Assignment
University
Rice UniversityCourse
Preparing for the AP Calculus AB ExamPages
2
Academic year
2022
Fantasticsa
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63
Related Rates. Problem Example 3 In this lecture I'll discuss three to four related examples of the rate formula, which you'll see in a variety of situations. This is a fairly straightforward problem.We assume that the spherical balloon is being filled with air at a constant rate, 500 cubic inches per minute. When the radius of the balloon is 30 inches, what is the rate of change of its radius? First, we must create the formula for the volume of a sphere. The formula is 4/3 × π × r3, where r represents the radius of the sphere. The Pythagorean theorem and other formulas such as the area of a circle or distance formula are important to know. However, if you are less experienced with these concepts, taking the derivative may be a challenge. If you have difficulty with problems involving the pythagorean theorem, the area of a circle or the distance formula, read the prompt carefully and develop a plan for solving it by breaking down each step. When we take the derivative of v and then "react," or apply a chain rule, we get dv dt, which is the same thing as 4/3 pi r cubed dr dt, because applying a chain rule does nothing more than rewrite an expression as another expression. So here's what we know: There is a time-dependent variable dv/dt, a time-independent variable r, and a time-independent variable dr/dt. Two of them have been given to us; one of them we need to find.
So, to find the volume of a cylindrical tank filled with sand at a rate of 500 cubic inches per minute, we must first find the derivative of volume with respect to time. The derivative of volume is equal to the instantaneous rate at which volume increases. Or, dv/dt = 500. If the rate of decrease was negative for some reason, it would be considered a negative rate. Next, we have the radius of 30. This is nice and easy; we just plug it into the equation for dr dt. So dr dt is the unknown. The variable Drdt is unknown because we are trying to find the rate of change of radius with respect to time. Drdt is defined as 500 times four pi to the thirty-second power, divided by r cubed. When we insert this formula into our equation, we get: 500 equals 4 pi 30 squared, dr dt. The area of the circle is pi times the square of its radius, or A = πr2. The volume of a sphere is 4/3πr3. Multiply these together, and you get 500πr3dr dt. Divide by 3,600πr, and it equals 5/36πr. The rate of change of the radius can be calculated by dividing the change in the radius by the elapsed time. It can be expressed as pi multiplied by the radius divided by time.
Problem Example 3. Related Rates
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