Unit 2: Introduction to Calculus PDF

# Unit 2: Introduction to Calculus ## 2.1.5 Maximum and Minimum Points The differentiation of various functions renders great services in solving the problems concerned with finding out the maximum and minimum values of quantities. In the various fields of engineering and technology, we have to fi...

# Unit 2: Introduction to Calculus ## 2.1.5 Maximum and Minimum Points The differentiation of various functions renders great services in solving the problems concerned with finding out the maximum and minimum values of quantities. In the various fields of engineering and technology, we have to find the maximum or minimum values of one quantity with respect to another quantity. For example, in finding the radius and height of a cylinder that is to be manufactured with the metal sheet of a given surface area, so that the capacity of the cylinder is maximum. Sometimes it is necessary to find the least cost for the transmission of given horsepower. We can find the dimensions of a plot of a given perimeter so that its area is maximum and so on. ## 2.1.5.1 Increasing and Decreasing Functions ### Activity 2.12 1. Find the real zeros of each of the following functions. - *f(x) = 4x - 8* - *f(x) = x² -x-12* - *f(x) = (1-√x)/(x-2)²* 2. Consider the graph of the following function: - **Description of the graph:** The function is a piecewise function that is constant on each interval. The graph starts at a low point, rises to a peak, then falls to a lower point. It then rises again to another peak before falling to a low point again. - **Identify the intervals in which the graph is rising and falling as you move from left to right starting at a.** - **Identify the intervals in which the graph is neither rising nor falling.** 3. Solve each of the following inequalities using sign chart method: - *2x²+3x-2≥0* - *x²+3x-4<0* ## Definition 2.8 Let f be a function on an interval I. If x₁ and x₂ are in I, - For X₁ < x₂ if f(x₁) ≤ f(x₂), then _f_ is said to be increasing on I. - For X₁ < x₂ if f(x₁) ≥ f(x₂), then _f_ is said to be decreasing on I. - For X₁ < x₂ if f(x₁) < f(x₂), then _f_ is strictly increasing on I. - For X₁ < x₂ if f(x₁) > f(x₂), then _f_ is strictly decreasing on I. **Geometrically, a function is increasing if its graph rises and decreasing if its graph falls while _x_ moves to the right.** **Description of the graph:** The graph on the left side shows an increasing function. The function is continuous and has a positive slope across the domain. The graph on the right side shows a decreasing function. The function is also continuous, but has a negative slope across the domain. ## Increasing and decreasing test Suppose that f(x₁) ≥ f(x₂), _f_ is differentiable in the interior of an interval I. - If f'(x) ≥ 0, for all _x_ in the interior of I, then _f_ is increasing on I. - If f'(x) ≤ 0, for all _x_ in the interior of I, then _f_ is decreasing on I. - If f'(x) > 0 and f'(x) = 0 on I, then _f_ is strictly increasing on I. - If f'(x) < 0 and f'(x) = 0 on I, then _f_ is strictly decreasing on I. **Note:** A function that is either increasing or decreasing is known as a **monotonic function**. ## Definition 2.9 A number _c_ in the domain of a function _f_ is said to be a **critical number** of _f_ if and only if either f'(c) = 0 or _f_ has no derivative at _c_. ### Example 1 Find the critical numbers of the given functions. - *f(x) = 4x³-5x² -8x+20* - *f(x)=2√x(6-x)* **Solution:** - *f'(x)=12x²-10x-8* - *f'(x) = 0 ⇒ 2(6x² – 5x-4) = 2(2x+1)(3x-4) = 0 ⇒ x = -1/2 or x = 4/3.* - Hence, the critical numbers of _f_ are *x = -1/2* and *x = 4/3*. - *f'(x)=2(6-x)/√x -2√x = (6-x)/√x -3(2-x)/√x ⇒ f'(x) = 0 ⇒ x = 2.