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Chapter 6 Unit 6 定积分的物理应用定积分的物理应用
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New Words Work 功 Pressure 压力 The universal gravitational constant 万有引力常数 Horizontal component 水平分力 Well-proportioned 均匀的 Perpendicular bisector 中垂线 Orthogonal triangle 直角三角形
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The definite integral has wide applications in mathematics, the physical science and engineering. Here, we will introduce some simple applications in the physical science. At first, we will use the definite integral to compute the work done by a varying force that moves an object along a straight line. Then we use the definite integral to compute the pressure of water and the force between two objects.
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To calculate a quantities by setting up a definite integral, the general procedure may be outlined as follows: Step 1: Chop up the desired quantity into very small parts. Step 2: Within each small part, calculate an approximation to the desired quantity. Step 3: Add up the results of all of the small parts approximation from step 2 Step 4: Obtain a definite integral by taking the limit of the sum in step 3 as the part get smaller and smaller.
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1. Work In this subsection we compute the work done by a varying force that moves an object along a straight line.
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But, if we are faced with a problem in which the force is variable, we cannot use the above formula. Instead, the definite integral is required. Example 1 A rocket has mass 2000kg. Find the work done in launching the rockets from the earth’s surface to an altitude of 100 km. (Use 6400km as the radius of the earth and ignore the weight of the fuel burned during the flight)
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Solution According to the Newton’s law of gravitation. The force F(x) needed to overcome the force of gravity is
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Example 2 A spring is stretched 0.5 meter longer than its rest length. The force required to keep it at that length is 3 newtons. Find the total work accomplished in stretching the spring 0.5 meter from its rest length Solution Since the force is proportional to x, it is of the form kx for some constant k. We know that F=3 when x=0.5, so
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To estimate the work involved in stretching the spring from x to x+dxx+dx The distance dx dx is small. As the end of the spring is stretched from x to x+dx, the force is almost constant. The work accomplished in stretching the spring from x to x+dx x+dx is then approximately The element of work Hence, the total work is
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Example 3 Solution A cylinder with base radius 3 meters and height 5 meters is full of work. How much work is required to pump out all water in the cylinder? See figure
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The element of work Hence, the total work is
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Example 4 Solution:See figure
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x x x+dx 0 30
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2. The pressure force of water Physics tells us the intensity of pressure of water with depth h is p=rh, where r is the weight of one cubic unit of water. If a board with area A is placed horizontally in water at a depth h, then the force on this board is
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F=pA. But, if the board is submerged vertically in water, we can not use the above formula. Instead, the definite integral is required. Example 5 Solution: A cylindrical water tank has radius R (See figure), and it is full of water by half. Find the force against one end of the tank, where r is the weight of one cubic unit water. See figure
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The area of small strip is
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Example 6 A orthogonal triangle board with side a and 2a 2a is
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Solution The typical area of the small piece is submerged vertically in water (see figure). Find the force against the board The element of force
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3. Gravitation If we want to calculate the gravitational attraction between a mass and a piece of stick,
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the definite integral is required, because the distance between them is not constant. Example 7 Solution See figure
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This section presented various physical applications of the definite integral. We showed that work, pressure force and gravitational attraction can be represented as definite integrals.
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