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Neet class 11 12 basic mathematics notes

The document provides an overview of various mathematical concepts including functions, differentiation, polynomial and trigonometric functions, and their graphical representations. It includes examples and definitions related to slopes, derivatives, and critical points, as well as transformation between degrees and radians. Additionally, it covers exponential and logarithmic functions along with rules for their differentiation and properties.

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Welcome to Orientation Sessions NOTES
Functions • Input variable is independent and output variable is dependent on it. • A function is a mathematical relationship between independent and dependent variables Algebraic Trigonometric Exponential Logarithmic Functions
Polynomials 𝑦 = 𝑓 𝑥 = 𝑎0 + 𝑎1𝑥1 + 𝑎2𝑥2 + 𝑎3𝑥3 + … … . 𝑎𝑛𝑥𝑛 Constants Variables = = Degree of a polynomial Highest power Maximum number of solutions Algebraic Function
The graphical representation of a linear function is a straight line 1 Linear Function 𝑦 = 𝑓 𝑥 = 𝑎0 + 𝑎1𝑥1 Constants Variables = = = Degree of a polynomial Highest power Maximum number of solutions Equation of line 𝑦 = 𝑚𝑥 + 𝑐 𝑐 → 𝑦-intercept of line = tan 𝜃 = 𝑦2−𝑦1 𝑥2−𝑥1 𝑚 → Slope of line
Rahul is standing 4 𝑚 East and 5 𝑚 North from Ashok. Taking Ashok’s position as origin, represent Rahul’s position on the coordinate plane given below. 𝑦(𝑚) 𝑥(𝑚) Coordinates of Rahul’s position
Ashok has now moved 1 𝑚 East and 1 𝑚 North while Rahul moved 1 𝑚 South. Calculate shortest distance between Rahul and Ashok. Shortest distance = 𝑥2 − 𝑥1 2 + 𝑦2 − 𝑦1 2 Shortest distance = 4 − 1 2 + 4 − 1 2 Shortest distance = 32 + 32 Shortest distance = 3 2 𝑚 𝑥 𝑦 1 1 Rahul→ (4,4) Ashok→ (1,1) 1
Determine direction of Rahul’s position (4,4) with respect to Ashok (1,1). Slope/Direction = tan 𝜃 = 𝑦2 − 𝑦1 𝑥2 − 𝑥1 tan 𝜃 = 4 − 1 4 − 1 = 1 tan 𝜃 = tan 45° 𝜃 = 45° from +𝑣𝑒 𝑥 −axis 𝑥 𝑦 Rahul→ (4,4) Ashok→ (1,1) 𝜃
Case ii : c = 0 Passing through origin 𝑥 𝑦 = 𝑥 𝑦 = −𝑥 𝑦 Case i : 𝑚 = 0 Parallel to x-axis 𝑦 = +𝑐 (0, 𝑐) 𝑦 𝑥 𝑐 Special cases
𝑦 = 𝑚𝑥 + 𝑐 Straight line equation Slope - intercept 𝑥 𝑦 𝜃 𝑐 𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1) Point - slope 𝑥 𝑦 𝜃 ( 𝑥1 , 𝑦1 ) 𝑦 − 𝑦1 = (𝑦2 − 𝑦1) (𝑥2 − 𝑥1) (𝑥 − 𝑥1) ( 𝑥 2 , 𝑦2 ) ( 𝑥1 , 𝑦1 ) Two- Point
𝒚 = 𝒂𝒙𝟐 + 𝒃𝒙 + 𝒄 (𝑎 ≠ 0) • Graph of 𝑦 = 𝑎𝑥2 + 𝑏𝑥 + 𝑐 is always parabolic. • If 𝑎 > 0, then parabola opens upward. • If a < 0, then parabola opens downward. Quadratic function: Root 1 : Root 2 : 𝛼 = −𝑏 − 𝑏2 − 4𝑎𝑐 2𝑎 𝛽 = −𝑏 + 𝑏2 − 4𝑎𝑐 2𝑎 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0 Sum of roots Difference of roots Product of roots 𝛼 + 𝛽 = − 𝑏 𝑎 𝛽 − 𝛼 = 𝑏2 − 4𝑎𝑐 𝑎 𝛼𝛽 = 𝑐 𝑎 Equation in terms of roots 𝑥2 − 𝛼 + 𝛽 𝑥 + 𝛼𝛽 = 0 Quadratic Function
𝑦 = 𝑎𝑥2 + 𝑐 Parabola(Quadratic Curve) 𝑦 = 𝑎𝑥2 𝑦 = 𝑎𝑥2 𝑎 > 0
Plane Angle in Radian 𝜽 = 𝒂𝒓𝒄 𝒍𝒆𝒏𝒈𝒕𝒉 𝒓𝒂𝒅𝒊𝒖𝒔 𝒐𝒇 𝒄𝒊𝒓𝒄𝒍𝒆 = 𝒍 𝒓 (𝒓𝒂𝒅) System - I Degree (°) Minute (′) Second (′′) System - II Radian (𝑟𝑎𝑑) 1° = 60′ 1′ = 60′′ 1° = 2𝜋 360 𝑟𝑎𝑑 𝒊) 540° = 2𝜋 × 540 360 = 3𝜋 𝑟𝑎𝑑 𝑑𝑒𝑔𝑟𝑒𝑒 𝑡𝑜 𝑟𝑎𝑑𝑖𝑎𝑛 × 2𝜋 360 𝑟𝑎𝑑𝑖𝑎𝑛 𝑡𝑜 𝑑𝑒𝑔𝑟𝑒𝑒 × 360 2𝜋 𝑖𝑖) 210° = 2𝜋 × 210 360 = 7 6 𝜋 𝑟𝑎𝑑 𝑖𝑖𝑖) 270° = 2𝜋 × 270 360 = 3 2 𝜋 𝑟𝑎𝑑
Trigonometric Ratios csc 𝜃 = 1 sin 𝜃 sec 𝜃 = 1 cos 𝜃 cot 𝜃 = 1 tan 𝜃 sin 𝜃 = 𝑜𝑝𝑝 ℎ𝑦𝑝 cos 𝜃 = 𝑎𝑑𝑗 ℎ𝑦𝑝 tan 𝜃 = 𝑜𝑝𝑝 𝑎𝑑𝑗 If A + B = 90°, then 𝟏. sin 𝑨 = cos 𝑩 𝟐. tan 𝑨 = cot 𝑩 𝟑. sec 𝑨 = csc 𝑩
Sine, Cosine and Tangent Curve Graph of 𝑦 = sin 𝑥 The period of sin 𝑥 is 2𝜋. −1 < sin 𝑥 < +1 Graph of 𝑦 = cos 𝑥 The period of cos 𝑥 is 2𝜋. −1 < cos 𝑥 < +1 Graph of 𝑦 = tan 𝑥 The period of tan 𝑥 is 𝜋. tan 𝑥 𝜖 (−∞, ∞)
sin(𝐴 ± 𝐵) = sin 𝐴 cos 𝐵 ± cos 𝐴 sin 𝐵 cos(𝐴 ± 𝐵) = cos 𝐴 cos 𝐵 ∓ sin 𝐴 sin 𝐵 tan(𝐴 ± 𝐵) = tan 𝐴 ± tan 𝐵 1 ∓ tan 𝐴 tan 𝐵 Additive Multiple angle sin 2𝐴 = 2 sin 𝐴 cos 𝐴 cos 2𝐴 = cos2 𝐴 − sin2 𝐴 = 1 − 2 sin2 𝐴 = 2 cos2 𝐴 − 1 tan 2𝐴 = 2 tan 𝐴 1 – tan2 𝐴
sin 90° − 𝜃 = + cos 𝜃 cos 90° − 𝜃 = + sin 𝜃 tan 90° − 𝜃 = + cot 𝜃 csc 90° − 𝜃 = + sec 𝜃 sec 90° − 𝜃 = + csc 𝜃 cot 90° − 𝜃 = + tan 𝜃 sin 180° − 𝜃 = + sin 𝜃 cos 180° − 𝜃 = − cos 𝜃 tan 180° − 𝜃 = − tan 𝜃 csc 180° − 𝜃 = + csc𝜃 sec 180° − 𝜃 = − sec 𝜃 cot 180° − 𝜃 = − cot 𝜃 Summary
𝑓(𝑥) = 𝑎𝑥 Where, • 𝑎 → base & 𝑎 > 0 & 𝑎 ≠ 1 • 𝑥 → any real number 𝑦 = 𝑒𝑥 Euler’s number 𝑒 ≈ 2.718 Special case: Natural Exponential Function Exponential Function
• y = log𝑎 𝑥 is logarithmic function where 𝑎 > 0 and 𝑎 ≠ 1 • Product Rule : log𝑎 𝑚𝑛 = log𝑎 𝑚 + log𝑎 𝑛 • Quotient Rule : log𝑎 𝑚 𝑛 = log𝑎 𝑚 − log𝑎 𝑛 • Power Rule : log𝑎 𝑚𝑛 = 𝑛 log𝑎 𝑚 • log𝑎 1 = 0 • log𝑎 0 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑 • log𝑎 𝑎 = 1 Logarithmic Function
Logarithmic Function base = e, 𝑓 𝑥 = 𝑙𝑜𝑔𝑒 𝑥 = 𝑙𝑛 𝑥 base = 10, 𝑓 𝑥 = 𝑙𝑜𝑔10 𝑥 𝑎 > 0 𝑎 ≠ 1 𝑥 > 0 𝑦 = 𝑙𝑜𝑔𝑎 𝑥 𝑥 = 𝑎𝑦 Common logarithm Natural Logarithm Graph of Logarithmic Function 𝑦 = 𝑙𝑜𝑔𝑏 𝑥 b > 1 0 < b <1
If log10 2 = 0.3010 and log10 3 = 0.4771 then find the approximate value of log10 36. log10 36 = log10(9 × 4) = log10 9 + log10 4 = log10 32 + log10 22 = 2 log10 3 + 2 log10 2 = 2 × 0.4771 + 2 × 0.3010 = 1.5562 log10 36 = 1.5562
Summary 𝑚 = tan 𝜃 Tangen t Secant • The slope of a line gives the measure of its steepness and direction. • Slope is generally denoted by m. • In the case of a secant, the line intersects the curve at two points. • A tangent is defined as a line touching the circle at only one point.. 𝑥 𝑦 −𝑥 −𝑦 θ L
Summary 𝑦 𝑥 𝐴 Δ𝑥 Δ𝑦 𝐵 𝜃 𝑦 𝑥 Tangent 𝜃 tan 𝜃 = 𝑦2 − 𝑦1 𝑥2 − 𝑥1 = Δ𝑦 Δ𝑥 Change in y w.r.t change in x, 𝑦2 − 𝑦1 → Difference in 𝑦 𝑥2 − 𝑥1 → Difference in 𝑥 𝑥 = independent variable 𝑦 = dependent variable 𝑑𝑦 𝑑𝑥 = lim ∆𝑥→0 Δ𝑦 Δ𝑥 , Where ∆𝑥 → 0 Slope = tan 𝜃 = 𝑑𝑦 𝑑𝑥 𝑑𝑦 𝑑𝑥 = Differentiation of 𝒚 w.r.t 𝒙
Differentiation • Power rule for differentiation 𝑑 𝑑𝑥 (𝑥𝑛) = 𝑛𝑥𝑛−1 • Differentiation of any constant is zero. 𝑥 = independent variable 𝑦 = dependent variable 𝑦 𝑑𝑦 𝑑𝑥 𝑥1 1 𝑥 1 2 𝑥 4 𝑥 1 4 𝑥− 3 4 1 𝑥 − 1 𝑥2 𝑦 𝑑𝑦 𝑑𝑥 1 𝑥3 − 3 𝑥4 1 𝑥7 − 7 𝑥8 1 𝑥16 − 16 𝑥17 5 𝑥 − 5 𝑥2