* *f'(x) does not exist when x=0*. Hence, the critical numbers of f are *x = 0* and *x = 2*. ### Example 2 Find the interval on which the function *f(x) = 3x² -4x³ -12x² +5* is increasing and decreasing. **Solution** *f'(x) = 12x³ -12x² - 24x = 12x(x²-x−2) = 12x(x-2)(x+1)* Hence, the critical numbers of f are *x = −1,0,2* | | x < -1 | -1 < x < 0 | 0 < x < 2 | x > 2 | |-------------|----------|---------------|------------|-------| | 12x | ----- | ------- | ++++ | +++++ | | x + 1 | ----- | ------- | ++++ | +++++ | | x - 2 | ----- | ------- | ----- | +++++ | | f'(x) | ----- | ++++ | ----- | +++++ | From the sign chart above, *f'(x) ≥ 0 on [-1,0] U [2,∞)* and *f'(x) ≤ 0 on (-∞, -1] U [0,2].* Therefore, the function is increasing on an interval *[-1,0]U[2,∞)* and decreasing on *(-∞,-1]U[0,2]*. ## Exercise 2.14 1. Find all the critical numbers of the following functions. - *f(x) = x²+6x² -24x* - *f(x) = (2x²)/(1-x²)* - *f(x) = 2x³ +9x²-24x* - *f(x) = 8x√(1-x²)* 2. Find the interval on which the following functions are increasing and decreasing. - *f(x) = x² + 6x²* - *f(x) = 2x² + 6x-10* - *f(x) = -2x² +3x² +12x+6* - *f(x)=x-6√x−1* - *f(x) = (x+3)²(x−1)²* ## 2.1.5.2 Minimum and maximum values of a function ### Activity 2.13 Consider the graph of the following function: - The graph has three peaks and two valleys. - **Where does the function attain maximum value and a minimum value?** - **What happens to the gradient at a maximum point and at a minimum point?** ## Definition 2.10 If there exists an open interval (a, b) containing _c_ such that f(x) < f(c) for all _x_ in (a, b) other than _c_ in the interval, then f(c) is a **relative (local) maximum value** of _f_ and a point (c, f(c)) is called **relative maximum point**. If f(x) > f(c) for all _x_ in (a,b) other than _c_, then f(c) is a **relative (local) minimum value** of _f_ and the point (c, f(c)) is known as **relative(local) minimum point** (See Figure 2.13 below). Functions may have any number of relative extrema. **Description of the graph:** The graph shows two peaks and one valley. The function is continuous and has a positive slope up to the first peak, a negative slope from the first peak to the first valley, a positive slope from the first valley to the second peak, a negative slope from the second peak to the second valley, and a positive slope from the second valley to the third peak. ## Definition 2.11 If _c_ is in the domain of _f_ and for all _x_ in the domain of the function, f(x) ≤ f(c), then (c, f(c)) is an **absolute maximum point** of the function _f_. If for all _x_ in the domain f(x) ≥ f(c) then (c, f(c)) is an **absolute minimum point** of the function _f_. (See Figure 2.14). The absolute maximum and minimum values of _f_ are called **absolute extreme values**. **Note:** - Absolute extrema are not necessarily unique. - At either local maximum or local minimum point the gradient is zero. **Description of the graph:** The graph shows many peaks and valleys. The function is continuous and has both a positive and negative slope across the domain. ## Procedure to find absolute extreme values on closed intervals We can find the absolute maximum and minimum values of a function _f_ on a closed interval [a, b] by the following steps: 1. Find all the critical numbers of a function _f_. 2. Evaluate _f_ at all critical numbers and at the end points _a_ and _b_. 3. Choose the largest and the smallest values obtained in step2. These values are the maximum and the minimum of a function _f_ on [a,b]. ### Example Find the absolute maximum and minimum values of the function *f(x) = x²-2x* on [-1,2]. **Solution** *f '(x) = 2x - 2 = 2(x−1)* *f '(x) = 0 ⇒ x = 1.* Thus, the critical number of _f_ is *x=1*. Thus, to determine the extreme value of the function we need to evaluate the value of the function at the boundary and critical number. *f(-1) = 1 + 2 = 3, f(1) =1-2=-1 and_f(2)=4− 4 = 0.* Therefore, the absolute maximum value is *f(-1) = 3* and the absolute minimum value is *f(1) = −1*. Thus, *(-1,3)* is the maximum point and *(1,-1)* is the minimum point of _f_. ## Exercise 2.15 Find the absolute maximum and minimum values of the following functions on a given interval. - *f(x) = x²+2x+3, [2,2]* - *f(x) = x² +3x²-9x+5,[-2,2]* - *f(x) = 3x²-26x³ – 60x² -11 on [1,5].