TLDR;
This YouTube video by Giri Tutorials covers applications of derivatives, focusing on finding equations of tangents and normals, determining values using given conditions, and solving word problems related to maxima and minima. The session includes detailed explanations, step-by-step solutions, and tips for exam preparation.
- Finding equations of tangents and normals
- Solving problems using derivatives
- Tips for exam preparation
Introduction [0:28]
The session will cover the application of derivatives, and the instructor asks viewers to share the video. He requests an hour of the viewers' time to cover the topic effectively.
Finding Equations of Tangents and Normals [2:46]
The instructor begins by solving a problem to find the equations of the tangent and normal to the curve y = x^2 + 2e^(x+2) at the point (0, 4). First, differentiate the equation to find dy/dx = 2x + 2e^x. Then, substitute the point (0, 4) into the derivative to find the slope of the tangent, which is 2. Since the normal is perpendicular to the tangent, its slope is -1/2. Using the point-slope form, the equation of the tangent is found to be 2x - y + 4 = 0, and the equation of the normal is x + 2y - 8 = 0.
Solving Problems with Given Conditions [11:58]
The next problem involves finding the values of A and B for the curve y^2 = Ax + B, given that the line y = 4x - 5 touches the curve. The slope of the line is 4. Differentiating the curve equation gives 2y(dy/dx) = A, so dy/dx = A/(2y). Equating the slopes of the line and the curve, 4 = A/(2y), which simplifies to A = 2. Substituting A = 2 back into the original equation and using the point of tangency (1, -3) allows solving for B, which is found to be -7.
Word Problem: Surface Area of a Spherical Balloon [19:41]
The surface area of a spherical balloon is increasing at a rate of 2 cm²/second. The task is to find the rate at which the volume of the balloon is increasing when the radius is 6 cm. Given dA/dt = 2 cm²/s and r = 6 cm, the goal is to find dV/dt. The surface area formula A = 4πr² is differentiated with respect to time, yielding dA/dt = 8πr(dr/dt). Solving for dr/dt gives dr/dt = 1/(24π). The volume formula V = (4/3)πr³ is differentiated with respect to time, yielding dV/dt = 4πr²(dr/dt). Substituting the known values, dV/dt = 6 cm³/second.
Approximate Value Problems [27:44]
The instructor explains how to find approximate values using derivatives. For √8.95, the function f(x) = √x is used. The nearest perfect square to 8.95 is 9, so a = 9 and h = -0.05. The derivative f'(x) = 1/(2√x) is evaluated at a = 9, giving f'(9) = 1/6. Using the approximation formula f(a + h) ≈ f(a) + h * f'(a), the approximate value is calculated as 3 + (-0.05) * (1/6) ≈ 2.9917.
Trigonometric Approximation [37:15]
To find the approximate value of sin(61°), the function f(x) = sin(x) is used. The nearest standard angle to 61° is 60°, so a = 60° and h = 1°. Convert h to radians: 1° = 0.01745 radians. The derivative f'(x) = cos(x) is evaluated at a = 60°, giving f'(60°) = 0.5. Using the approximation formula, the approximate value is calculated as sin(60° + 1°) ≈ sin(60°) + 1° * cos(60°) ≈ 0.866 + 0.01745 * 0.5 ≈ 0.8747.
Rolls Theorem and LMVT [41:58]
The instructor briefly explains Rolle's Theorem and the Mean Value Theorem (LMVT). For Rolle's Theorem, three conditions must be met: f(x) is continuous on [a, b], f(x) is differentiable on (a, b), and f(a) = f(b). If these conditions are met, there exists a c in (a, b) such that f'(c) = 0. For LMVT, the conditions are f(x) is continuous on [a, b] and f(x) is differentiable on (a, b). Then, there exists a c in (a, b) such that f'(c) = (f(b) - f(a))/(b - a).
Increasing and Decreasing Functions [44:35]
To find the values of x for which the function f(x) = 2x³ - 3x² - 12x + 6 is strictly increasing, first find the derivative f'(x) = 6x² - 6x - 12. Set f'(x) > 0 for increasing functions, which simplifies to x² - x - 2 > 0. Factor the quadratic to get (x - 2)(x + 1) > 0. Analyze the intervals to find that the function is increasing for x < -1 or x > 2.
Maxima and Minima: Word Problems [54:37]
The instructor explains the steps to solve maxima and minima problems. First, define the function f(x). Then, find the first derivative f'(x) and the second derivative f''(x). Set the first derivative equal to zero to find critical points. Use the second derivative test to determine if the critical point is a maximum (f''(x) < 0) or a minimum (f''(x) > 0).
Divide the Number 30 [56:27]
Divide the number 30 into two parts such that their product is maximum. Let the two parts be x and y, so x + y = 30, and y = 30 - x. The product P = x * y = x(30 - x) = 30x - x². To maximize the product, find the derivative P'(x) = 30 - 2x and set it to zero, giving x = 15. The second derivative P''(x) = -2, which is negative, indicating a maximum. Thus, the two parts are x = 15 and y = 15.
Wire of Length 36 Meter [1:03:29]
A wire of length 36 meters is bent into the form of a rectangle. Find its dimensions if the area of the rectangle is maximum. Let the length be x and the breadth be y. The perimeter is 2x + 2y = 36, so x + y = 18, and y = 18 - x. The area A = x * y = x(18 - x) = 18x - x². To maximize the area, find the derivative A'(x) = 18 - 2x and set it to zero, giving x = 9. The second derivative A''(x) = -2, which is negative, indicating a maximum. Thus, the dimensions are x = 9 meters and y = 9 meters.
Conclusion and Tomorrow's Topics [1:08:49]
The session concludes with a summary of the topics covered and an announcement of the next session, which will cover mathematical logic and matrices. The instructor encourages viewers to share the video and practice the concepts discussed.