Trigonometric, Exponential and Logarithmic Functions (sin 𝑥) = cos 𝑥; 𝑑 𝑑𝑥 (1) (tan 𝑥) = sec2 𝑥; 𝑑 𝑑𝑥 (3) (sec 𝑥) = sec 𝑥 tan 𝑥 𝑑 𝑑𝑥 (5) (𝑐𝑜𝑠𝑒𝑐 𝑥) = − cosec 𝑥 cot 𝑥 𝑑 𝑑𝑥 (6) (cot 𝑥) = − cosec2 𝑥 𝑑 𝑑𝑥 (4) (cos 𝑥) = − sin 𝑥; 𝑑 𝑑𝑥 (2) Exponential Functions Trigonometric 𝑑 𝑑𝑥 𝑒𝑥 = 𝑒𝑥 Logarithmic 𝑑 𝑑𝑥 ln 𝑥 = 1 𝑥 𝑑 𝑑𝑥 𝑐 = 0 𝑑 𝑑𝑥 𝑘𝑥 = 𝑘 Constant
Summary • Derivative of constant times a function 𝑑 𝑑𝑥 𝑎𝑓 𝑥 = 𝑎 𝑑𝑓 𝑥 𝑑𝑥 • Derivative of a sum or a difference of two functions • Derivative of a constant term 𝒅 𝒅𝒙 𝒂 = 𝟎 𝒅 𝒅𝒙 𝒇 𝒙 ± 𝒈(𝒙) = 𝒅𝒇 𝒙 𝒅𝒙 ± 𝒅𝒈 𝒙 𝒅𝒙
360° 270° 180° 90° -90° -180° -270° -360° -2π 2π 3π/2 π π/2 0 -π/2 -π -3π/2 0 1 0. 5 -0.5 -1 𝑦 = sin 𝑥 x 𝐴 𝐵 360° 270° 180° 90° -90° -180° -270° -360° 𝑦 = cos 𝑥 x -2π 2π 3π/2 π π/2 0 -π/2 -π -3π/2 1 0. 5 -0.5 -1 0 𝐴 𝐵 𝑦 = sin 𝑥 𝑑𝑦 𝑑𝑥 = cos 𝑥
A C D 2 𝑐𝑜𝑠 𝑥 − 6 sec tan 𝑥 2 sin 𝑥 − 6 sec 𝑥 tan 𝑥 2 sin 𝑥 − 6 sec 𝑥 tan 𝑥 Differentiate f (𝑥) = 2 cos (𝑥) − 6 sec (𝑥) + 3 𝑑 𝑑𝑥 cot 𝑥 = − csc2 𝑥 𝑑 𝑑𝑥 (sec 𝑥) = sec 𝑥 tan 𝑥 𝑑 𝑑𝑥 csc 𝑥 = − csc 𝑥 cot 𝑥 𝑑 𝑑𝑥 (sin 𝑥) = cos 𝑥 𝑑 𝑑𝑥 (tan 𝑥) = sec2 𝑥 𝑑 𝑑𝑥 cos 𝑥 = − sin 𝑥 B −2 sin 𝑥 − 6 sec 𝑥 tan 𝑥
𝑑 𝑓 𝑥 . 𝑔(𝑥) 𝑑𝑥 = 𝑑𝑓 𝑥 𝑑𝑥 𝑔 𝑥 + 𝑑𝑔 𝑥 𝑑𝑥 𝑓 𝑥 = 𝑓′ 𝑥 𝑔 𝑥 + 𝑔′ 𝑥 𝑓(𝑥) Product Rule Example: 𝑄1. 𝑦 = 𝑥 sin 𝑥 𝑑𝑦 𝑑𝑥 = 𝑥 sin 𝑥 ′ + sin 𝑥 (𝑥)′ 𝑑𝑦 𝑑𝑥 = 𝑥 cos 𝑥 + sin 𝑥 𝑄2. 𝑦 = 𝑥3 sin 𝑥 𝑑𝑦 𝑑𝑥 = 𝑥3 sin 𝑥 ′ + sin 𝑥 (𝑥3 )′ 𝑑𝑦 𝑑𝑥 = 𝑥3 cos 𝑥 + 3𝑥2 sin 𝑥 𝑄3. 𝑦 = 𝑥2 𝑒𝑥 𝑑𝑦 𝑑𝑥 = 𝑥2 𝑒𝑥 ′ + 𝑒𝑥 (𝑥2 )′ 𝑑𝑦 𝑑𝑥 = 𝑥2 𝑒𝑥 + 2𝑥𝑒𝑥 𝑄4. 𝑦 = cos 𝑥 . log𝑒 𝑥 𝑑𝑦 𝑑𝑥 = cos 𝑥 log𝑒 𝑥 ′ + log𝑒 𝑥 (cos 𝑥)′ 𝑑𝑦 𝑑𝑥 = cos 𝑥 𝑥 − sin 𝑥 . log𝑒 𝑥
Differentiate 𝑓 𝑥 = tan 𝑥 sec 𝑥 w.r.t 𝑥. 𝑑 𝑑𝑥 cot 𝑥 = − csc2 𝑥 𝑑 𝑑𝑥 (sec 𝑥) = sec 𝑥 tan 𝑥 𝑑 𝑑𝑥 csc𝑥 = − csc𝑥 cot 𝑥 𝑑 𝑑𝑥 (sin 𝑥) = cos 𝑥 𝑑 𝑑𝑥 (tan 𝑥) = sec2 𝑥 𝑑 𝑑𝑥 cos 𝑥 = − sin 𝑥 𝑑𝑦 𝑑𝑥 = 𝑡𝑎𝑛𝑥 s𝑒𝑐 𝑥 ′ + s𝑒𝑐 𝑥 (𝑡𝑎𝑛𝑥)′ 𝑑𝑦 𝑑𝑥 = (𝑠𝑒𝑐𝑥 𝑡𝑎𝑛𝑥) tan 𝑥 + sec 𝑥( 𝑠𝑒𝑐𝑥2 ) 𝑑𝑦 𝑑𝑥 = (𝑠𝑒𝑐𝑥 𝑡𝑎𝑛𝑥2 + 𝑠𝑒𝑐𝑥3 ) A sec 𝑥 tan2 𝑥 B 𝑥 tan2 𝑥 + sec3 𝑥 C sec 𝑥 tan2 𝑥 + sec3 𝑥 D sec 𝑥 + sec3 𝑥
𝑑 𝑑𝑥 𝑓 𝑥 𝑔 𝑥 = 𝑑𝑓 𝑥 𝑑𝑥 𝑔 𝑥 − 𝑑𝑔 𝑥 𝑑𝑥 𝑓 𝑥 (𝑔 𝑥 )2 Division Rule Derivative of tan 𝑥 w.r.t 𝑥 𝑑 𝑑𝑥 tan 𝑥 = 𝑑 𝑑𝑥 sin 𝑥 cos 𝑥 = cos 𝑥. 𝑑 𝑑𝑥 sin 𝑥 − sin 𝑥. 𝑑 𝑑𝑥 cos 𝑥 cos2𝑥 = cos2 𝑥 + sin2 𝑥 cos2𝑥 = 1 + tan2 𝑥 = sec2 𝑥 Example: Differentiate: 𝑓 𝑥 = 1 1−𝑥2 w.r.t to 𝑥 using division rule. 𝑓 𝑥 = 1 1 − 𝑥2 𝑓′ 𝑥 = 1 − 𝑥2 0 − 1 −2𝑥 1 − 𝑥2 2 = 2𝑥 1−𝑥2 2
⇒ 𝑑𝑓 𝑑𝑥 = 𝑑𝑓 𝑑𝑔 . 𝑑𝑔 𝑑𝑥 ⇒ 𝑓′(𝑥) = 𝑓′(𝑔) 𝑔′(𝑥) Example of a Composite function • 𝑓 𝑔 = sin 𝑔 • 𝑔 𝑥 = 𝑥2 • 𝑓 𝑔 𝑥 = sin(𝑥2 ) If 𝑓 = 𝑓(𝑔); 𝑔 = 𝑔(𝑥), then it can be written as 𝑓(𝑔(𝑥)) Chain Rule Differentiate: 𝑓 𝑥 = 𝑒sin ln 𝑥 w.r.t to 𝑥 using to chain rule. 𝑓 𝑥 = 𝑒sin ln 𝑥 𝑓′ (𝑥) = 𝑑 𝑑𝑥 𝑒sin ln 𝑥 = 𝑒sin ln 𝑥 . cos ln 𝑥 . 1 𝑥 Differentiate: 𝑔 𝑡 = 1 4𝑡2−3𝑡+2 2w.r.t to 𝑥 using chain rule. 𝑔 𝑡 = 1 4𝑡2 − 3𝑡 + 2 2 𝑔′ (𝑡) = 𝑑 𝑑𝑡 4𝑡2 − 3𝑡 + 2 −2 = − (16𝑡 − 6) 4𝑡2 − 3𝑡 + 2 3
𝑦 = cos 𝑥2 𝑑𝑦 𝑑𝑥 = 𝑑 𝑑𝑥 cos 𝑥2 = −sin 𝑥2 ∙ 2𝑥 ⇒ 𝑑2 𝑦 𝑑𝑥2 = −2 sin 𝑥2 ∙ 𝑑 𝑑𝑥 𝑥 + 𝑥. 𝑑 𝑑𝑥 sin 𝑥2 ⇒ 𝑑2 𝑦 𝑑𝑥2 = −2 sin 𝑥2 ∙ 1 + cos 𝑥2 ∙ 2𝑥 ∙ 𝑥 ⇒ 𝑑𝑦 𝑑𝑥 = −2𝑥 sin 𝑥2 ⇒ 𝑑2 𝑦 𝑑𝑥2 = −4𝑥2 cos 𝑥2 − 2 sin 𝑥2 Successive or Double Differentiation 𝑦 = cos 𝑥2 𝑑𝑦 𝑑𝑥 = 𝑑 𝑑𝑥 cos 𝑥2 = −2𝑥 sin 𝑥2 𝑑2 𝑦 𝑑𝑥2 = −2 𝑑 𝑑𝑥 𝑥 sin 𝑥2 = −2[sin 𝑥2 + 2𝑥2 cos 𝑥2 ] ⇒ 𝑑2 𝑦 𝑑𝑥2 = −2[sin 𝑥2 + 2𝑥2 cos 𝑥2 ]
Applications of Differentiation Variation of one quantity with respect to another 1 Slope of curve 2 Maxima and minima of any function 3
If the motion of a particle is represented by, 𝑠 = 𝑡3 + 𝑡2 − 𝑡 + 2 𝑚. Find the magnitude of position, velocity & acceleration of the particle at 𝑡 = 2 𝑠. At 𝑡 = 2 𝑠 Position = 12𝑚 Position 𝑠 = 𝑡3 + 𝑡2 − 𝑡 + 2 Velocity = 𝑑𝑠 𝑑𝑡 = 3𝑡2 + 2𝑡 − 1 Acceleration = 𝑑2𝑠 𝑑𝑡2 = 6𝑡 + 2 Velocity = 15 𝑚/𝑠 Acceleration = 14 𝑚/𝑠2
• for 𝑥1 < 𝑥2, We observe 𝑓 𝑥1 < 𝑓(𝑥2) 𝑓(𝑥1) 𝑓(𝑥2) 𝑥2 𝑥1 𝑥 𝑦 Increasing function • 𝑑𝑦 𝑑𝑥 > 0 𝑓(𝑥2) 𝑓(𝑥1) 𝑥2 𝑥1 𝑥 𝑦 Decreasing function • for 𝑥1 < 𝑥2, We observe 𝑓 𝑥1 > 𝑓(𝑥2) • 𝑑𝑦 𝑑𝑥 < 0 Increasing and Decreasing Functions
Critical Points 𝑦 𝑥 Critical Points Slope of Tangent at critical points tan 𝜃 = tan 0° = 0
Find the critical points of the function 𝑓 𝑥 = 𝑥4 − 8𝑥2 . A B C D −2, −16 , 0,0 , (2, −16) −2, −16 , 0,0 , (4,128) −4,128 , 0,0 , (2, −16) −4,128 , 0,0 , (4,128) 𝑓′ 𝑥 = 0 ⇒ 4𝑥3 − 16𝑥 = 0 ⇒ 4𝑥 𝑥2 − 4 = 0 ⇒ 4𝑥 𝑥 − 2 (𝑥 + 2) = 0 ⇒ 𝑥 = 0,2, −2 𝑓 −2 = −16 𝑓 0 = 0 𝑓 2 = −16 So, critical points are: −2, −16 , 0,0 , (2, −16)
At Maxima • 𝑑𝑦 𝑑𝑥 = 0 • 𝑑2𝑦 𝑑𝑥2 < 0 At Minima • 𝑑𝑦 𝑑𝑥 = 0 • 𝑑2𝑦 𝑑𝑥2 > 0 Critical Points
Find the maxima and minima for function 𝑦 = 𝑥3 − 3𝑥 + 2 1. Put 𝑑𝑦 𝑑𝑥 = 0 𝑑𝑦 𝑑𝑥 = 3𝑥2 − 3 = 0 ⇒ 𝑥 = +1 & 𝑥 = −1 𝑑2 𝑦 𝑑𝑥2 = 6𝑥 For 𝑥 = +1 𝑑2𝑦 𝑑𝑥2 = +6 > 0 For 𝑥 = −1 𝑑2𝑦 𝑑𝑥2 = −6 < 0 ⇒ Minima ⇒ Maxima 2. Find 𝑑2𝑦 𝑑𝑥2 for each value of 𝑥
A ball is thrown into air. Its height at any instant of time 𝑡 is given by ℎ = 3 + 14𝑡 − 5𝑡2. What is the maximum height attained by the ball?( 𝑡 is in second) Given : height ℎ at time 𝑡, ℎ = 3 + 14𝑡 − 5𝑡2 Say at some time 𝑡 the ball achieves maximum height, ⇒ 𝑑ℎ 𝑑𝑡 = 14 − 10𝑡 = 0 ⇒ 𝑡 = 7/5 𝑠𝑒𝑐𝑜𝑛𝑑 Second derivative of ℎ, 𝑑2 ℎ 𝑑𝑡2 = −10, − 𝑣𝑒 Hence, at 𝑡 = 7/5 𝑠𝑒𝑐𝑜𝑛𝑑 height will be maximum ℎ𝑚𝑎𝑥 = 3 + 14 7 5 − 5 7 5 2 = 𝟏𝟐. 𝟖 𝑼𝒏𝒊𝒕
Integration Mathematically Integration and differentiation are inverse of each other. • 𝑑 𝑑𝑥 𝑐 = 0 • 𝑑 𝑑𝑥 [𝑘𝑥] = 𝑘 • 𝑑 𝑑𝑥 sin 𝑥 = cos 𝑥 • 𝑑 𝑑𝑥 cos 𝑥 = − sin 𝑥 • 𝑑 𝑑𝑥 tan 𝑥 = sec2 𝑥 • ‫׬‬ 0 𝑑𝑥 = 𝑐 • ‫׬‬ 𝑘 𝑑𝑥 = 𝑘𝑥 + 𝑐 • ‫׬‬cos 𝑥 𝑑𝑥 = sin 𝑥 + 𝑐 • ‫׬‬ − sin 𝑥 𝑑𝑥 = cos 𝑥 + 𝑐 • ‫׬‬ sec2 𝑥 𝑑𝑥 = tan 𝑥 + 𝑐
Integration of polynomial function Properties of Integration • ‫׬‬ 𝑥𝑛 𝑑𝑥 = 𝑥𝑛+1 𝑛+1 + 𝑐 𝑛 ≠ −1 • ‫׬‬ 𝑘𝑥𝑛 𝑑𝑥 = 𝑘 𝑥𝑛+1 𝑛+1 + 𝑐 𝑛 ≠ −1 • ‫׬‬ 1 𝑥 𝑑𝑥 = ln 𝑥 + 𝑐 • ‫׬‬ 𝑎 𝑓(𝑥) 𝑑𝑥 = 𝑎 ‫׬‬ 𝑓(𝑥) 𝑑𝑥 • ‫׬‬ 𝑓 𝑥 ± 𝑔 𝑥 𝑑𝑥 = ‫׬‬ 𝑓(𝑥) 𝑑𝑥 ± ‫׬‬ 𝑔 𝑥 𝑑𝑥 𝑎 𝑓1 𝑥 = 𝑥6 ⇒ 𝐼1= න 𝑓1(𝑥) 𝑑𝑥 = 𝑥7 7 + 𝑐 𝑏 𝑓2 𝑥 = −10𝑥3 ⇒ 𝐼2= න 𝑓2(𝑥) 𝑑𝑥 = − 10𝑥4 4 + 𝑐 𝑐 𝑓3 𝑥 = 1 𝑥 ⇒ 𝐼3= න 𝑓3(𝑥) 𝑑𝑥 = ln(𝑥) + 𝑐 𝑑 𝑓4 𝑥 = 5 ⇒ 𝐼4= න 𝑓4(𝑥) 𝑑𝑥 = 5𝑥 + 𝑐
න 𝑥3𝑑𝑥 = ? A B C 2 3 𝑥5 + 𝐶 2 5 𝑥5 + 𝐶 D 2 3 𝑥3 + 𝐶 2 5 𝑥3 + 𝐶 න 𝑥3𝑑𝑥 = 𝑥 3 2 +1 3 2 + 1 + 𝑐 = 2 5 𝑥5 + 𝐶
