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Application of Derivatives Class 12 Maths CBSE 2023 Complete Revision | Shimon Sir | V Master Tamil|

Vedantu Master Tamil・35 minutes read

A Saturday morning session will cover 3D geometry, focusing on common exam questions and providing help even after the exam. Understanding functions, limits, and derivatives is crucial for effective problem-solving and finding critical points for maximizing volume.

Insights

  • Understanding calculus concepts like critical points, derivatives, and rate of change is crucial for solving math problems effectively and determining maxima or minima in functions.
  • Practical applications of calculus, such as calculating rates of change for geometric shapes like cubes and rectangles, demonstrate the real-world relevance of mathematical principles and their impact on everyday scenarios.

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Recent questions

  • How can I determine the rate of change for the volume of a cube?

    The rate of change of volume for a cube is 3a^2 * dA/dt.

  • What is the formula for the volume of a sphere and its rate of change?

    The volume of a sphere is 4/3 * π * r^3, and the rate of change of volume is 4πr^2 * dR/dt.

  • How can I identify critical points and determine if they are minima or maxima?

    Critical points are where the first derivative is zero, and the second derivative test helps determine if they are minima or maxima.

  • What is the relationship between arithmetic mean and geometric mean?

    Arithmetic mean is greater than or equal to geometric mean.

  • How can I determine the rate of change of perimeter and area for a rectangle?

    The rate of change of perimeter of a rectangle is -2 centimeters per minute, and the rate of change of area of a rectangle is +2 centimeters square per minute.

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Summary

00:00

Math Session: 3D Geometry, Derivatives, Limits

  • The session will cover 3D geometry and will take place on Saturday morning at 5 o'clock just before the exam.
  • The session will focus on common questions from previous years, so it's important to pay attention and follow along.
  • Make sure to have your pen, paper, and book ready for the math session.
  • The teacher will provide help even after the board exam, showing dedication to students' progress.
  • The rate of change of volume for a cube is 3a^2 * dA/dt, and for the surface area, it's 12a * dA/dt.
  • The volume of a sphere is 4/3 * π * r^3, and the rate of change of volume is 4πr^2 * dR/dt.
  • Understanding increasing and decreasing functions involves analyzing the first derivative being positive or negative.
  • Critical points are where the first derivative is zero, and the second derivative test helps determine if they are minima or maxima.
  • The first derivative test involves the slope changing from positive to zero to negative for maxima and negative to zero to positive for minima.
  • Standard limits and step-by-step explanations are crucial for solving questions effectively.

35:59

Finding Minima and Maxima in Calculus

  • To find local Minima, first derivative is calculated by multiplying the first into minus B by X square.
  • Critical points for Maxima or Minima are found by equating the first derivative to zero, resulting in X equal to plus or minus root of B by A.
  • The second derivative test is used to confirm Minima, with the second derivative being minus B by X square differential plus two B by X cube.
  • The minimum value of the function is found to be 2 root A B.
  • Arithmetic mean is greater than or equal to geometric mean, with the arithmetic mean of AX plus B by X being greater than or equal to twice the square root of A times B.
  • The function is strictly increasing between -3 to 2 and 4 to infinity, and strictly decreasing from minus infinity to -3 and 2 to 4.
  • The function F is increasing between -2 and -1.
  • The rate of change of perimeter of a rectangle is found to be -2 centimeters per minute.
  • The rate of change of area of a rectangle is calculated to be +2 centimeters square per minute.
  • The rate of increase of surface area of a cube is determined to be 3.6 centimeters square per second when the side length is 10 centimeters.

01:06:10

Maximizing Cuboid Volume: Critical Point Analysis

  • The volume of the cuboid is calculated as 18 minus 2x length, 18 minus 2x breadth, and x height.
  • To maximize the volume, the equation is simplified to 18 minus 2x whole square into x.
  • The critical points for maximizing the volume are x equal to 9 and x equal to 3, with x equal to 3 being the answer.
  • The second derivative is calculated to ensure the maximum volume is achieved at x equal to 3.
  • The maximum volume of the box is determined to be 432 centimeter cube.
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