Applications of Derivatives in Plane Curves PDF

# Applications of Derivatives in Plane Curves ## 4.1 Tangent and Normals A tangent to a curve is a straight line that touches the curve at one point and has the same slope as the curve at that point. ### Equation of Tangent Line $y-y_1=m(x-x_1)$ A normal to a curve is a straight line which is pe...

# Applications of Derivatives in Plane Curves ## 4.1 Tangent and Normals A tangent to a curve is a straight line that touches the curve at one point and has the same slope as the curve at that point. ### Equation of Tangent Line $y-y_1=m(x-x_1)$ A normal to a curve is a straight line which is perpendicular to a tangent to the curve. The word “normal” is used in advanced mathematics to mean “perpendicular.” It is in this sense rather than in the sense of "natural" or "usual" that the word is used here. ### Equation of Normal Line $y-y_1 = \frac{1}{m}(x-x_1)$ ### Example Find the equations of the tangent and the normal line of $y=3x^2-2x+1$ at point $(1,2)$. #### Solution: $y=3x^2-2x+1$ $y' = 6x - 2$ $y' = 6(1) - 2$ $y' = 4$ $y- y_1 = m(x-x_1)$ #### Tangent Line: $y -2 = 4(x-1)$ $y-2 = 4x-4$ $4x - y +2 -4 = 0$ $4x - y - 2 = 0$ #### Normal Line: $y - y_1 = \frac{1}{m}(x-x_1)$ $y - 2 = \frac{1}{4}(x-1)$ $4y - 8 = -x + 1$ $x + 4y - 8 -1=0$ $x + 4y -9 = 0$ ## Example Find the equation of the tangent and the normal line of $y=2+4x-x^3$ at $x=-1$ #### Solution If $x=-1$ $y=2+4(-1)-(-1)^3$ $y = 2-4+1$ $y = -1$ $y'= 4-3x^2$ $y'= 4-3(-1)^2$ $y' = 4-3(1)$ $y' = 1$ #### Tangent Line: $y - y_1 = m(x-x_1)$ $y - (-1) = 1(x+1)$ $y + 1=x+1$ $x-y-1+1=0$ $x-y=0$ #### Normal Line: $y - y_1 = \frac{1}{m}(x -x_1)$ $y - (-1) = \frac{1}{1}(x-(-1))$ $y+1 = x+1$ $y+1 = -x-1$ $x + y + 1 + 1 = 0$ $x+y+2 = 0$ ## Example Find the equations of the tangent lines and the normal lines of *y=x² - 7x + 6* at its points of intersection with the line *y=0*. #### Solution: $y=x^2-7x+6$ $0 = x^2 - 7x + 6$ $x^2 - 7x + 6 = 0$ $(x-1)(x-6) = 0$ $x-1=0$ ; $x-6 = 0$ $x= 1$ ; $x=6$ #### If x=1 $y= x^2-7x+6$ $y = (1)^2 - 7(1) + 6$ $y= 1-7 + 6$ $y=0$ #### If x=6 $y=x^2-7x+6$ $y = (6)^2 - 7(6) + 6$ $y = 36 - 42+ 6$ $y= 0$ #### Slope @ (1,0): $y' = 2x-7$ $y' = 2(1) -7$ $y' = -5$ #### Tangent Line @ (1,0): $y-y_1 = m(x-x_1)$ $y - 0 = -5(x-1)$ $y = -5x+5$ $5x + y - 5 = 0$ #### Normal Line @( 1, 0): $y - y_1 = \frac{1}{m}(x-x_1)$ $y -0 = \frac{1}{-5}(x-1)$ $-5y = -x + 1$ $x - 5y - 1 = 0$ ## Example Find the tangent lines as directed to the ellipse *x² - xy + 2y² - 4x + 2y + 2 = 0* parallel to the line *x - 4y = 2*. #### Solution: $x^2 - xy + 2y^2 - 4x + 2y + 2 = 0$ $2x - xy' - y + 4yy' - 4 + 2y' = 0$ $-xy' + 4yy' + 2y' = -2x + y + 4$ $y'(-x + 4y +2) = -2x + y + 4$ $y' =\frac{-2x + y + 4}{-x + 4y +2}$ #### To find m $x - 4y = 2$ $y = mx + b$ $-4y = -x + 2$ $\frac{-4y}{-4} = \frac{-x}{-4} + \frac{2}{-4}$ $y = \frac{1}{4}x - \frac{1}{2}$ $m = \frac{1}{4}$ #### Substitute m in $y' = \frac{-2x + y + 4}{-x + 