* - *f(x) = (x²-2x+4)/(x-2), [-3,1]* ## The first derivative test Let _c_ be a critical number of a function _f_ on an open interval I containing _c_. - If the sign of _f_’ changes from negative to positive at _c_, then _f_ has a relative minimum at _c_. - If the sign of _f_’ changes from positive to negative at _c_, then _f_ has a relative maximum at _c_. - If the sign of _f_’ does not change at _c_, then _f_ has no relative maximum or minimum at _c_. ## Procedure to find relative extreme value To determine relative extreme value of a function using the first derivative test, one can follow the following steps. 1. Find all the critical numbers of _f_. 2. Form the sign chart for f’(x) using the critical number as poles. 3. Apply first derivative test and decide the extreme values if any. ### Example 1 Find the relative maximum and minimum values of - *f(x) = 2x² +3x² -12x-3* - *f(x) = (x−1)2(x-3)²* **Solution:** - *f'(x) = 6x² + 6x-12 = 6(x²+x-2) = 6(x+2)(x-1)* Hence, the critical numbers are *x = −2,1* | | x < -2 | -2 < x < 1 | x > 1 | |-------------|---------|---------------|-------| | 6(x+2) | ----- | ++++ | +++++ | | x - 1 | ----- | ----- | +++++ | | f'(x) | ----- | ++++ | +++++ | From the sign chart above we see that, f' changes its sign from positive to negative at x = -2 and changes from negative to positive x = 1, therefore, the function has a local maximum value at x = -2 and a local minimum value at x=1. This implies the point (-2, f (-2)) = (-2,17) is a relative maximum and (1, f (1)) = (1,−10) is a relative minimum point of the function. Thus, 17 is a relative maximum value and – 10 is a relative minimum value. - *f'(x) = 2(x-1)(x-3)² + 2(x−1)2(x-3) = 2(x-1)(x-3)(x-3+x-1) = 2(x-1)(x-3)(2x-4) = 4(x-1)(x-2)(x-3)* Hence, the critical numbers are *x = 1,2,3*. | | x < 1 | 1 < x < 2 | 2 < x < 3 | x > 3 | |-------------|---------|------------|------------|--------| | 4(x - 1) | ----- | ++++ | ++++ | +++++ | | x - 2 | ----- | ----- | ++++ | +++++ | | x - 3 | ----- | ----- | ----- | +++++ | | f'(x) | ----- | ++++ | ----- | +++++ | From the sign chart, we see that f' changes its sign from negative to positive at x=1 andx=3 and changes from positive to negative at x = 2 Therefore, (1, f (1)) = (1, 0) and (3, f(3)) = (3, 0) are relative minimum points and (2, f (2)) = (2,1) is the relative maximum point of the function. Thus, 0 is a relative maximum value and 1 is a relative minimum value. ### Example 2 If *f(x) = ax³ +bx-5* has a relative minimum value of -6 at *x = 1/2*, find *a* and *b*. **Solution:** Given *f(1/2) = a(1/2)³ + b(1/2) -5 = -6 ↔ a + 4b = -8* and *f'(1/2) = 0, f'(x) = 3ax² + b ⇒ 3a + b = 0 ⇔ 3a + 4b = 0* (i) and (ii) imply ⇒ a = 4 and b = -3. ## Exercise 2.16 1. Find the relative extreme value of the following functions (if any). - *f(x) = 2x³-9x²-24* - *f(x) = x³-3x* - *f(x) = (x²+x+1)/(x²-x+1)* 2. If f(x) = x² + ax + b has a relative minimum value of 4 at x = 2, find *a* and *b*. ## 2.1.6 Equation of Tangents and Normal Lines to Curves ### Activity 2.14 1. Find the equation of a line containing the points (1,3) and (-2,3). 2. Find the equation of a line perpendicular to the line found in question 1 above containing the point (1,3). 3. Write the equation of a tangent line to a curve if its slope is 3 at the point (-2,5). 