Summary 𝑑 𝑑𝑥 𝑒𝑥 = 𝑒𝑥 න 𝑒𝑥 𝑑𝑥 = 𝑒𝑥 + 𝑐 𝑑 𝑑𝑥 sin 𝑥 = cos 𝑥 න cos 𝑥 𝑑𝑥 = sin 𝑥 + 𝑐 𝑑 𝑑𝑥 cos 𝑥 = − sin 𝑥 න sin 𝑥 𝑑𝑥 = − cos 𝑥 + 𝑐 𝑑 𝑑𝑥 tan 𝑥 = sec2 𝑥 𝑑 𝑑𝑥 cosec 𝑥 = −cosec 𝑥 cot 𝑥 𝑑 𝑑𝑥 sec 𝑥 = sec 𝑥 tan 𝑥 𝑑 𝑑𝑥 cot 𝑥 = −cosec2 𝑥 න −cosec 𝑥 cot 𝑥 𝑑𝑥 = cosec 𝑥 + 𝑐 න sec 𝑥 tan 𝑥 𝑑𝑥 = sec 𝑥 + 𝑐 න −cosec2 𝑥 𝑑𝑥 = cot 𝑥 + 𝑐 න sec2 𝑥 𝑑𝑥 = tan 𝑥 + 𝑐
Summary If 𝐼 = ‫׬‬ 𝑓 𝑎𝑥 + 𝑏 𝑑𝑥, then assume 𝑋 = 𝑎𝑥 + 𝑏 and integrate further. ⇒ 𝐼 = 1 𝑎 𝐹 𝑋 Now 𝐹(𝑋) = ‫׬‬ 𝑓 𝑋 𝑑𝑥,then Replace 𝑋 with 𝑎𝑥 + 𝑏 𝐼 = 1 𝑎 𝐹(𝑎𝑥 + 𝑏) 𝐼1 = න 2𝑥 + 3 2 𝑑𝑥 = (𝟐𝒙 + 𝟑)3 6 + c 𝐼2 = න sin 2𝑥 + 3 𝑑𝑥 = − cos 𝟐𝒙 + 𝟑 2 + c 𝐼3 = න 1 2𝑥 + 3 𝑑𝑥 = log𝑒 2𝑥 + 3 2 + 𝑐
Integrate the function 𝑓 𝑥 = 𝑥6 − 10𝑥3 + 1 𝑥 + 2 sin 2𝑥 A B C 4𝑥6 − 70𝑥3 + 28(log𝑒 𝑥 − cos 2𝑥 4 + 𝑐 4𝑥7 − 70𝑥4 + 28(log𝑒 𝑥 − cos 2𝑥 28 + 𝑐 D 4𝑥7 − 70𝑥4 + 28(log𝑒 𝑥 − cos 2𝑥 24 + 𝑐 4𝑥7 − 7𝑥4 + 28(log𝑒 𝑥 − cos 2𝑥 24 + 𝑐
Summary 𝐼 = න 2𝑥 𝑥2 + 1 𝑑𝑥 Let 𝑡 = 𝑥2 + 1 ⇒ 𝑑𝑡 𝑑𝑥 = 2𝑥 ⇒ 𝑑𝑡 = 2𝑥 𝑑𝑥 𝐼 = න 𝑑𝑡 𝑡 = log𝑒 𝑡 + 𝑐 ⇒ 𝐼 = log𝑒(𝑥2 + 1) + 𝑐 𝐼 = න sin 𝑥 cos 𝑥 𝑑𝑥 Let 𝑡 = cos 𝑥 ⇒ 𝑑𝑡 𝑑𝑥 = − sin 𝑥 ⇒ −𝑑𝑡 = sin 𝑥 𝑑𝑥 𝐼 = − න 𝑑𝑡 𝑡 = −2 𝑡 + 𝑐 ⇒ 𝐼 = −2 cos 𝑥 + 𝑐
𝐼 = න 𝑥2 sin 𝑥3 𝑑𝑥 =? 𝑡 = 𝑥3 ⇒ 𝑑𝑡 𝑑𝑥 = 3𝑥2 Let ⇒ 1 3 𝑑𝑡 = 𝑥2 𝑑𝑥 𝐼 = න sin 𝑡 . 1 3 𝑑𝑡 𝐼 = − 1 3 cos 𝑡 + 𝑐 𝐼 = − 1 3 cos 𝑥3 + 𝑐
Integration 𝑏 ∶ Upper limit of Integration 𝑎 ∶ Lower limit of Integration
Find the value of A B C D ln 2 + 𝐶 ln 2 ln 2 + 1 + 𝐶 ln 2 + 1 න 1 2 1 𝑥 𝑑𝑥 න 1 2 1 𝑥 𝑑𝑥 = [ln 𝑥 ]1 2 = ln(2)
• Divide the whole area under the curve into infinitely small strips of width 𝑑𝑥. We take a strip of width 𝑑𝑥 at 𝑥 = 𝑥. • Area of shown small strip is 𝑑𝐴 = 𝑓(𝑥)𝑑𝑥 • Total area between the curve and 𝑥 – axis. 𝐴 = න 𝑎 𝑏 𝑓(𝑥)𝑑𝑥 Geometrical Meaning of Integration 𝑓 𝑥 𝑥 𝑑𝑥 𝑎 𝑏
𝐴𝑟𝑒𝑎 𝑜𝑓 𝑇𝑟𝑎𝑝𝑒𝑧𝑖𝑢𝑚 = ℎ(𝑎 + 𝑏) 2 = 4(2 + 6) 2 = 16 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑇𝑟𝑎𝑝𝑒𝑧𝑖𝑢𝑚 = න 2 6 𝑥 𝑑𝑥 = 𝑥2 2 = 16 6 2 Find the area bounded by the lines 𝑦 = 𝑥, 𝑦 = 0 , 𝑥 = 2 and 𝑥 = 6 𝑦 = 𝑥
1) න 𝑎 𝑐 𝑓 𝑥 𝑑𝑥 = න 𝑎 𝑏 𝑓 𝑥 𝑑𝑥 + න 𝑏 𝑐 𝑓 𝑥 𝑑𝑥 Properties of definite Integration 2) න 𝑎 𝑎 𝑓 𝑥 𝑑𝑥 = 0 3) න 𝑎 𝑏 𝑓 𝑥 𝑑𝑥 = − න 𝑏 𝑎 𝑓 𝑥 𝑑𝑥 𝑓 𝑥 𝑥 𝑦 𝑎 𝑏 𝑐
Evaluate the integral: න ൗ −𝜋 2 ൗ 𝜋 2 cos 𝑥 𝑑𝑥 A 2 B 4 C 0 D 3 න 1 4 1 2 𝑥 2 𝑑𝑥 + න 4 1 1 2 𝑥 2 𝑑𝑥 =? A 0 B 4 C 2 D 3
Vector A physical quantity is a vector when, • It has a magnitude and a direction • It obeys laws of vector algebra
Vector Representation Vector 𝑨 Magnitude Direction 𝐴 or Ԧ 𝐴 መ 𝐴
Vector divided by its own magnitude is a vector with unit magnitude and in direction along the parent vector. መ 𝐴 𝐴 Unit Vector 𝐴 |𝐴 | = መ 𝐴 Ԧ 𝐴 = |𝐴 | መ 𝐴
If a car has a velocity of 20 m/s in the east direction as shown in 𝐴 ? (b) How to represent velocity vector of a girl running with 5 m/s in opposite direction 𝐵 = − 05 መ 𝐴 𝑚/𝑠 𝐴 = 20 መ 𝐴 𝑚/𝑠 If the car’s velocity of 20 m/s along east is represented by Ԧ 𝐴, then how can we represent velocity of a girl moving at 5 𝑚/𝑠 in the opposite direction? If the car’s velocity of 20 𝑚/𝑠 along east is represented by Ԧ 𝐴, then how can we represent velocity of a girl moving at 5 𝑚/𝑠 in the opposite direction?
Cartesian Co-ordinate System Representation of Unit Vector along Coordinate Axes
• A vector can be displaced parallel to itself as it does not change its magnitude and direction. • If a vector is rotated through an angle other than multiple of 2𝜋 (or 360°) the vector changes. Properties of a Vector Angle between Vectors Step 1: Place vectors tail to tail or head to head. Step 2: Measure the smaller angle i.e. angle which is less than 180° between them.
(i) 𝐴 and 𝐵 is 60° (ii) 𝐴 and 𝐶 is 45° (iii) 𝐵 and 𝐶 is 105° Three Vectors are arranged along sides of triangle as shown. Find the angle between the vectors. (i) (ii) (iii)
Multiplication of a Vector with a Scalar
Equal Vectors • Ԧ 𝐴 = 𝐵 = Ԧ 𝐶 Equal magnitude • መ 𝐴 = ෠ 𝐵 = መ 𝐶 Same direction Parallel vectors Angle between any two vectors is 0° or 180° then they are along the same direction or exactly opposite directions. • 𝜃 = 0° or 180° • Ԧ 𝐴 ≠ 𝐵 • Ԧ 𝐴 ∥ 𝐵 Null vector • Ԧ 𝐴 = 0 If two vectors are co-linear or parallel then one can be expressed in the terms of another as, • Ԧ 𝑎 = 𝜆𝑏 (where 𝜆 is constant) Collinear vectors Types of Vectors
Three (or more) vectors are called coplanar vectors if they lie in the same plane. Two (free) vectors can be made coplanar. Coplanar vectors Same in magnitude but opposite in direction. • መ 𝐶 = −෡ 𝐷 • Ԧ 𝐶 = 𝐷 Negative of a vectors Types of Vectors
Magnitude of a Vector and Displacement Vector Displacement Vector is the change in position vector. Ԧ 𝑠 = 𝑥2 − 𝑥1 Ƹ 𝑖 + 𝑦2 − 𝑦1 Ƹ 𝑗 + (𝑧2 − 𝑧1)෠ 𝑘 Ԧ 𝑟 = 𝑎 Ƹ 𝑖 + 𝑏 Ƹ 𝑗 Ԧ 𝑟 =Length of line 𝑂𝐴 Ԧ 𝑟 = 𝑎2 + 𝑏2
Position vector gives us the position of a point in magnitude and direction. Ԧ 𝑟1 = 𝑎 Ƹ 𝑖 + 𝑏 Ƹ 𝑗 Position Vector in 2D and 3D Position vector Ԧ 𝑟1 = 𝑥1 Ƹ 𝑖 + 𝑦1 Ƹ 𝑗 + 𝑧1 ෠ 𝑘 Ԧ 𝑟2 = 𝑥2 Ƹ 𝑖 + 𝑦2 Ƹ 𝑗 + 𝑧2 ෠ 𝑘 Magnitude Ԧ 𝑟1 = 𝑥1 2 + 𝑦1 2 + 𝑧1 2 Ԧ 𝑟2 = 𝑥2 2 + 𝑦2 2 + 𝑧2 2 Ԧ 𝑟1 = 𝑎2 + 𝑏2
Magnitude 𝑅 = 𝑅𝑥 Ƹ 𝑖 + 𝑅𝑦 Ƹ 𝑗 Direction 𝑅 = 𝑅𝑥 2 + 𝑅𝑦 2 tan 𝜃 = 𝑅𝑦 𝑅𝑥 𝑦 𝑥 𝑅 𝑂 𝜃 Magnitude and Direction 𝑅 = 𝑅𝑥 + 𝑅𝑦 𝑅 = 𝑅𝑥 Ƹ 𝑖 + 𝑅𝑦 Ƹ 𝑗 𝑅 = 𝑅 cos 𝜃 Ƹ 𝑖 + 𝑅 sin 𝜃 Ƹ 𝑗
𝑦 𝑥 𝐴 = 30 𝑂 𝑦 𝐵 = 80 𝑂 Fill in the Blanks Ԧ 𝐴 = Ƹ 𝑖 + Ƹ 𝑗 −15 2 15 2 𝐵 = Ƹ 𝑖 + Ƹ 𝑗 −40 −40 3 𝑥
Two friends 𝐴 and 𝐵 are running parallel to each other. Velocity of 𝐴 is 3 Ƹ 𝑖 + 4 Ƹ 𝑗 𝑚/𝑠 and speed of 𝐵 is 20 𝑚/𝑠. Find the velocity of 𝐵? 𝐵 = 𝐵 ෠ 𝐵 𝐵 = 20 × 3 Ƹ 𝑖 + 4 Ƹ 𝑗 5 Ԧ 𝐴 = 32 + 42 = 5 Unit vector along Ԧ 𝐴, መ 𝐴 = 3 Ƹ 𝑖 + 4 Ƹ 𝑗 5 𝐵 = 12 Ƹ 𝑖 + 16 Ƹ 𝑗 መ 𝐴 = ෠ 𝐵 መ 𝐴 = Ԧ 𝐴 Ԧ 𝐴
Steps for vector Addition 𝑦 𝑥 Ԧ 𝑃 𝜃 ∅ 𝑃𝑥 𝑄𝑦 𝑄𝑥 𝑃𝑦 𝑄 Ԧ 𝑃 = 𝑃𝑥 Ƹ 𝑖 + 𝑃𝑦 Ƹ 𝑗 𝑄 = 𝑄𝑥 Ƹ 𝑖 + 𝑄𝑦 Ƹ 𝑗 Ԧ 𝑃 = 𝑃 cos 𝜃 Ƹ 𝑖 + 𝑃 sin 𝜃 Ƹ 𝑗 𝑄 = 𝑄 cos 𝜙 Ƹ 𝑖 + 𝑄 sin 𝜙 Ƹ 𝑗 Ԧ 𝑃 + 𝑄 = 𝑃𝑥 + 𝑄𝑥 Ƹ 𝑖 + 𝑃𝑦 + 𝑄𝑦 Ƹ 𝑗 Ԧ 𝑃 + 𝑄 = 𝑃 cos 𝜃 + 𝑄 cos 𝜙 Ƹ 𝑖 + 𝑃 sin 𝜃 + 𝑄 sin 𝜙 Ƹ 𝑗
Ԧ 𝑃 + 𝑄 = 9 2 Ƹ 𝑗 + 9 3 2 Ƹ 𝑖 + 3 Ƹ 𝑗 − 3 3 Ƹ 𝑖 Ԧ 𝑃 + 𝑄 = 9 2 Ƹ 𝑗 + 3 Ƹ 𝑗 + 9 3 2 Ƹ 𝑖 − 3 3 Ƹ 𝑖 Ԧ 𝑃 + 𝑄 = 15 2 Ƹ 𝑗 + 3 3 2 Ƹ 𝑖 30° 𝑦 𝑥 6 30° 3 3 3 9 9 3 2 9 2 If Ԧ 𝑃 = 9 units and 𝑄 = 6 units, then find Ԧ 𝑃 + 𝑄.