4y +2}$ $\frac{1}{4} = \frac{-2x + y + 4}{-x + 4y +2}$ $-x + 4y +2 = 4(-2x + y + 4)$ $-x + 4y +2 = -8x + 4y +16$ $-x+4y + 2 + 8x - 4y - 16 =0$ $7x - 14 = 0$ $7x = 14$ $x = 2$ #### Substitute x in *x^2 - xy + 2y^2 - 4x + 2y + 2 = 0* $(2)^2 - (2)y + 2y^2 - 4(2)+2y+2 = 0$ $4 - 2y + 2y^2 - 8 + 2y + 2 = 0$ $2y^2 - 2 = 0$ $2y^2 = 2$ $y^2 = 1$ $y = \pm 1$ #### If y=1, x=2 #### If y=-1, x=2 #### Tangent Line @ (2,1): $y - y_1 = m(x-x_1)$ $y - 1 = \frac{1}{4}(x-2)$ #### Tangent Line @ (2,-1): $y - y_1 = m(x-x_1)$ $y - (-1)= \frac{1}{4}(x-2)$ ## Example Make the parabola *y=ax^2+bx+c* pass through (3,13) and be tangent to the line 8x-y=15 at (2,1). Determine the coefficients a, b, c to satisfy the stipulated conditions. **Solution:** * **@ (3,13)** $y=ax^2 + bx + c$ $13 = a(3)^2 + b(3) + c$ $13 = 9a + 3b + c -> eq. 1$ * **@ the line** $8x - y = 15$ $-y = -8x + 15$ $y = 8x - 15$ $m = 8$ #### Subtract eq. 1 ^ eq. 2 $13= 9a + 3b + c$ $-1 = -4a - 2b - c$ $12 = 5a + b -> eq. 4$ * **@ (2, 1)** $y=ax^2 + bx + c$ $1 =a (2)^2 + b(2) + c$ $1 = 4a + 2b + c -> eq. 2$ #### Subtract eq. 3 ^ eq. 4 $8 = 4a + b$ $-12 = -5a - b$ $-4 = -a$ $a = 4$ #### Substitute a in eq. 3 or eq. 4 (eq. 3) $8 = 4a + b$ $8 = 4(4) + b$ $8 = 16 + b$ $b = -8$ #### Substitute a and b in eq. 1 or eq. 2 (eq. 1) $13 = 9a + 3b + c$ $13 = 9(4) + 3(-8) + c$ $13 = 36 - 24 + c$ $13 = 12 + c$ $c= 1$ #### Substitute a, b, and c $y = ax^2 + bx + c$ $y = 4x^2 - 8x + 1$ ## Example Make the parabola *y=ax^2+bx+c* pass through (3,2) and be tangent to the line *y=4x+2* at (2,4). Determine the coefficients a, b, c to satisfy the stipulated conditions. **Solution:** * **@ (3,2)** $y = ax^2 + bx + c$ #$2 = a(3)^2 + b(3) + c$ $2 = 9a+ 3b + c -> eq. 1$ * **@ (2,4)** $y= ax^2 + bx + c$ $4 = a(2)^2 + b(2) + c$ $4 = 4a + 2b + c -> eq. 2$ * **@ the line** $y = 4x +2$ $m=4$ #### Subtract eq. 1 ^ eq. 2 $2 = 9a + 3b + c$ $-4 = -4a - 2b - c$ $-2 = 5a + b -> eq. 4$ * **@ (2, 1)** $y = ax^2 + bx + c$ $y' = 2ax + b$ $4 = 2a(2) + b$ $4= 4a + b - eq. 3$ #### Subtract eq. 3 ^ eq. 4 $4 = 4a + b$ $2 = -5a -b$ $6 = -a$ $a = -6$ #### Substitute a in eq. 3 or eq. 4 (eq. 3) $4 = 4a + b$ $4 = 4(-6) + b$ $4 = -24+ b$ $b= 28$ #### Substitute a and b in eq. 1 or eq. 2 (eq. 1) $2 = 9a + 3b + c$ $2 = 9(-6) + 3(28) + c$ $2 = -54 + 84 + c$ $2 = 30 + c$ $c = -28$ #### Substitute a, b, and c $y=ax^2 + bx + c$ $y = -6x^2 + 28x -28 $ ## 4.2 Increasing and Decreasing Functions Consider a function *y=f(x)* which has a continuous derivative on some range of x values. We know that the derivative *y'* is the rate of change of y with respect to x. If * y’>0* in some *x* interval, the rate of change of *y* is positive, so that *y* increases as *x* increases in that interval. If *y’<0* in some interval, * y* decreases as *x* increases in that interval. The same conclusions are exhibited vividly by the graph of the function because * y’* is also the slope of the curve. In the figure shown, as *x* increases, the curves rises if the slope is positive, as on the arc AB; it falls if the slope is negative, as along BD: If * y’>0*, *y* increases; If *y’<0*, *y* decreases. The above results are useful not only as employed in this chapter but also in demonstrating the validity of certain inequalities which are valuable in many phases of advanced mathematics. ## Maxima and Minima At a point such as B in figure above, where the function is algebraically greater than at any neighboring point, the function is said to have a maximum value, and the point is called a maximum point. Similarly, at D the function has a minimum value. At such points the tangent is parallel to Ox; i.e., *y’=0*. But the vanishing of the derivative does not mean that the function is necessarily a maximum or a minimum; the tangent is parallel to Ox at F, yet the function is neither a maximum nor a minimum there. From the figure we deduce the following test: At a point where *y’=0*, if *y*’ changes from positive to negative (as *x* increases), *y* is a maximum; if *y’* changes from negative to positive, *y* is minimum; if *y’* does not change sign, *y* is neither a maximum nor a minimum. The points at which *y’=0* are called critical points, and the corresponding values of *x* are the critical values of *x*: in Figure 1, B, D, F are critical points. Maxima and minima collectively are called extremes: in the figure, B and D are extremes. An extreme is not necessarily the greatest (or least) value that the function attains anywhere in its range. The ordinate of F, for example, is greater than that of B. An extreme is merely greater (or less) than any neighboring value. The greatest value that the function can assume anywhere in its range (if such a value exists) is the absolute maximum; a maximum (such as at B) that is greater than any other in the neighborhood is a relative maximum. In the majority of applications, we are concerned with the absolute maximum or minimum. When the function is a polynomial there can never be an absolute extreme if *x* is unrestricted; but even in the case of polynomials such extremes frequently occur in practical problems, owing to the fact that *x* is limited in range. ## Example Locate the critical points and determine if it is maximum or minimum $y=4- 6x + x^2$. **Solution:** $y= 4 - 6x + x^2$ $y' = -6 + 2x$ $0 = -6 + 2x$ $2x = 6$ $x = 3$ If $x = 3$ $y = 4 - 6x + x^2$ $y = 4 - 6(3) + (3)^2$ $y = -5$ **Test @ (3,-5)** $y' = -6 + 2x$ $y' = 2 > 0$ **critical point (3,-5), minimum** ## Example Locate the critical points and determine if it is maximum or minimum $y = -4(x+2)^2$. **Solution:** $y = -4(x+2)^2$ $y' = (-4)(2)(x+2) + (x+2)^2(0)$ $y' = -8x -16$ $0 = -8x - 16$ $8x = -16$ $x = -2$ If $x=-2$ $y = -4(x+2)^2$ $y = -4(-2+2)^2$ $y = 0$ **Test @ (-2,0)** $y' = -8x - 16$ $y' = 0 $ **critical point (-2, 0), maximum** ## Example Locate the critical points and determine if it is maximum or minimum $y = x^3 - 3x^2 + 4x + 5$ **Solution:** $y = x^3 - 3x^2 + 4x +5$ $y' = 3x^2 - 6x + 4$ $0 = 3x^2 - 6x + 4$ $ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4 (3)(4) }}{2(3)}$ $x = \frac{6 \pm \sqrt{-12}}{6}$ **No critical point** ## Example Locate the critical points and determine if it is maximum or minimum $y =x^3 - 6x^2 + 12x$ **Solution:** $y = x^3 - 6x^2 + 12x$ $y' = 3x^2 - 12x + 12$ $0 = 3x^2 - 12x + 12$ $0 = x^2 - 4x + 4$ $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $x = \frac{ - (-4) \pm \sqrt{(-4)^2 - 4 (1)(4) }}{2(1)}$ $x = \frac{4 \pm 0}{2}$ $x = 2$ If $x=2$ $y = x^3 - 6x^2 + 12x$ $y = (2)^3 - 6(2)^2 + 12(2)$ $y=8$ **critical point (2,8), it fails** **Test @ (2,8)** $y' = 3x^2- 12x + 12$ $y' = 6x - 12$ $y' = 6(2) -12$ $y' = 12 - 12$ $y' = 0 = 0$ ## Example Make the curve *y=ax^3 + bx^2 + cx + d* have a critical point at (0,3) & (1,2). Determining the coefficients a, b, c, etc. so that the curve will satisfy the stipulated conditions **Solution:** $y =ax^3 + bx^2 + cx + d$ $y' = 3ax^2 + 2bx + c$ $0 = 3ax^2 + 2bx + c -> eq. 1$ **@ (0,3)** $y =ax^3 + bx^2 + cx + d$ $3 = a(0)^3 + b(0)^2 + c(0) + d$ $d = 3$ **@ (1,2)** $y= ax^3 + bx^2 + cx + d$ $2 = a(1)^3 + b(1)^2 + c(1) + d$ $2 = a + b + c + d -> eq. 2$ **@ (0,3) eq 1** $0 = 3ax^2 + 2bx + c$ $0 = 3a(0)^2 + 2b(0) + c$ $c= 0$ **@ (1,2) eq 3** $0 = 3ax^2 + 2bx + c$ $0 = 3a(1)^2 + 2b(1) + c$ $0 = 3a + 2b + c -> eq. 3$ #### Substitute d in eq. 2 $2 = a + b + c + d$ $2 = a + b + c + 3$ $2-3 = a + b + c$ $-1 = a + b + c -> eq. 4$ #### Subtract eq. 3 and eq. 4 $0 = 3a + 2b + c$ $1 = -a -b - c$ $1 = 2a + b -> eq. 5$ #### Substitute c & d in eq.2 $2 = a + b + c + d$ $2 = a + b + 0 + 3$ $2 - 1 = a + b$ $1 = a + b -> eq. 6$ #### Subtract eq. 5 and eq. 6 $1 = 2a + b$ $-1 = -a -b$ $a=0$ #### Substitute a, c, and d in eq. 2 $2 = 0 + b + 0 + 3$ $2 - 3 = b$ $b = -1$ #### Substitute a, b, c, and d in $y=ax^3 + bx^2 + cx^2 + d$ $y = (0)x^3 + (-1)x^2 +(0)x + 3$ $y = -x^2 + 3$ ## 4.3 Points of Inflection A point of inflection is a point at which the curve changes from concave upward to concave downward, or vice versa (the points C, E, F in the figure shown. At a point of inflection the tangent reverses the sense in which it turns, which means that *y’* changes from an increasing to a decreasing function, or vice versa. Hence at such a point * y’ ’ * changes sign and, if it is continuous, must vanish. Conversely, a point at which * y’’* vanishes is a point of inflection, provided *y’’* changes sign at that point. Since y’’, the rate of change of the slope, is zero at a point of inflection, the tangent is sometimes said to be stationary for an instant at such a point, and in the neighborhood of the point it turns very slowly. Hence the inflectional tangent agrees more closely with the curve near its point of contact than does an ordinary tangent; it is therefore especially useful in tracing the curve to draw the tangent at each point of inflection. A