4. What is the relationship between the slopes of a tangent line and a normal line to a curve at a point of tangency? Is there any relation ship between the first derivative of a function at a point of tengency and the slope of a tangent line to the curve of the function? We have seen in section 2.1.5 that the first derivative of a function at a given point is interpreted as the slope of the tangent line at that point. This implies f' is the slope of the line tangent to the graph of f at a point (a, f(a)). Thus, the equation of the tangent line to the curve y = f(x) at x = a is given by y = f'(a)(x-a)+f(a). A line that is perpendicular to the tangent line at the point of tangency is known as the **normal line** to the curve. Since the product of the slope of two non-vertical perpendicular lines is -1, the slope of the normal line at the point of tangency (a, f (a)) is -1/f'(a) and its equation is given by y = (1/f'(a))(x-a)+f(a). **Note:** - If the first derivative of f at x = a is zero, then ƒ has a horizontal tangent line with equation y = f(a) and vertical normal line given by x = a. - If the first derivative of f at x = a undefined, then f has a vertical tangent line with equation x = a and a horizontal normal line given by y = f(a). ### Example 1 Find the equation of the line tangent and the normal line to the graph of the following functions at the given point. - *f(x) = x²-2x; (1,−1)* - *f(x) = √20-x²; (2,4)* - *f(x) = {x, if x > 3; x² x²-6, if x ≤ 3}* - *f(x) = x√x² +16; (0,0)* **Solution:** - *f'(x) = 3x²-2⇒ f'(1)=1* - Thus, the equation of the tangent line to the graph of f at (1,-1) is *y = f '(1)(x−1) + f (1) = 1(x−1)−1=x−2. ⇔ y = x−2.* - The equation of the normal line to the graph of f at (1,-1) is *y=-- -(x−1) + f (1) f'(1) =-1(x-1)-1 = -x. ⇔y=-x ⇒ - *f'(x) = -x/√20-x² ⇒ f'(2)= -1/2* - The equation of the tangent line to the graph of f at (2,4) is *y = f'(2)(x - 2) + f(2)=--=(x-2)+4=-x+5 1 1 2 2* - Since the slope of a normal line is negative reciprocal of that of the tangent line, the slope of the normal line in this problem is 2. Hence, the equation of the normal line to the graph of f at (2,4) is *y=-1 -(x - 2) + f (2) = 2(x - 2) + 4 = 2x. f'(2) ⇔ y = 2x.* - *f'(x) = 2x, if x > 3 ⇒ f'(1) = 2.* - The equation of the tangent line to the graph of f at (1,-5) is *y = f'(1)(x-1) + f (2) = 2(x−1)− 5 = 2x−´ -5=2x-7.* - The equation of the normal line to the graph of f at (1,-5) is *y=-1 -(x-1)+f(1) ===(x-1)-5=2x-2 f'(1) 9 ⇒ 1 9 y=-x- 2 2* - *f '(x) = √x² +16 + x² /(2√x² +16) ⇒ f'(0) = 4* - The equation of the tangent line to the graph of f at (0,0) is *y = f'(0)(x−0) + f (0) = 4(x−0)+0 = 4x. y = 4x.* - The equation of the normal line to the graph of f at (0,0) is *y=-1 -(x-0)+f(0)=(x-0)=x. f'(0) -1 4 y = -x. 4* ### Example 2 Find all points at where *f(x) = √x²-6x* has vertical tangent line. **Solution** The function will have a vertical tangent line at its first derivative is undefined in its domain and *f'(x)= (2x-6) /(2√x²-6x) = x-3 √x²-6x * is undefined at the points where x(x−6) = 0 ⇒ x = 0 or x = 6. Hence, f(x) has a vertical tangent line at (6,0) and (0,0). ## Exercise 2.17 1. Find the equation of the line tangent and the normal line to the graph of the following functions at the given point. - *f(x) = x² +x+2; (0,2)* - *f(x)=(1-x³)√x+2; (−1,2).* 2. Find all points at where the following functions has vertical tangent line. - *f(x) = √x²-x* - *f(x) = 2√x* 3. If *f(x) = x² + ax+band g(x) = x² -c* have the same tangent line at (1,2) find the values of the constants *a,b* and *c* and the equation of their normal lines. ## 2.2 Application of Derivative ### Activity 2.15 1. Given a set S = {3,4,5,6,7} and *f(x) = 3x + 2* - Find K = {f(x)\x∈S}. - What is the smallest element of K? - What is the largest element of K? 2. Given a set S on an open interval (3,7) and *f(x) = 3x + 2* - Find T = {f(x) | x ∈ S}. - Can you list all the elements of T? - Can you guess the smallest element of T? - Can you guess the largest element of T? 3. Given a set S as a closed interval [3,7] and *f (x) = 3x + 2* - Find H = {f(x) | x ∈ S}. - Can you list all the elements of H? - Can you guess the smallest element of H? - Can you guess the largest element of H? Applications of derivatives are varied not only in mathematics but also in real life. To give an example, derivatives have various important applications in mathematics