Laws of Vector Addition/Subtraction Laws of Vector Addition/Subtraction Triangle Law Parallelogram Law Polygon Law
𝑅 = 𝐴2 + 𝐵2 + 2𝐴𝐵 cos 𝜃 𝑦 𝜃 𝐵 𝑅 𝐴 Ƹ 𝑖 𝑥 Ԧ 𝐴 𝐵 sin 𝜃 Ƹ 𝑗 𝑅 𝐵 cos 𝜃 Ƹ 𝑖 𝑅 = Ԧ 𝐴 + 𝐵 = 𝐴 + 𝐵 cos 𝜃 Ƹ 𝑖 + 𝐵 sin 𝜃 Ƹ 𝑗 𝑅 = Ԧ 𝐴 + 𝐵 𝑅 = 𝐴 + 𝐵 cos 𝜃 2 + 𝐵 sin 𝜃 2 𝑅 = 𝐴2 + 𝐵2 cos2 𝜃 + 2𝐴𝐵 cos 𝜃 + 𝐵2 sin2 𝜃 Triangle Law
When two vectors with common origin represent two adjacent sides of a parallelogram in magnitude and direction, then the resultant vector is represented both in magnitude and direction by the diagonal passing through that point. 𝑦 𝑅 = 𝐴2 + 𝐵2 + 2𝐴𝐵 cos 𝜃 Here 𝛼 is the angle between Ԧ 𝐴 and 𝑅 tan 𝛼 = 𝐵 sin 𝜃 Ԧ 𝐴 + 𝐵 cos 𝜃 Parallelogram Law
𝑅3 𝐷 Ԧ 𝐴 𝐵 Ԧ 𝐶 • Join vectors in same order (head to tail). • Resultant Vector – closing side of polygon in opposite order 𝑅3 = Ԧ 𝐴 + 𝐵 + Ԧ 𝐶 + 𝐷 Polygon Law of Vector Addition
• Ԧ 𝐴 + 𝐵 : Ԧ 𝐴 and 𝐵 must be same physical quantities. • Commutative Property : Ԧ 𝐴 + 𝐵 = 𝐵 + Ԧ 𝐴 • Associative Property : ( Ԧ 𝐴 + 𝐵) + Ԧ 𝐶 = Ԧ 𝐴 + (𝐵 + Ԧ 𝐶) Properties of Vector Addition
𝜃 𝐵 𝐴 Ԧ 𝐴. 𝐵 = 𝐴𝐵 cos 𝜃 The dot product of two vectors 𝐴 and 𝐵 is equal to the products of their magnitude times the cosine of angle between them. Scalar / Dot Product Ԧ 𝐴. 𝐵 = Ԧ 𝐴 𝐵 cos 𝜃 𝐴 = 𝐴𝑥 Ƹ 𝑖 + 𝐴𝑦 Ƹ 𝑗 + 𝐴𝑧 ෠ 𝑘 , 𝐵 = 𝐵𝑥 Ƹ 𝑖 + 𝐵𝑦 Ƹ 𝑗 + 𝐵𝑧 ෠ 𝑘 𝐴 . 𝐵 = 𝐴𝑥𝐵𝑥 + 𝐴𝑦𝐵𝑦 + 𝐴𝑧𝐵𝑧
60° 𝐵 𝐴 If 𝐴 = 2, 𝐵 = 3 and the angle between 𝐴 and 𝐵 is 60°. Find 𝐴 . 𝐵 𝐴 .𝐵 = |𝐴 ||𝐵 |cos 𝜃 = 2 × 3 × cos 60° = 2 × 3 × 1 2 = 3 Ԧ 𝐴. 𝐵 = 3
• Dot product of two vectors is always (commutative), i.e. 𝐴 . 𝐵 = 𝐵 . 𝐴 • Dot product of two vectors is always distributive, i.e. 𝐴 . 𝐵 + 𝐶 = 𝐴 . 𝐵 + 𝐴 . 𝐶 • Dot product of two vectors will be maximum when they are parallel (i.e., angle between them is zero). Properties of Dot Product 𝐴 . 𝐵 is positive 𝜽 < 𝟗𝟎° ⇒ 𝐂𝐨𝐬 𝜽 > 𝟎 Ԧ 𝐴. 𝐵 = 𝐴𝐵 cos 𝜃 𝜽 = 𝟗𝟎° 𝐴 . 𝐵 is zero 𝐴 . 𝐵 is negative 𝜽 > 𝟗𝟎° ⇒ 𝐂𝐨𝐬 𝜽 < 𝟎 • 𝐴 .𝐴 = 𝐴2 • ො 𝑛 . ො 𝑛 = 1 • Ƹ 𝑖 . Ƹ 𝑗 = Ƹ 𝑗 . ෠ 𝑘 = ෠ 𝑘 . Ƹ 𝑖 = 0
𝐴 .𝐵 = 0 ⟹ 𝐴 𝐵 cos 𝜃 = 0 ⟹ cos 𝜃 = 0 𝜃 = cos−1 0 = 90° 𝐴 𝐵 90° Thus, 𝐴 .𝐵 = 0 is a condition for two vectors to be orthogonal. If the scalar product of two non-zero vectors becomes zero. What is the angle between them?
The cross product of two vectors 𝐴 and 𝐵 is equal to the products of their magnitude times the sine of angle between them and direction perpendicular to the plane containing the two vectors. Ԧ 𝐴 × 𝐵 = Ԧ 𝐴 𝐵 sin 𝜃 ො 𝑛 Ԧ 𝐴 × 𝐵 = Ԧ 𝐴 𝐵 sin 𝜃 Vector / Cross Product Here, ො 𝑛 is a unit vector perpendicular to both Ԧ 𝐴 and 𝐵. Ԧ 𝐴 × 𝐵 = Ԧ 𝐴 𝐵 sin 𝜃 ො 𝑛
To find the direction of Ԧ 𝐴 × 𝐵: • Draw Ԧ 𝐴 and 𝐵 tail-to-tail. • Place the stretched right palm such that the fingers are along Ԧ 𝐴 and when the fingers are closed, they go towards 𝐵. • The direction in which thumb points gives the direction of Ԧ 𝐶 = Ԧ 𝐴 × 𝐵. Right Hand Thumb Rule Ԧ 𝐴 𝐵 Ԧ 𝐴 𝐵 Ԧ 𝐶
Ԧ 𝐴 × 𝐵 = Ԧ 𝐴 𝐵 sin 𝜃 = 5 × 4 × sin 30° = 10 Using the right-hand thumb rule, the direction of Ԧ 𝐴 × 𝐵 is along −𝑧 axis Find the magnitude and direction of Ԧ 𝐴 × 𝐵 and 𝐵 × Ԧ 𝐴, where Ԧ 𝐴 & 𝐵 lie in 𝑥 − 𝑦 plane as shown. The direction of 𝐵 × Ԧ 𝐴 is along +𝑧 axis. ⟹ Ԧ 𝐴 × 𝐵 = −10෠ 𝑘 ⟹ 𝐵 × Ԧ 𝐴 = 10෠ 𝑘 Ԧ 𝐴 𝐵 30° 4 5
In the right-handed coordinate system, the coordinate axes 𝑥, 𝑦 and 𝑧 are chosen such that bending the fingers of the right hand from 𝑥 to 𝑦 will lead the thumb along the 𝑧 −axis. Ƹ 𝑖 Ƹ 𝑗 ෠ 𝑘 Ƹ 𝑖 × Ƹ 𝑗 = ෠ 𝑘, Ƹ 𝑗 × Ƹ 𝑖 = − ෠ 𝑘 Ƹ 𝑗 × ෠ 𝑘 = Ƹ 𝑖, ෠ 𝑘 × Ƹ 𝑗 = − Ƹ 𝑖 ෠ 𝑘 × Ƹ 𝑖 = Ƹ 𝑗, Ƹ 𝑖 × ෠ 𝑘 = − Ƹ 𝑗 Cross Product of Orthogonal Unit Vectors
Cross Product in Component Form Ԧ 𝐴 × 𝐵 = 𝐴𝑥 Ƹ 𝑖 + 𝐴𝑦 Ƹ 𝑗 + 𝐴𝑧 ෠ 𝑘 × 𝐵𝑥 Ƹ 𝑖 + 𝐵𝑦 Ƹ 𝑗 + 𝐵𝑧 ෠ 𝑘 Ԧ 𝐴 × 𝐵 = 𝐴𝑥𝐵𝑥 Ƹ 𝑖 × Ƹ 𝑖 + 𝐴𝑦𝐵𝑥 Ƹ 𝑗 × Ƹ 𝑖 + 𝐴𝑧𝐵𝑥 ෠ 𝑘 × Ƹ 𝑖 +𝐴𝑥𝐵𝑦 Ƹ 𝑖 × Ƹ 𝑗 + 𝐴𝑦𝐵𝑦 Ƹ 𝑗 × Ƹ 𝑗 + 𝐴𝑧𝐵𝑦 ෠ 𝑘 × Ƹ 𝑗 +𝐴𝑥𝐵𝑧 Ƹ 𝑖 × ෠ 𝑘 + 𝐴𝑦𝐵𝑧 Ƹ 𝑗 × ෠ 𝑘 + 𝐴𝑧𝐵𝑧 ෠ 𝑘 × ෠ 𝑘 Ԧ 𝐴 × 𝐵 = (𝐴𝑦𝐵𝑧 − 𝐴𝑧𝐵𝑦) Ƹ 𝑖 − 𝐴𝑥𝐵𝑧 − 𝐴𝑧𝐵𝑥 Ƹ 𝑗 + (𝐴𝑥𝐵𝑦 − 𝐴𝑦𝐵𝑥)෠ 𝑘 Ƹ 𝑖 Ƹ 𝑗 ෠ 𝑘 Ԧ 𝐴 = 𝐴𝑥 Ƹ 𝑖 + 𝐴𝑦 Ƹ 𝑗 + 𝐴𝑧 ෠ 𝑘 𝐵 = 𝐵𝑥 Ƹ 𝑖 + 𝐵𝑦 Ƹ 𝑗 + 𝐵𝑧 ෠ 𝑘
Ԧ 𝐴 × 𝐵 = (𝐴𝑦𝐵𝑧 − 𝐴𝑧𝐵𝑦) Ƹ 𝑖 − 𝐴𝑥𝐵𝑧 − 𝐴𝑧𝐵𝑥 Ƹ 𝑗 + (𝐴𝑥𝐵𝑦 − 𝐴𝑦𝐵𝑥)෠ 𝑘 Ԧ 𝐴 × 𝐵 = 2 Ƹ 𝑖 − 5 Ƹ 𝑗 + 3෠ 𝑘 × 3 Ƹ 𝑖 + 4 Ƹ 𝑗 − 9෠ 𝑘 = 6 Ƹ 𝑖 × Ƹ 𝑖 − 15 Ƹ 𝑗 × Ƹ 𝑖 + 9 ෠ 𝑘 × Ƹ 𝑖 +8 Ƹ 𝑖 × Ƹ 𝑗 − 20 Ƹ 𝑗 × Ƹ 𝑗 + 12 ෠ 𝑘 × Ƹ 𝑗 − 18 Ƹ 𝑖 × ෠ 𝑘 + 45 Ƹ 𝑗 × ෠ 𝑘 − 27 (෠ 𝑘 × ෠ 𝑘) Ԧ 𝐴 × 𝐵 = Ƹ 𝑖 45 − 12 − Ƹ 𝑗 −18 − 9 + ෠ 𝑘 8 − (−15) Ԧ 𝐴 × 𝐵 = 33 Ƹ 𝑖 + 27 Ƹ 𝑗 + 23෠ 𝑘 Find Ԧ 𝐴 × 𝐵 , where Ԧ 𝐴 = 2 Ƹ 𝑖 − 5 Ƹ 𝑗 + 3෠ 𝑘 and 𝐵 = 3 Ƹ 𝑖 + 4 Ƹ 𝑗 − 9෠ 𝑘.

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Neet class 11 12 basic mathematics notes

  • 1.
  • 2.
    Functions • Input variableis independent and output variable is dependent on it. • A function is a mathematical relationship between independent and dependent variables Algebraic Trigonometric Exponential Logarithmic Functions
  • 3.
    Polynomials 𝑦 = 𝑓𝑥 = 𝑎0 + 𝑎1𝑥1 + 𝑎2𝑥2 + 𝑎3𝑥3 + … … . 𝑎𝑛𝑥𝑛 Constants Variables = = Degree of a polynomial Highest power Maximum number of solutions Algebraic Function
  • 4.