point at which *y’’* vanishes without changing sign is not a point of inflection; the result means that near that point the tangent turns even more slowly than near a point of inflection. As noted in the past lesson, a point where *y’=0* is a maximum or a minimum, provided *y’’≠0*. If *y’* and *y’’* both equal to zero, the point is in general a point of inflection with a horizontal tangent (the point F in Figure); but if *y’’* vanishes without changing sign, the point is a maximum or minimum. Theorem: If *x=a* is a root of odd order –¿simple, triple, etc. -¿ of the equation *y’=0*, then *x=a* is a maximum or minimum; if *x=a* is a root of even order, *x=a* is a point of inflection with horizontal tangent. ## Example Find the point of inflection of *y=x^3 - 3x^2*. **Solution:** $y=x^3-3x^2$ $y' = 3x^2 - 6x$ $y'' = 6x-6$ $0 = 6x-6$ $6x = 6$ $x = 1$ If $x = 1$ $y = x^3 -3x^2$ $y = (1)^3 - 3(1)^2$ $y = -2$ ## Example Find the point of inflection of *y=36+12x-x^3* **Solution:** $y = 36 + 12x-x^3$ $y' = 12 - 3x^2$ $y'' = -6x$ $0 = -6x$ $x = 0$ If $x=1$ $y = 36 + 12x -x^3$ $y = 36 + 12(0) - (0)^3$ $y = 36$ ## Example Find the point of inflection of *y=2x^3 + 3x^2 - 12x + 7* **Solution:** $y= 2x^3+3x^2-12x + 7$ $y' = 6x^2 + 6x - 12$ $y'' = 12x + 6$ $0 = 12x + 6$ $12x = -6$ $x = \frac{-1}{2}$ If $x = \frac{-1}{2}$ $y = 2x^3 + 3x^2 - 12x + 7$ $y = 2 (\frac{-1}{2})^3 + 3 (\frac{-1}{2})^2 - 12 (\frac{-1}{2}) + 7$ $y = \frac{-1}{4} + \frac{3}{4} + 6 + 7$ $y = 27$ ## Example Find the point of inflection of *y=x^4 - 4x^3 + 8x* **Solution:** $y = x^4 - 4x^3 + 8x$ $y' = 4x^3-12x^2 + 8$ $y'' = 12x^2 - 24x$ $0 = 12x^2 - 24x$ $12x^2 = 24x$ $x = 2$ If $x=2$ $y = x^4 - 4x^3 + 8x$ $y = (2)^4 - 4(2)^3 + 8(2)$ $y = 0$ ## Example Find the point of inflection of *y = x^4 - 6x^2 - 7* **Solution:** $y = x^4 - 6x^2 - 7$ $y' = 4x^3 -12x$ $y'' = 12x^2 - 12$ $0 = 12x^2 - 12$ $12x^2 = 12$ $ x^2 = 1$ $x = \pm 1$ #### If x=1 $y = x^4 -6x^2 - 7$ $y = (1)^4 - 6 (1)^2 - 7$ $y = -12$ #### If x=-1 $y = x^4 - 6x^2 - 7$ $y = (-1)^4 - 6(-1)^2 - 7$ $y = -12$ ## 4.4 Critical Points We say that *x=c* is a critical point of the function f(x) if f(c) exists and if either of the following are true. *f’(c) = 0* or *f’(c)* doesn’t exist ## Example Determine all the critical points for the function *f(x) =6x^5+33x^4-30x^3+100* **Solution:** $f(x) = 6x^5 +33x^4 - 30x^3 + 100$ $f'(x) = 30x^4 + 132x^3 - 90x^2$ $f'(x) = 6x^2 (5x^2+12x-15)$ $f'(x)= 6x^2 (5x-3)(x+5)$ The only critical points will be those values of x which makes the derivative zero. $6x^2 (5x - 3)(x+5) = 0$ **Critical points** $x=0, x=-5, x = \frac{3}{5}$ ## Example Determine all the critical points for the function *g(t)=√(t^²(2t-1))* **Solution:** $g(t) = \sqrt{t^2(2t-1)}$ $g(t) = t \sqrt{(2t-1)}$ $g'(t) = \frac{1}{2} (2t-1)^{-\frac{1}{2}}(2)t + (2t-1)^{\frac{1}{2}}(1)$ $g'(t) = \frac{t}{\sqrt{2t-1}} + \sqrt{2t-1}$ $g'(t) = \frac{t+ 2t -1}{\sqrt{2t-1}}$ $g'(t) = \frac{3t - 