such as finding the rate of change of a quantity, finding the approximate value, finding the equation of the tangent line and the normal line to a curve, and finding the minimum and maximum values of algebraic expressions. Derivatives also has a wide range of applications in real-life usage. For instance, in business, derivatives are used to find profit and loss for the future of the investment using graphs. It is used to calculate the rate of change of distance of a moving body with respect to time. Derivatives can have different interpretations in each of the sciences. For instance; chemists who study a chemical reaction may be interested in the rate of change in the concentration of a reactant with respect to time (the rate of reaction). A biologist is interested in the rate of change of the population of a colony of bacteria with respect to time. In section 2.1, you have studied derivatives and have developed methods to find derivatives of various functions. Once you have developed the properties of the mathematical concepts once and for all, you can then turn around and apply these results to all of the sciences. You have already investigated some of the applications of derivatives, but now that you know the differentiation rules, you are in a better position to pursue the applications of differentiation in a greater depth. In this section, you will study problems involving variables that are changing with respect to time. If two or more such variables are related to each other, then their rates of change with respect to time are also related. ## 2.2.1 Applications of Derivatives in Finding Rate of Change ### Example 1 If the radius of a circle is increasing at a rate of 1.5 cm/sec., find the rate at which the area is increasing at the instant when the diameter is 12 cm. **Solution** We know that area of a circle with radius r is given by *A = πr²* and we are given *dr/dt=1.5cm/sec*, to find *dA /dt* at diameter *d=12cm*. *dA dt =d(r²) dt * = π *2r dr dt * = 2πr dr dt *. Besides when the diameter *d = 12cm, r = 6cm*. Hence, *dA dt = 2π(6)(1.5) = 18π cm² / sec*. ### Example 2 An object is moving along the path *3y = x²*. If the x-coordinate of the object is increasing at a rate of 2 ft/min at the instant when x = 6, find the rate at which the y -coordinate is increasing. **Solution** Given *3y = x²* and *dx dt = 2ft/min*. *3y = x² ⇒ 3 dy = 2x dx dt dt ⇒ dy dt = 2x dx 3 dt *. Thus, at *x = 6, dy dt = 2 3 (6)(2 ft/min)=8 ft/min*. ### Example 3 Suppose the side of an equilateral triangle is increasing at a rate of 2 cm/sec. Find the rate at which the area is increasing at the instant when the side is 4 cm. **Solution** The area of an equilateral triangle having one side of lenth s is given by *A = √3/4 *s²* We are given, *ds dt = 2cm/sec*. To find *dA dt* when *s = 4cm*. *A = √3/4 *s² ⇒ dA dt = √3/2 *s ds dt* *dA dt |s=4 = 2x √3/4 *4 = 4√3 cm²/sec* ### Example 4 Air is being pumped into a spherical balloon at the rate of 4 cm³ / min. Find the rate of change of the radius when the radius is 2 cm. **Solution** Let r be the radius of the sphere. The volume V of the sphere is given by *V = 4/3 πν³* *V = 4/3 πν³ ⇒ dV dt = 4πr² dr dt ⇒ dr dt = 1 4πr² dV dt * = 1 4π (2)² 4 cm³/min = 1 4π **Example 5** The displacement of a particle at any time t is given by *S = t³ – 3t² + 5t + 7*. Find its velocity and acceleration at the end of 2 hrs, where S is in kilometers and t is in hrs. **Solution** Given *S = t³ – 3t² + 5t + 7* *Velocity v = ds dt = 3t2 - 6t + 5* *Acceleration a = dv dt = 6t-6* Hence, the velocity at the end of 2 hrs is *ν(2) = ds dt |t=2hrs = 3(2)² -6(2) + 5 = 5 km/hr*. Acceleration at the end of 2 hrs is *a = dv dt = 6(2) - 6 = 6 km/hr²*. ## Exercise 2.18 1. A circular plate of metal is being expanded by heating. The radius of plate increases at the rate of 0.3 cm per second. Find the rate of increase of area when its radius becomes 15 cm.