    The graphical representationof a linear function is a straight line 1 Linear Function 𝑦 = 𝑓 𝑥 = 𝑎0 + 𝑎1𝑥1 Constants Variables = = = Degree of a polynomial Highest power Maximum number of solutions Equation of line 𝑦 = 𝑚𝑥 + 𝑐 𝑐 → 𝑦-intercept of line = tan 𝜃 = 𝑦2−𝑦1 𝑥2−𝑥1 𝑚 → Slope of line
  • 5.
    Rahul is standing4 𝑚 East and 5 𝑚 North from Ashok. Taking Ashok’s position as origin, represent Rahul’s position on the coordinate plane given below. 𝑦(𝑚) 𝑥(𝑚) Coordinates of Rahul’s position
  • 6.
    Ashok has nowmoved 1 𝑚 East and 1 𝑚 North while Rahul moved 1 𝑚 South. Calculate shortest distance between Rahul and Ashok. Shortest distance = 𝑥2 − 𝑥1 2 + 𝑦2 − 𝑦1 2 Shortest distance = 4 − 1 2 + 4 − 1 2 Shortest distance = 32 + 32 Shortest distance = 3 2 𝑚 𝑥 𝑦 1 1 Rahul→ (4,4) Ashok→ (1,1) 1
  • 7.
    Determine direction ofRahul’s position (4,4) with respect to Ashok (1,1). Slope/Direction = tan 𝜃 = 𝑦2 − 𝑦1 𝑥2 − 𝑥1 tan 𝜃 = 4 − 1 4 − 1 = 1 tan 𝜃 = tan 45° 𝜃 = 45° from +𝑣𝑒 𝑥 −axis 𝑥 𝑦 Rahul→ (4,4) Ashok→ (1,1) 𝜃
  • 8.
    Case ii :c = 0 Passing through origin 𝑥 𝑦 = 𝑥 𝑦 = −𝑥 𝑦 Case i : 𝑚 = 0 Parallel to x-axis 𝑦 = +𝑐 (0, 𝑐) 𝑦 𝑥 𝑐 Special cases
  • 9.
    𝑦 = 𝑚𝑥+ 𝑐 Straight line equation Slope - intercept 𝑥 𝑦 𝜃 𝑐 𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1) Point - slope 𝑥 𝑦 𝜃 ( 𝑥1 , 𝑦1 ) 𝑦 − 𝑦1 = (𝑦2 − 𝑦1) (𝑥2 − 𝑥1) (𝑥 − 𝑥1) ( 𝑥 2 , 𝑦2 ) ( 𝑥1 , 𝑦1 ) Two- Point
  • 10.
    𝒚 = 𝒂𝒙𝟐 +𝒃𝒙 + 𝒄 (𝑎 ≠ 0) • Graph of 𝑦 = 𝑎𝑥2 + 𝑏𝑥 + 𝑐 is always parabolic. • If 𝑎 > 0, then parabola opens upward. • If a < 0, then parabola opens downward. Quadratic function: Root 1 : Root 2 : 𝛼 = −𝑏 − 𝑏2 − 4𝑎𝑐 2𝑎 𝛽 = −𝑏 + 𝑏2 − 4𝑎𝑐 2𝑎 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0 Sum of roots Difference of roots Product of roots 𝛼 + 𝛽 = − 𝑏 𝑎 𝛽 − 𝛼 = 𝑏2 − 4𝑎𝑐 𝑎 𝛼𝛽 = 𝑐 𝑎 Equation in terms of roots 𝑥2 − 𝛼 + 𝛽 𝑥 + 𝛼𝛽 = 0 Quadratic Function
  • 11.
    𝑦 = 𝑎𝑥2 +𝑐 Parabola(Quadratic Curve) 𝑦 = 𝑎𝑥2 𝑦 = 𝑎𝑥2 𝑎 > 0
  • 12.
    Plane Angle inRadian 𝜽 = 𝒂𝒓𝒄 𝒍𝒆𝒏𝒈𝒕𝒉 𝒓𝒂𝒅𝒊𝒖𝒔 𝒐𝒇 𝒄𝒊𝒓𝒄𝒍𝒆 = 𝒍 𝒓 (𝒓𝒂𝒅) System - I Degree (°) Minute (′) Second (′′) System - II Radian (𝑟𝑎𝑑) 1° = 60′ 1′ = 60′′ 1° = 2𝜋 360 𝑟𝑎𝑑 𝒊) 540° = 2𝜋 × 540 360 = 3𝜋 𝑟𝑎𝑑 𝑑𝑒𝑔𝑟𝑒𝑒 𝑡𝑜 𝑟𝑎𝑑𝑖𝑎𝑛 × 2𝜋 360 𝑟𝑎𝑑𝑖𝑎𝑛 𝑡𝑜 𝑑𝑒𝑔𝑟𝑒𝑒 × 360 2𝜋 𝑖𝑖) 210° = 2𝜋 × 210 360 = 7 6 𝜋 𝑟𝑎𝑑 𝑖𝑖𝑖) 270° = 2𝜋 × 270 360 = 3 2 𝜋 𝑟𝑎𝑑
  • 13.
    Trigonometric Ratios csc 𝜃= 1 sin 𝜃 sec 𝜃 = 1 cos 𝜃 cot 𝜃 = 1 tan 𝜃 sin 𝜃 = 𝑜𝑝𝑝 ℎ𝑦𝑝 cos 𝜃 = 𝑎𝑑𝑗 ℎ𝑦𝑝 tan 𝜃 = 𝑜𝑝𝑝 𝑎𝑑𝑗 If A + B = 90°, then 𝟏. sin 𝑨 = cos 𝑩 𝟐. tan 𝑨 = cot 𝑩 𝟑. sec 𝑨 = csc 𝑩
  • 14.
    Sine, Cosine andTangent Curve Graph of 𝑦 = sin 𝑥 The period of sin 𝑥 is 2𝜋. −1 < sin 𝑥 < +1 Graph of 𝑦 = cos 𝑥 The period of cos 𝑥 is 2𝜋. −1 < cos 𝑥 < +1 Graph of 𝑦 = tan 𝑥 The period of tan 𝑥 is 𝜋. tan 𝑥 𝜖 (−∞, ∞)
  • 15.
    sin(𝐴 ± 𝐵)= sin 𝐴 cos 𝐵 ± cos 𝐴 sin 𝐵 cos(𝐴 ± 𝐵) = cos 𝐴 cos 𝐵 ∓ sin 𝐴 sin 𝐵 tan(𝐴 ± 𝐵) = tan 𝐴 ± tan 𝐵 1 ∓ tan 𝐴 tan 𝐵 Additive Multiple angle sin 2𝐴 = 2 sin 𝐴 cos 𝐴 cos 2𝐴 = cos2 𝐴 − sin2 𝐴 = 1 − 2 sin2 𝐴 = 2 cos2 𝐴 − 1 tan 2𝐴 = 2 tan 𝐴 1 – tan2 𝐴
  • 16.
    sin 90° −𝜃 = + cos 𝜃 cos 90° − 𝜃 = + sin 𝜃 tan 90° − 𝜃 = + cot 𝜃 csc 90° − 𝜃 = + sec 𝜃 sec 90° − 𝜃 = + csc 𝜃 cot 90° − 𝜃 = + tan 𝜃 sin 180° − 𝜃 = + sin 𝜃 cos 180° − 𝜃 = − cos 𝜃 tan 180° − 𝜃 = − tan 𝜃 csc 180° − 𝜃 = + csc𝜃 sec 180° − 𝜃 = − sec 𝜃 cot 180° − 𝜃 = − cot 𝜃 Summary
  • 17.
    𝑓(𝑥) = 𝑎𝑥 Where, •𝑎 → base & 𝑎 > 0 & 𝑎 ≠ 1 • 𝑥 → any real number 𝑦 = 𝑒𝑥 Euler’s number 𝑒 ≈ 2.718 Special case: Natural Exponential Function Exponential Function
  • 18.
    • y =log𝑎 𝑥 is logarithmic function where 𝑎 > 0 and 𝑎 ≠ 1 • Product Rule : log𝑎 𝑚𝑛 = log𝑎 𝑚 + log𝑎 𝑛 • Quotient Rule : log𝑎 𝑚 𝑛 = log𝑎 𝑚 − log𝑎 𝑛 • Power Rule : log𝑎 𝑚𝑛 = 𝑛 log𝑎 𝑚 • log𝑎 1 = 0 • log𝑎 0 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑 • log𝑎 𝑎 = 1 Logarithmic Function
  • 19.
    Logarithmic Function base =e, 𝑓 𝑥 = 𝑙𝑜𝑔𝑒 𝑥 = 𝑙𝑛 𝑥 base = 10, 𝑓 𝑥 = 𝑙𝑜𝑔10 𝑥 𝑎 > 0 𝑎 ≠ 1 𝑥 > 0 𝑦 = 𝑙𝑜𝑔𝑎 𝑥 𝑥 = 𝑎𝑦 Common logarithm Natural Logarithm Graph of Logarithmic Function 𝑦 = 𝑙𝑜𝑔𝑏 𝑥 b > 1 0 < b <1
  • 20.
    If log10 2= 0.3010 and log10 3 = 0.4771 then find the approximate value of log10 36. log10 36 = log10(9 × 4) = log10 9 + log10 4 = log10 32 + log10 22 = 2 log10 3 + 2 log10 2 = 2 × 0.4771 + 2 × 0.3010 = 1.5562 log10 36 = 1.5562
  • 21.
    Summary 𝑚 = tan𝜃 Tangen t Secant • The slope of a line gives the measure of its steepness and direction. • Slope is generally denoted by m. • In the case of a secant, the line intersects the curve at two points. • A tangent is defined as a line touching the circle at only one point.. 𝑥 𝑦 −𝑥 −𝑦 θ L
  • 22.
    Summary 𝑦 𝑥 𝐴 Δ𝑥 Δ𝑦 𝐵 𝜃 𝑦 𝑥 Tangent 𝜃 tan 𝜃 = 𝑦2− 𝑦1 𝑥2 − 𝑥1 = Δ𝑦 Δ𝑥 Change in y w.r.t change in x, 𝑦2 − 𝑦1 → Difference in 𝑦 𝑥2 − 𝑥1 → Difference in 𝑥 𝑥 = independent variable 𝑦 = dependent variable 𝑑𝑦 𝑑𝑥 = lim ∆𝑥→0 Δ𝑦 Δ𝑥 , Where ∆𝑥 → 0 Slope = tan 𝜃 = 𝑑𝑦 𝑑𝑥 𝑑𝑦 𝑑𝑥 = Differentiation of 𝒚 w.r.t 𝒙
  • 23.
    Differentiation • Power rulefor differentiation 𝑑 𝑑𝑥 (𝑥𝑛) = 𝑛𝑥𝑛−1 • Differentiation of any constant is zero. 𝑥 = independent variable 𝑦 = dependent variable 𝑦 𝑑𝑦 𝑑𝑥 𝑥1 1 𝑥 1 2 𝑥 4 𝑥 1 4 𝑥− 3 4 1 𝑥 − 1 𝑥2 𝑦 𝑑𝑦 𝑑𝑥 1 𝑥3 − 3 𝑥4 1 𝑥7 − 7 𝑥8 1 𝑥16 − 16 𝑥17 5 𝑥 − 5 𝑥2
  • 24.
    Trigonometric, Exponential and LogarithmicFunctions (sin 𝑥) = cos 𝑥; 𝑑 𝑑𝑥 (1) (tan 𝑥) = sec2 𝑥; 𝑑 𝑑𝑥 (3) (sec 𝑥) = sec 𝑥 tan 𝑥 𝑑 𝑑𝑥 (5) (𝑐𝑜𝑠𝑒𝑐 𝑥) = − cosec 𝑥 cot 𝑥 𝑑 𝑑𝑥 (6) (cot 𝑥) = − cosec2 𝑥 𝑑 𝑑𝑥 (4) (cos 𝑥) = − sin 𝑥; 𝑑 𝑑𝑥 (2) Exponential Functions Trigonometric 𝑑 𝑑𝑥 𝑒𝑥 = 𝑒𝑥 Logarithmic 𝑑 𝑑𝑥 ln 𝑥 = 1 𝑥 𝑑 𝑑𝑥 𝑐 = 0 𝑑 𝑑𝑥 𝑘𝑥 = 𝑘 Constant
  • 25.
    Summary • Derivative ofconstant times a function 𝑑 𝑑𝑥 𝑎𝑓 𝑥 = 𝑎 𝑑𝑓 𝑥 𝑑𝑥 • Derivative of a sum or a difference of two functions • Derivative of a constant term 𝒅 𝒅𝒙 𝒂 = 𝟎 𝒅 𝒅𝒙 𝒇 𝒙 ± 𝒈(𝒙) = 𝒅𝒇 𝒙 𝒅𝒙 ± 𝒅𝒈 𝒙 𝒅𝒙
  • 26.
    360° 270° 180° 90° -90° -180° -270° -360° -2π 2π 3π/2 π π/2 0 -π/2 -π -3π/2 0 1 0. 5 -0.5 -1 𝑦 =sin 𝑥 x 𝐴 𝐵 360° 270° 180° 90° -90° -180° -270° -360° 𝑦 = cos 𝑥 x -2π 2π 3π/2 π π/2 0 -π/2 -π -3π/2 1 0. 5 -0.5 -1 0 𝐴 𝐵 𝑦 = sin 𝑥 𝑑𝑦 𝑑𝑥 = cos 𝑥
  • 27.
    A C D 2 𝑐𝑜𝑠𝑥 − 6 sec tan 𝑥 2 sin 𝑥 − 6 sec 𝑥 tan 𝑥 2 sin 𝑥 − 6 sec 𝑥 tan 𝑥 Differentiate f (𝑥) = 2 cos (𝑥) − 6 sec (𝑥) + 3 𝑑 𝑑𝑥 cot 𝑥 = − csc2 𝑥 𝑑 𝑑𝑥 (sec 𝑥) = sec 𝑥 tan 𝑥 𝑑 𝑑𝑥 csc 𝑥 = − csc 𝑥 cot 𝑥 𝑑 𝑑𝑥 (sin 𝑥) = cos 𝑥 𝑑 𝑑𝑥 (tan 𝑥) = sec2 𝑥 𝑑 𝑑𝑥 cos 𝑥 = − sin 𝑥 B −2 sin 𝑥 − 6 sec 𝑥 tan 𝑥
  • 28.