1}{\sqrt{2t-1}}$ **Critical points** $t=0, t=\frac{1}{3}$ ## Practice Problem 4.4 Determine the critical points of the following functions: 1. *f(x) = 8x^3 + 81x^2 - 42 -8 = 0* 2. *R(t) = 1 + 80t^3 + 5t^4 - 2t^5* 3. *g(w) = 2w^3 - 7w^2 - 2w - 2* 4. *g(x) = x - 2x^5 + 8x^4* 5. *h(z) = 4z^3 - 3x^2 + 9x + 12* 6. *Q(x) = (2-8x)(x^2-9)^3* 7. *f(z) = \frac{z+4}{2z^2+z+8}* 8. *R(x) = \frac{1 - x}{x^2 + 2x - 15}* 9. *r(y) = y^2 - 6y* 10. *h(t) = 15 - (3t)(t^2 - 8t + 7)* 11. *s(z) = 4cos(z) - z* 12. *f(y) = sin(\frac{y^3}{3}) + 2y* 13. *v(t) = sin(3t) +1* 14. *f(x) = 5xe^(9-2x)* 15. *g(w) = e^(w^3 - 2w^2 - 7w)* ## 4.5 Minimum and Maximum Valves Many of our applications in this chapter will revolve around minimum and maximum values of a function. While we can all visualize the minimum and maximum values of a function we want to be a little more specific in our work here. In particular, we want to differentiate between two types of minimum or maximum values. The following definition gives the types of minimums and/or maximums values that we'll be looking at. 1. We say that f(x) has an absolute (or global) maximum at x = c if f(x) ≤ f(c) for every x in the domain we are working on. 2. We say that f(x) has a relative (or local) maximum at x = c if f(x) ≤ f(c) for every x in some open interval around x = c. 3. We say that f(x) has an absolute (or global) minimum at x = c if f(x) ≥ f(c) for every x in the domain we are working on. 4. We say that f(x) has a relative (or local) minimum at x = c if f(x) ≥ f(c) for every x in some open interval around x = c. It’s usually easier to get a feel for the definitions by taking a quick look at a graph. ## Example Below is the graph of some function, (x). Identify all of the relative extrema and absolute extrema of the function. ## Example Below is the graph of some function, f(x)f(x). Identify all of the relative extrema and absolute extrema of the function. ## Refences ### A) Book/Printed Resources * AL., L. G. (1989). Calculus for Business Economics and the Social Life Science, 3rd Edition. McGraw_hill Book Corporation. * Anton, H. (1992). Multivariable Calculus, 4th Edition. New York. * Anton, H. (n.d.). Multivariable Calculus, 4th Edition. 1992. New York: John Wiley and Sons Incorporated. * Berkey, D. D. (1990). Calculus for Management and Social Sciences, Second Edition. Saunders College Publishing. * Farlow, S. J. (1993). Calculus and its Application. Prentice-Hall Incorporated. * Leithold, L. (2001). The Calculus, 7th Edition. Addison-Wesley. * Margaret B. Cozzens. (1987). Mathematics with Calculus. D.C. Health. * Robert Ellis et. AL. (1990). Calculus with AnalyticGeometry. Hardcourt Brace Jovanovich Incorported. ### B) e-Resources * http://www.soton.ac.uk/~cjg/eng1/modules/modules.html * http://www.mit.opencourseware.com * http://www.mathalino.com/reviewer/advance-engineering-mathematics/advance-engineering-mathematics * http://ecereviewcourse.blogspot.com/p/math.html