of area are given, so you can use the equation of the area of a circle, A = π 𝑟 2 ● Differentiate both sides of the chosen equation, with respect to time. a. dA/dt = 2πr(dr/dt) ● Plug in the given values into the differentiated equation a. 16π 𝑖𝑛2/ sec = 2π(4)(dr/dt) ● Isolate variables to find the variable that you’re looking for a. dr/dt = (16π 𝑖𝑛2 / sec) / 8π = 2 in/ sec ● Linear approximation: ○ You can use the equation of the tangent line of a graph to approximate the value of the function at a certain point. ○ Ex. Approximate f(2.1), when f(x) = 𝑥 2 ■ First, find the equation of the tangent line, which is done by taking the derivative of the function at x =2 → f’(x) = 2x ; f’(2) = 4, and then using (x,y) to determine the equation. In this case (2,4) is a point on the graph. ● Using y = mx + b, solve for b using (2,4), which gives you the final tangent line equation of y (or f’(x)) = 4x - 4 ■ Once you have solved for the equation of the tangent line, plug in the value that the equation wants you to approximate into the equation ● f’(2.1) = 4(2.1) - 4 = 4.4 ○ To determine if the approximation is an overapproximation or an underapproximation, you need to know whether the original function is concave up or down. If the graph is concave up, then the linear approximation is an underapproximation, and if the graph is concave down, then the linear approximation is an overapproximation ● L’Hospital’s Rule: ○ If the lim x → c gives you either 0/0 or ∞/∞, then you can use L’Hospital’s rule to determine the limit ○ Ex. 𝑙𝑖𝑚( 𝑥 2 - 4) / (4x - 8) gives you 0/0, so you have to use L’Hospital’s Rule, 𝑥→2 and basically just find the limit as x → 2 of the derivative of the numerator and the derivative of the denominator ■ 𝑙𝑖𝑚 (2x) / 4 = 4/4 = 1 𝑥→2