    𝑑 𝑓 𝑥. 𝑔(𝑥) 𝑑𝑥 = 𝑑𝑓 𝑥 𝑑𝑥 𝑔 𝑥 + 𝑑𝑔 𝑥 𝑑𝑥 𝑓 𝑥 = 𝑓′ 𝑥 𝑔 𝑥 + 𝑔′ 𝑥 𝑓(𝑥) Product Rule Example: 𝑄1. 𝑦 = 𝑥 sin 𝑥 𝑑𝑦 𝑑𝑥 = 𝑥 sin 𝑥 ′ + sin 𝑥 (𝑥)′ 𝑑𝑦 𝑑𝑥 = 𝑥 cos 𝑥 + sin 𝑥 𝑄2. 𝑦 = 𝑥3 sin 𝑥 𝑑𝑦 𝑑𝑥 = 𝑥3 sin 𝑥 ′ + sin 𝑥 (𝑥3 )′ 𝑑𝑦 𝑑𝑥 = 𝑥3 cos 𝑥 + 3𝑥2 sin 𝑥 𝑄3. 𝑦 = 𝑥2 𝑒𝑥 𝑑𝑦 𝑑𝑥 = 𝑥2 𝑒𝑥 ′ + 𝑒𝑥 (𝑥2 )′ 𝑑𝑦 𝑑𝑥 = 𝑥2 𝑒𝑥 + 2𝑥𝑒𝑥 𝑄4. 𝑦 = cos 𝑥 . log𝑒 𝑥 𝑑𝑦 𝑑𝑥 = cos 𝑥 log𝑒 𝑥 ′ + log𝑒 𝑥 (cos 𝑥)′ 𝑑𝑦 𝑑𝑥 = cos 𝑥 𝑥 − sin 𝑥 . log𝑒 𝑥
  • 29.
    Differentiate 𝑓 𝑥= tan 𝑥 sec 𝑥 w.r.t 𝑥. 𝑑 𝑑𝑥 cot 𝑥 = − csc2 𝑥 𝑑 𝑑𝑥 (sec 𝑥) = sec 𝑥 tan 𝑥 𝑑 𝑑𝑥 csc𝑥 = − csc𝑥 cot 𝑥 𝑑 𝑑𝑥 (sin 𝑥) = cos 𝑥 𝑑 𝑑𝑥 (tan 𝑥) = sec2 𝑥 𝑑 𝑑𝑥 cos 𝑥 = − sin 𝑥 𝑑𝑦 𝑑𝑥 = 𝑡𝑎𝑛𝑥 s𝑒𝑐 𝑥 ′ + s𝑒𝑐 𝑥 (𝑡𝑎𝑛𝑥)′ 𝑑𝑦 𝑑𝑥 = (𝑠𝑒𝑐𝑥 𝑡𝑎𝑛𝑥) tan 𝑥 + sec 𝑥( 𝑠𝑒𝑐𝑥2 ) 𝑑𝑦 𝑑𝑥 = (𝑠𝑒𝑐𝑥 𝑡𝑎𝑛𝑥2 + 𝑠𝑒𝑐𝑥3 ) A sec 𝑥 tan2 𝑥 B 𝑥 tan2 𝑥 + sec3 𝑥 C sec 𝑥 tan2 𝑥 + sec3 𝑥 D sec 𝑥 + sec3 𝑥
  • 30.
    𝑑 𝑑𝑥 𝑓 𝑥 𝑔 𝑥 = 𝑑𝑓𝑥 𝑑𝑥 𝑔 𝑥 − 𝑑𝑔 𝑥 𝑑𝑥 𝑓 𝑥 (𝑔 𝑥 )2 Division Rule Derivative of tan 𝑥 w.r.t 𝑥 𝑑 𝑑𝑥 tan 𝑥 = 𝑑 𝑑𝑥 sin 𝑥 cos 𝑥 = cos 𝑥. 𝑑 𝑑𝑥 sin 𝑥 − sin 𝑥. 𝑑 𝑑𝑥 cos 𝑥 cos2𝑥 = cos2 𝑥 + sin2 𝑥 cos2𝑥 = 1 + tan2 𝑥 = sec2 𝑥 Example: Differentiate: 𝑓 𝑥 = 1 1−𝑥2 w.r.t to 𝑥 using division rule. 𝑓 𝑥 = 1 1 − 𝑥2 𝑓′ 𝑥 = 1 − 𝑥2 0 − 1 −2𝑥 1 − 𝑥2 2 = 2𝑥 1−𝑥2 2
  • 31.
    ⇒ 𝑑𝑓 𝑑𝑥 = 𝑑𝑓 𝑑𝑔 . 𝑑𝑔 𝑑𝑥 ⇒ 𝑓′(𝑥) =𝑓′(𝑔) 𝑔′(𝑥) Example of a Composite function • 𝑓 𝑔 = sin 𝑔 • 𝑔 𝑥 = 𝑥2 • 𝑓 𝑔 𝑥 = sin(𝑥2 ) If 𝑓 = 𝑓(𝑔); 𝑔 = 𝑔(𝑥), then it can be written as 𝑓(𝑔(𝑥)) Chain Rule Differentiate: 𝑓 𝑥 = 𝑒sin ln 𝑥 w.r.t to 𝑥 using to chain rule. 𝑓 𝑥 = 𝑒sin ln 𝑥 𝑓′ (𝑥) = 𝑑 𝑑𝑥 𝑒sin ln 𝑥 = 𝑒sin ln 𝑥 . cos ln 𝑥 . 1 𝑥 Differentiate: 𝑔 𝑡 = 1 4𝑡2−3𝑡+2 2w.r.t to 𝑥 using chain rule. 𝑔 𝑡 = 1 4𝑡2 − 3𝑡 + 2 2 𝑔′ (𝑡) = 𝑑 𝑑𝑡 4𝑡2 − 3𝑡 + 2 −2 = − (16𝑡 − 6) 4𝑡2 − 3𝑡 + 2 3
  • 32.
    𝑦 = cos𝑥2 𝑑𝑦 𝑑𝑥 = 𝑑 𝑑𝑥 cos 𝑥2 = −sin 𝑥2 ∙ 2𝑥 ⇒ 𝑑2 𝑦 𝑑𝑥2 = −2 sin 𝑥2 ∙ 𝑑 𝑑𝑥 𝑥 + 𝑥. 𝑑 𝑑𝑥 sin 𝑥2 ⇒ 𝑑2 𝑦 𝑑𝑥2 = −2 sin 𝑥2 ∙ 1 + cos 𝑥2 ∙ 2𝑥 ∙ 𝑥 ⇒ 𝑑𝑦 𝑑𝑥 = −2𝑥 sin 𝑥2 ⇒ 𝑑2 𝑦 𝑑𝑥2 = −4𝑥2 cos 𝑥2 − 2 sin 𝑥2 Successive or Double Differentiation 𝑦 = cos 𝑥2 𝑑𝑦 𝑑𝑥 = 𝑑 𝑑𝑥 cos 𝑥2 = −2𝑥 sin 𝑥2 𝑑2 𝑦 𝑑𝑥2 = −2 𝑑 𝑑𝑥 𝑥 sin 𝑥2 = −2[sin 𝑥2 + 2𝑥2 cos 𝑥2 ] ⇒ 𝑑2 𝑦 𝑑𝑥2 = −2[sin 𝑥2 + 2𝑥2 cos 𝑥2 ]
  • 33.
    Applications of Differentiation Variationof one quantity with respect to another 1 Slope of curve 2 Maxima and minima of any function 3
  • 34.
    If the motionof a particle is represented by, 𝑠 = 𝑡3 + 𝑡2 − 𝑡 + 2 𝑚. Find the magnitude of position, velocity & acceleration of the particle at 𝑡 = 2 𝑠. At 𝑡 = 2 𝑠 Position = 12𝑚 Position 𝑠 = 𝑡3 + 𝑡2 − 𝑡 + 2 Velocity = 𝑑𝑠 𝑑𝑡 = 3𝑡2 + 2𝑡 − 1 Acceleration = 𝑑2𝑠 𝑑𝑡2 = 6𝑡 + 2 Velocity = 15 𝑚/𝑠 Acceleration = 14 𝑚/𝑠2
  • 35.
    • for 𝑥1< 𝑥2, We observe 𝑓 𝑥1 < 𝑓(𝑥2) 𝑓(𝑥1) 𝑓(𝑥2) 𝑥2 𝑥1 𝑥 𝑦 Increasing function • 𝑑𝑦 𝑑𝑥 > 0 𝑓(𝑥2) 𝑓(𝑥1) 𝑥2 𝑥1 𝑥 𝑦 Decreasing function • for 𝑥1 < 𝑥2, We observe 𝑓 𝑥1 > 𝑓(𝑥2) • 𝑑𝑦 𝑑𝑥 < 0 Increasing and Decreasing Functions
  • 36.
    Critical Points 𝑦 𝑥 Critical Points Slopeof Tangent at critical points tan 𝜃 = tan 0° = 0
  • 37.
    Find the criticalpoints of the function 𝑓 𝑥 = 𝑥4 − 8𝑥2 . A B C D −2, −16 , 0,0 , (2, −16) −2, −16 , 0,0 , (4,128) −4,128 , 0,0 , (2, −16) −4,128 , 0,0 , (4,128) 𝑓′ 𝑥 = 0 ⇒ 4𝑥3 − 16𝑥 = 0 ⇒ 4𝑥 𝑥2 − 4 = 0 ⇒ 4𝑥 𝑥 − 2 (𝑥 + 2) = 0 ⇒ 𝑥 = 0,2, −2 𝑓 −2 = −16 𝑓 0 = 0 𝑓 2 = −16 So, critical points are: −2, −16 , 0,0 , (2, −16)
  • 38.
    At Maxima • 𝑑𝑦 𝑑𝑥 =0 • 𝑑2𝑦 𝑑𝑥2 < 0 At Minima • 𝑑𝑦 𝑑𝑥 = 0 • 𝑑2𝑦 𝑑𝑥2 > 0 Critical Points
  • 39.
    Find the maximaand minima for function 𝑦 = 𝑥3 − 3𝑥 + 2 1. Put 𝑑𝑦 𝑑𝑥 = 0 𝑑𝑦 𝑑𝑥 = 3𝑥2 − 3 = 0 ⇒ 𝑥 = +1 & 𝑥 = −1 𝑑2 𝑦 𝑑𝑥2 = 6𝑥 For 𝑥 = +1 𝑑2𝑦 𝑑𝑥2 = +6 > 0 For 𝑥 = −1 𝑑2𝑦 𝑑𝑥2 = −6 < 0 ⇒ Minima ⇒ Maxima 2. Find 𝑑2𝑦 𝑑𝑥2 for each value of 𝑥
  • 40.
    A ball isthrown into air. Its height at any instant of time 𝑡 is given by ℎ = 3 + 14𝑡 − 5𝑡2. What is the maximum height attained by the ball?( 𝑡 is in second) Given : height ℎ at time 𝑡, ℎ = 3 + 14𝑡 − 5𝑡2 Say at some time 𝑡 the ball achieves maximum height, ⇒ 𝑑ℎ 𝑑𝑡 = 14 − 10𝑡 = 0 ⇒ 𝑡 = 7/5 𝑠𝑒𝑐𝑜𝑛𝑑 Second derivative of ℎ, 𝑑2 ℎ 𝑑𝑡2 = −10, − 𝑣𝑒 Hence, at 𝑡 = 7/5 𝑠𝑒𝑐𝑜𝑛𝑑 height will be maximum ℎ𝑚𝑎𝑥 = 3 + 14 7 5 − 5 7 5 2 = 𝟏𝟐. 𝟖 𝑼𝒏𝒊𝒕
  • 41.
    Integration Mathematically Integration and differentiationare inverse of each other. • 𝑑 𝑑𝑥 𝑐 = 0 • 𝑑 𝑑𝑥 [𝑘𝑥] = 𝑘 • 𝑑 𝑑𝑥 sin 𝑥 = cos 𝑥 • 𝑑 𝑑𝑥 cos 𝑥 = − sin 𝑥 • 𝑑 𝑑𝑥 tan 𝑥 = sec2 𝑥 • ‫׬‬ 0 𝑑𝑥 = 𝑐 • ‫׬‬ 𝑘 𝑑𝑥 = 𝑘𝑥 + 𝑐 • ‫׬‬cos 𝑥 𝑑𝑥 = sin 𝑥 + 𝑐 • ‫׬‬ − sin 𝑥 𝑑𝑥 = cos 𝑥 + 𝑐 • ‫׬‬ sec2 𝑥 𝑑𝑥 = tan 𝑥 + 𝑐
  • 42.
    Integration of polynomialfunction Properties of Integration • ‫׬‬ 𝑥𝑛 𝑑𝑥 = 𝑥𝑛+1 𝑛+1 + 𝑐 𝑛 ≠ −1 • ‫׬‬ 𝑘𝑥𝑛 𝑑𝑥 = 𝑘 𝑥𝑛+1 𝑛+1 + 𝑐 𝑛 ≠ −1 • ‫׬‬ 1 𝑥 𝑑𝑥 = ln 𝑥 + 𝑐 • ‫׬‬ 𝑎 𝑓(𝑥) 𝑑𝑥 = 𝑎 ‫׬‬ 𝑓(𝑥) 𝑑𝑥 • ‫׬‬ 𝑓 𝑥 ± 𝑔 𝑥 𝑑𝑥 = ‫׬‬ 𝑓(𝑥) 𝑑𝑥 ± ‫׬‬ 𝑔 𝑥 𝑑𝑥 𝑎 𝑓1 𝑥 = 𝑥6 ⇒ 𝐼1= න 𝑓1(𝑥) 𝑑𝑥 = 𝑥7 7 + 𝑐 𝑏 𝑓2 𝑥 = −10𝑥3 ⇒ 𝐼2= න 𝑓2(𝑥) 𝑑𝑥 = − 10𝑥4 4 + 𝑐 𝑐 𝑓3 𝑥 = 1 𝑥 ⇒ 𝐼3= න 𝑓3(𝑥) 𝑑𝑥 = ln(𝑥) + 𝑐 𝑑 𝑓4 𝑥 = 5 ⇒ 𝐼4= න 𝑓4(𝑥) 𝑑𝑥 = 5𝑥 + 𝑐
  • 43.
    න 𝑥3𝑑𝑥 =? A B C 2 3 𝑥5 + 𝐶 2 5 𝑥5 + 𝐶 D 2 3 𝑥3 + 𝐶 2 5 𝑥3 + 𝐶 න 𝑥3𝑑𝑥 = 𝑥 3 2 +1 3 2 + 1 + 𝑐 = 2 5 𝑥5 + 𝐶
  • 44.
    Summary 𝑑 𝑑𝑥 𝑒𝑥 = 𝑒𝑥 න 𝑒𝑥 𝑑𝑥= 𝑒𝑥 + 𝑐 𝑑 𝑑𝑥 sin 𝑥 = cos 𝑥 න cos 𝑥 𝑑𝑥 = sin 𝑥 + 𝑐 𝑑 𝑑𝑥 cos 𝑥 = − sin 𝑥 න sin 𝑥 𝑑𝑥 = − cos 𝑥 + 𝑐 𝑑 𝑑𝑥 tan 𝑥 = sec2 𝑥 𝑑 𝑑𝑥 cosec 𝑥 = −cosec 𝑥 cot 𝑥 𝑑 𝑑𝑥 sec 𝑥 = sec 𝑥 tan 𝑥 𝑑 𝑑𝑥 cot 𝑥 = −cosec2 𝑥 න −cosec 𝑥 cot 𝑥 𝑑𝑥 = cosec 𝑥 + 𝑐 න sec 𝑥 tan 𝑥 𝑑𝑥 = sec 𝑥 + 𝑐 න −cosec2 𝑥 𝑑𝑥 = cot 𝑥 + 𝑐 න sec2 𝑥 𝑑𝑥 = tan 𝑥 + 𝑐
  • 45.
    Summary If 𝐼 =‫׬‬ 𝑓 𝑎𝑥 + 𝑏 𝑑𝑥, then assume 𝑋 = 𝑎𝑥 + 𝑏 and integrate further. ⇒ 𝐼 = 1 𝑎 𝐹 𝑋 Now 𝐹(𝑋) = ‫׬‬ 𝑓 𝑋 𝑑𝑥,then Replace 𝑋 with 𝑎𝑥 + 𝑏 𝐼 = 1 𝑎 𝐹(𝑎𝑥 + 𝑏) 𝐼1 = න 2𝑥 + 3 2 𝑑𝑥 = (𝟐𝒙 + 𝟑)3 6 + c 𝐼2 = න sin 2𝑥 + 3 𝑑𝑥 = − cos 𝟐𝒙 + 𝟑 2 + c 𝐼3 = න 1 2𝑥 + 3 𝑑𝑥 = log𝑒 2𝑥 + 3 2 + 𝑐
  • 46.
    Integrate the function𝑓 𝑥 = 𝑥6 − 10𝑥3 + 1 𝑥 + 2 sin 2𝑥 A B C 4𝑥6 − 70𝑥3 + 28(log𝑒 𝑥 − cos 2𝑥 4 + 𝑐 4𝑥7 − 70𝑥4 + 28(log𝑒 𝑥 − cos 2𝑥 28 + 𝑐 D 4𝑥7 − 70𝑥4 + 28(log𝑒 𝑥 − cos 2𝑥 24 + 𝑐 4𝑥7 − 7𝑥4 + 28(log𝑒 𝑥 − cos 2𝑥 24 + 𝑐
  • 47.
    Summary 𝐼 = න 2𝑥 𝑥2+ 1 𝑑𝑥 Let 𝑡 = 𝑥2 + 1 ⇒ 𝑑𝑡 𝑑𝑥 = 2𝑥 ⇒ 𝑑𝑡 = 2𝑥 𝑑𝑥 𝐼 = න 𝑑𝑡 𝑡 = log𝑒 𝑡 + 𝑐 ⇒ 𝐼 = log𝑒(𝑥2 + 1) + 𝑐 𝐼 = න sin 𝑥 cos 𝑥 𝑑𝑥 Let 𝑡 = cos 𝑥 ⇒ 𝑑𝑡 𝑑𝑥 = − sin 𝑥 ⇒ −𝑑𝑡 = sin 𝑥 𝑑𝑥 𝐼 = − න 𝑑𝑡 𝑡 = −2 𝑡 + 𝑐 ⇒ 𝐼 = −2 cos 𝑥 + 𝑐
  • 48.
    𝐼 = න𝑥2 sin 𝑥3 𝑑𝑥 =? 𝑡 = 𝑥3 ⇒ 𝑑𝑡 𝑑𝑥 = 3𝑥2 Let ⇒ 1 3 𝑑𝑡 = 𝑥2 𝑑𝑥 𝐼 = න sin 𝑡 . 1 3 𝑑𝑡 𝐼 = − 1 3 cos 𝑡 + 𝑐 𝐼 = − 1 3 cos 𝑥3 + 𝑐
  • 49.
    Integration 𝑏 ∶ Upperlimit of Integration 𝑎 ∶ Lower limit of Integration
  • 50.
    Find the valueof A B C D ln 2 + 𝐶 ln 2 ln 2 + 1 + 𝐶 ln 2 + 1 න 1 2 1 𝑥 𝑑𝑥 න 1 2 1 𝑥 𝑑𝑥 = [ln 𝑥 ]1 2 = ln(2)
  • 51.
    • Divide thewhole area under the curve into infinitely small strips of width 𝑑𝑥. We take a strip of width 𝑑𝑥 at 𝑥 = 𝑥. • Area of shown small strip is 𝑑𝐴 = 𝑓(𝑥)𝑑𝑥 • Total area between the curve and 𝑥 – axis. 𝐴 = න 𝑎 𝑏 𝑓(𝑥)𝑑𝑥 Geometrical Meaning of Integration 𝑓 𝑥 𝑥 𝑑𝑥 𝑎 𝑏
  • 52.
    𝐴𝑟𝑒𝑎 𝑜𝑓 𝑇𝑟𝑎𝑝𝑒𝑧𝑖𝑢𝑚= ℎ(𝑎 + 𝑏) 2 = 4(2 + 6) 2 = 16 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑇𝑟𝑎𝑝𝑒𝑧𝑖𝑢𝑚 = න 2 6 𝑥 𝑑𝑥 = 𝑥2 2 = 16 6 2 Find the area bounded by the lines 𝑦 = 𝑥, 𝑦 = 0 , 𝑥 = 2 and 𝑥 = 6 𝑦 = 𝑥
  • 53.
    1) න 𝑎 𝑐 𝑓 𝑥𝑑𝑥 = න 𝑎 𝑏 𝑓 𝑥 𝑑𝑥 + න 𝑏 𝑐 𝑓 𝑥 𝑑𝑥 Properties of definite Integration 2) න 𝑎 𝑎 𝑓 𝑥 𝑑𝑥 = 0 3) න 𝑎 𝑏 𝑓 𝑥 𝑑𝑥 = − න 𝑏 𝑎 𝑓 𝑥 𝑑𝑥 𝑓 𝑥 𝑥 𝑦 𝑎 𝑏 𝑐
  • 54.
    Evaluate the integral:න ൗ −𝜋 2 ൗ 𝜋 2 cos 𝑥 𝑑𝑥 A 2 B 4 C 0 D 3 න 1 4 1 2 𝑥 2 𝑑𝑥 + න 4 1 1 2 𝑥 2 𝑑𝑥 =? A 0 B 4 C 2 D 3
  • 55.
    Vector A physical quantityis a vector when, • It has a magnitude and a direction • It obeys laws of vector algebra
  • 56.
    Vector Representation Vector 𝑨 MagnitudeDirection 𝐴 or Ԧ 𝐴 መ 𝐴
  • 57.
    Vector divided byits own magnitude is a vector with unit magnitude and in direction along the parent vector. መ 𝐴 𝐴 Unit Vector 𝐴 |𝐴 | = መ 𝐴 Ԧ 𝐴 = |𝐴 | መ 𝐴
  • 58.
    If a carhas a velocity of 20 m/s in the east direction as shown in 𝐴 ? (b) How to represent velocity vector of a girl running with 5 m/s in opposite direction 𝐵 = − 05 መ 𝐴 𝑚/𝑠 𝐴 = 20 መ 𝐴 𝑚/𝑠 If the car’s velocity of 20 m/s along east is represented by Ԧ 𝐴, then how can we represent velocity of a girl moving at 5 𝑚/𝑠 in the opposite direction? If the car’s velocity of 20 𝑚/𝑠 along east is represented by Ԧ 𝐴, then how can we represent velocity of a girl moving at 5 𝑚/𝑠 in the opposite direction?
  • 59.
    Cartesian Co-ordinate System Representationof Unit Vector along Coordinate Axes
  • 60.
    • A vectorcan be displaced parallel to itself as it does not change its magnitude and direction. • If a vector is rotated through an angle other than multiple of 2𝜋 (or 360°) the vector changes. Properties of a Vector Angle between Vectors Step 1: Place vectors tail to tail or head to head. Step 2: Measure the smaller angle i.e. angle which is less than 180° between them.
  • 61.
    (i) 𝐴 and𝐵 is 60° (ii) 𝐴 and 𝐶 is 45° (iii) 𝐵 and 𝐶 is 105° Three Vectors are arranged along sides of triangle as shown. Find the angle between the vectors. (i) (ii) (iii)
  • 62.
    Multiplication of aVector with a Scalar
  • 63.
    Equal Vectors • Ԧ 𝐴= 𝐵 = Ԧ 𝐶 Equal magnitude • መ 𝐴 = ෠ 𝐵 = መ 𝐶 Same direction Parallel vectors Angle between any two vectors is 0° or 180° then they are along the same direction or exactly opposite directions. • 𝜃 = 0° or 180° • Ԧ 𝐴 ≠ 𝐵 • Ԧ 𝐴 ∥ 𝐵 Null vector • Ԧ 𝐴 = 0 If two vectors are co-linear or parallel then one can be expressed in the terms of another as, • Ԧ 𝑎 = 𝜆𝑏 (where 𝜆 is constant) Collinear vectors Types of Vectors
  • 64.
    Three (or more)vectors are called coplanar vectors if they lie in the same plane. Two (free) vectors can be made coplanar. Coplanar vectors Same in magnitude but opposite in direction. • መ 𝐶 = −෡ 𝐷 • Ԧ 𝐶 = 𝐷 Negative of a vectors Types of Vectors
  • 65.
    Magnitude of aVector and Displacement Vector Displacement Vector is the change in position vector. Ԧ 𝑠 = 𝑥2 − 𝑥1 Ƹ 𝑖 + 𝑦2 − 𝑦1 Ƹ 𝑗 + (𝑧2 − 𝑧1)෠ 𝑘 Ԧ 𝑟 = 𝑎 Ƹ 𝑖 + 𝑏 Ƹ 𝑗 Ԧ 𝑟 =Length of line 𝑂𝐴 Ԧ 𝑟 = 𝑎2 + 𝑏2
  • 66.
    Position vector givesus the position of a point in magnitude and direction. Ԧ 𝑟1 = 𝑎 Ƹ 𝑖 + 𝑏 Ƹ 𝑗 Position Vector in 2D and 3D Position vector Ԧ 𝑟1 = 𝑥1 Ƹ 𝑖 + 𝑦1 Ƹ 𝑗 + 𝑧1 ෠ 𝑘 Ԧ 𝑟2 = 𝑥2 Ƹ 𝑖 + 𝑦2 Ƹ 𝑗 + 𝑧2 ෠ 𝑘 Magnitude Ԧ 𝑟1 = 𝑥1 2 + 𝑦1 2 + 𝑧1 2 Ԧ 𝑟2 = 𝑥2 2 + 𝑦2 2 + 𝑧2 2 Ԧ 𝑟1 = 𝑎2 + 𝑏2
  • 67.
    Magnitude 𝑅 = 𝑅𝑥Ƹ 𝑖 + 𝑅𝑦 Ƹ 𝑗 Direction 𝑅 = 𝑅𝑥 2 + 𝑅𝑦 2 tan 𝜃 = 𝑅𝑦 𝑅𝑥 𝑦 𝑥 𝑅 𝑂 𝜃 Magnitude and Direction 𝑅 = 𝑅𝑥 + 𝑅𝑦 𝑅 = 𝑅𝑥 Ƹ 𝑖 + 𝑅𝑦 Ƹ 𝑗 𝑅 = 𝑅 cos 𝜃 Ƹ 𝑖 + 𝑅 sin 𝜃 Ƹ 𝑗
  • 68.
    𝑦 𝑥 𝐴 = 30 𝑂 𝑦 𝐵= 80 𝑂 Fill in the Blanks Ԧ 𝐴 = Ƹ 𝑖 + Ƹ 𝑗 −15 2 15 2 𝐵 = Ƹ 𝑖 + Ƹ 𝑗 −40 −40 3 𝑥
  • 69.
    Two friends 𝐴and 𝐵 are running parallel to each other. Velocity of 𝐴 is 3 Ƹ 𝑖 + 4 Ƹ 𝑗 𝑚/𝑠 and speed of 𝐵 is 20 𝑚/𝑠. Find the velocity of 𝐵? 𝐵 = 𝐵 ෠ 𝐵 𝐵 = 20 × 3 Ƹ 𝑖 + 4 Ƹ 𝑗 5 Ԧ 𝐴 = 32 + 42 = 5 Unit vector along Ԧ 𝐴, መ 𝐴 = 3 Ƹ 𝑖 + 4 Ƹ 𝑗 5 𝐵 = 12 Ƹ 𝑖 + 16 Ƹ 𝑗 መ 𝐴 = ෠ 𝐵 መ 𝐴 = Ԧ 𝐴 Ԧ 𝐴
  • 70.
    Steps for vectorAddition 𝑦 𝑥 Ԧ 𝑃 𝜃 ∅ 𝑃𝑥 𝑄𝑦 𝑄𝑥 𝑃𝑦 𝑄 Ԧ 𝑃 = 𝑃𝑥 Ƹ 𝑖 + 𝑃𝑦 Ƹ 𝑗 𝑄 = 𝑄𝑥 Ƹ 𝑖 + 𝑄𝑦 Ƹ 𝑗 Ԧ 𝑃 = 𝑃 cos 𝜃 Ƹ 𝑖 + 𝑃 sin 𝜃 Ƹ 𝑗 𝑄 = 𝑄 cos 𝜙 Ƹ 𝑖 + 𝑄 sin 𝜙 Ƹ 𝑗 Ԧ 𝑃 + 𝑄 = 𝑃𝑥 + 𝑄𝑥 Ƹ 𝑖 + 𝑃𝑦 + 𝑄𝑦 Ƹ 𝑗 Ԧ 𝑃 + 𝑄 = 𝑃 cos 𝜃 + 𝑄 cos 𝜙 Ƹ 𝑖 + 𝑃 sin 𝜃 + 𝑄 sin 𝜙 Ƹ 𝑗
  • 71.
    Ԧ 𝑃 + 𝑄= 9 2 Ƹ 𝑗 + 9 3 2 Ƹ 𝑖 + 3 Ƹ 𝑗 − 3 3 Ƹ 𝑖 Ԧ 𝑃 + 𝑄 = 9 2 Ƹ 𝑗 + 3 Ƹ 𝑗 + 9 3 2 Ƹ 𝑖 − 3 3 Ƹ 𝑖 Ԧ 𝑃 + 𝑄 = 15 2 Ƹ 𝑗 + 3 3 2 Ƹ 𝑖 30° 𝑦 𝑥 6 30° 3 3 3 9 9 3 2 9 2 If Ԧ 𝑃 = 9 units and 𝑄 = 6 units, then find Ԧ 𝑃 + 𝑄.
  • 72.
    Laws of Vector Addition/Subtraction Lawsof Vector Addition/Subtraction Triangle Law Parallelogram Law Polygon Law
  • 73.
    𝑅 = 𝐴2+ 𝐵2 + 2𝐴𝐵 cos 𝜃 𝑦 𝜃 𝐵 𝑅 𝐴 Ƹ 𝑖 𝑥 Ԧ 𝐴 𝐵 sin 𝜃 Ƹ 𝑗 𝑅 𝐵 cos 𝜃 Ƹ 𝑖 𝑅 = Ԧ 𝐴 + 𝐵 = 𝐴 + 𝐵 cos 𝜃 Ƹ 𝑖 + 𝐵 sin 𝜃 Ƹ 𝑗 𝑅 = Ԧ 𝐴 + 𝐵 𝑅 = 𝐴 + 𝐵 cos 𝜃 2 + 𝐵 sin 𝜃 2 𝑅 = 𝐴2 + 𝐵2 cos2 𝜃 + 2𝐴𝐵 cos 𝜃 + 𝐵2 sin2 𝜃 Triangle Law
  • 74.
    When two vectorswith common origin represent two adjacent sides of a parallelogram in magnitude and direction, then the resultant vector is represented both in magnitude and direction by the diagonal passing through that point. 𝑦 𝑅 = 𝐴2 + 𝐵2 + 2𝐴𝐵 cos 𝜃 Here 𝛼 is the angle between Ԧ 𝐴 and 𝑅 tan 𝛼 = 𝐵 sin 𝜃 Ԧ 𝐴 + 𝐵 cos 𝜃 Parallelogram Law
  • 75.
    𝑅3 𝐷 Ԧ 𝐴 𝐵 Ԧ 𝐶 • Join vectorsin same order (head to tail). • Resultant Vector – closing side of polygon in opposite order 𝑅3 = Ԧ 𝐴 + 𝐵 + Ԧ 𝐶 + 𝐷 Polygon Law of Vector Addition
  • 76.
    • Ԧ 𝐴 +𝐵 : Ԧ 𝐴 and 𝐵 must be same physical quantities. • Commutative Property : Ԧ 𝐴 + 𝐵 = 𝐵 + Ԧ 𝐴 • Associative Property : ( Ԧ 𝐴 + 𝐵) + Ԧ 𝐶 = Ԧ 𝐴 + (𝐵 + Ԧ 𝐶) Properties of Vector Addition
  • 77.
    𝜃 𝐵 𝐴 Ԧ 𝐴. 𝐵 =𝐴𝐵 cos 𝜃 The dot product of two vectors 𝐴 and 𝐵 is equal to the products of their magnitude times the cosine of angle between them. Scalar / Dot Product Ԧ 𝐴. 𝐵 = Ԧ 𝐴 𝐵 cos 𝜃 𝐴 = 𝐴𝑥 Ƹ 𝑖 + 𝐴𝑦 Ƹ 𝑗 + 𝐴𝑧 ෠ 𝑘 , 𝐵 = 𝐵𝑥 Ƹ 𝑖 + 𝐵𝑦 Ƹ 𝑗 + 𝐵𝑧 ෠ 𝑘 𝐴 . 𝐵 = 𝐴𝑥𝐵𝑥 + 𝐴𝑦𝐵𝑦 + 𝐴𝑧𝐵𝑧
  • 78.
    60° 𝐵 𝐴 If 𝐴 =2, 𝐵 = 3 and the angle between 𝐴 and 𝐵 is 60°. Find 𝐴 . 𝐵 𝐴 .𝐵 = |𝐴 ||𝐵 |cos 𝜃 = 2 × 3 × cos 60° = 2 × 3 × 1 2 = 3 Ԧ 𝐴. 𝐵 = 3
  • 79.
    • Dot productof two vectors is always (commutative), i.e. 𝐴 . 𝐵 = 𝐵 . 𝐴 • Dot product of two vectors is always distributive, i.e. 𝐴 . 𝐵 + 𝐶 = 𝐴 . 𝐵 + 𝐴 . 𝐶 • Dot product of two vectors will be maximum when they are parallel (i.e., angle between them is zero). Properties of Dot Product 𝐴 . 𝐵 is positive 𝜽 < 𝟗𝟎° ⇒ 𝐂𝐨𝐬 𝜽 > 𝟎 Ԧ 𝐴. 𝐵 = 𝐴𝐵 cos 𝜃 𝜽 = 𝟗𝟎° 𝐴 . 𝐵 is zero 𝐴 . 𝐵 is negative 𝜽 > 𝟗𝟎° ⇒ 𝐂𝐨𝐬 𝜽 < 𝟎 • 𝐴 .𝐴 = 𝐴2 • ො 𝑛 . ො 𝑛 = 1 • Ƹ 𝑖 . Ƹ 𝑗 = Ƹ 𝑗 . ෠ 𝑘 = ෠ 𝑘 . Ƹ 𝑖 = 0
  • 80.
    𝐴 .𝐵 =0 ⟹ 𝐴 𝐵 cos 𝜃 = 0 ⟹ cos 𝜃 = 0 𝜃 = cos−1 0 = 90° 𝐴 𝐵 90° Thus, 𝐴 .𝐵 = 0 is a condition for two vectors to be orthogonal. If the scalar product of two non-zero vectors becomes zero. What is the angle between them?
  • 81.
    The cross productof two vectors 𝐴 and 𝐵 is equal to the products of their magnitude times the sine of angle between them and direction perpendicular to the plane containing the two vectors. Ԧ 𝐴 × 𝐵 = Ԧ 𝐴 𝐵 sin 𝜃 ො 𝑛 Ԧ 𝐴 × 𝐵 = Ԧ 𝐴 𝐵 sin 𝜃 Vector / Cross Product Here, ො 𝑛 is a unit vector perpendicular to both Ԧ 𝐴 and 𝐵. Ԧ 𝐴 × 𝐵 = Ԧ 𝐴 𝐵 sin 𝜃 ො 𝑛
  • 82.
    To find thedirection of Ԧ 𝐴 × 𝐵: • Draw Ԧ 𝐴 and 𝐵 tail-to-tail. • Place the stretched right palm such that the fingers are along Ԧ 𝐴 and when the fingers are closed, they go towards 𝐵. • The direction in which thumb points gives the direction of Ԧ 𝐶 = Ԧ 𝐴 × 𝐵. Right Hand Thumb Rule Ԧ 𝐴 𝐵 Ԧ 𝐴 𝐵 Ԧ 𝐶
  • 83.
    Ԧ 𝐴 × 𝐵= Ԧ 𝐴 𝐵 sin 𝜃 = 5 × 4 × sin 30° = 10 Using the right-hand thumb rule, the direction of Ԧ 𝐴 × 𝐵 is along −𝑧 axis Find the magnitude and direction of Ԧ 𝐴 × 𝐵 and 𝐵 × Ԧ 𝐴, where Ԧ 𝐴 & 𝐵 lie in 𝑥 − 𝑦 plane as shown. The direction of 𝐵 × Ԧ 𝐴 is along +𝑧 axis. ⟹ Ԧ 𝐴 × 𝐵 = −10෠ 𝑘 ⟹ 𝐵 × Ԧ 𝐴 = 10෠ 𝑘 Ԧ 𝐴 𝐵 30° 4 5
  • 84.
    In the right-handedcoordinate system, the coordinate axes 𝑥, 𝑦 and 𝑧 are chosen such that bending the fingers of the right hand from 𝑥 to 𝑦 will lead the thumb along the 𝑧 −axis. Ƹ 𝑖 Ƹ 𝑗 ෠ 𝑘 Ƹ 𝑖 × Ƹ 𝑗 = ෠ 𝑘, Ƹ 𝑗 × Ƹ 𝑖 = − ෠ 𝑘 Ƹ 𝑗 × ෠ 𝑘 = Ƹ 𝑖, ෠ 𝑘 × Ƹ 𝑗 = − Ƹ 𝑖 ෠ 𝑘 × Ƹ 𝑖 = Ƹ 𝑗, Ƹ 𝑖 × ෠ 𝑘 = − Ƹ 𝑗 Cross Product of Orthogonal Unit Vectors
  • 85.
    Cross Product inComponent Form Ԧ 𝐴 × 𝐵 = 𝐴𝑥 Ƹ 𝑖 + 𝐴𝑦 Ƹ 𝑗 + 𝐴𝑧 ෠ 𝑘 × 𝐵𝑥 Ƹ 𝑖 + 𝐵𝑦 Ƹ 𝑗 + 𝐵𝑧 ෠ 𝑘 Ԧ 𝐴 × 𝐵 = 𝐴𝑥𝐵𝑥 Ƹ 𝑖 × Ƹ 𝑖 + 𝐴𝑦𝐵𝑥 Ƹ 𝑗 × Ƹ 𝑖 + 𝐴𝑧𝐵𝑥 ෠ 𝑘 × Ƹ 𝑖 +𝐴𝑥𝐵𝑦 Ƹ 𝑖 × Ƹ 𝑗 + 𝐴𝑦𝐵𝑦 Ƹ 𝑗 × Ƹ 𝑗 + 𝐴𝑧𝐵𝑦 ෠ 𝑘 × Ƹ 𝑗 +𝐴𝑥𝐵𝑧 Ƹ 𝑖 × ෠ 𝑘 + 𝐴𝑦𝐵𝑧 Ƹ 𝑗 × ෠ 𝑘 + 𝐴𝑧𝐵𝑧 ෠ 𝑘 × ෠ 𝑘 Ԧ 𝐴 × 𝐵 = (𝐴𝑦𝐵𝑧 − 𝐴𝑧𝐵𝑦) Ƹ 𝑖 − 𝐴𝑥𝐵𝑧 − 𝐴𝑧𝐵𝑥 Ƹ 𝑗 + (𝐴𝑥𝐵𝑦 − 𝐴𝑦𝐵𝑥)෠ 𝑘 Ƹ 𝑖 Ƹ 𝑗 ෠ 𝑘 Ԧ 𝐴 = 𝐴𝑥 Ƹ 𝑖 + 𝐴𝑦 Ƹ 𝑗 + 𝐴𝑧 ෠ 𝑘 𝐵 = 𝐵𝑥 Ƹ 𝑖 + 𝐵𝑦 Ƹ 𝑗 + 𝐵𝑧 ෠ 𝑘
  • 86.
    Ԧ 𝐴 × 𝐵= (𝐴𝑦𝐵𝑧 − 𝐴𝑧𝐵𝑦) Ƹ 𝑖 − 𝐴𝑥𝐵𝑧 − 𝐴𝑧𝐵𝑥 Ƹ 𝑗 + (𝐴𝑥𝐵𝑦 − 𝐴𝑦𝐵𝑥)෠ 𝑘 Ԧ 𝐴 × 𝐵 = 2 Ƹ 𝑖 − 5 Ƹ 𝑗 + 3෠ 𝑘 × 3 Ƹ 𝑖 + 4 Ƹ 𝑗 − 9෠ 𝑘 = 6 Ƹ 𝑖 × Ƹ 𝑖 − 15 Ƹ 𝑗 × Ƹ 𝑖 + 9 ෠ 𝑘 × Ƹ 𝑖 +8 Ƹ 𝑖 × Ƹ 𝑗 − 20 Ƹ 𝑗 × Ƹ 𝑗 + 12 ෠ 𝑘 × Ƹ 𝑗 − 18 Ƹ 𝑖 × ෠ 𝑘 + 45 Ƹ 𝑗 × ෠ 𝑘 − 27 (෠ 𝑘 × ෠ 𝑘) Ԧ 𝐴 × 𝐵 = Ƹ 𝑖 45 − 12 − Ƹ 𝑗 −18 − 9 + ෠ 𝑘 8 − (−15) Ԧ 𝐴 × 𝐵 = 33 Ƹ 𝑖 + 27 Ƹ 𝑗 + 23෠ 𝑘 Find Ԧ 𝐴 × 𝐵 , where Ԧ 𝐴 = 2 Ƹ 𝑖 − 5 Ƹ 𝑗 + 3෠ 𝑘 and 𝐵 = 3 Ƹ 𝑖 + 4 Ƹ 𝑗 − 9෠ 𝑘.