Example Problem 4. Related Rates

Related Rates. Example Problem 4 ● The rate of change of height, or how fast something rises or drops over a given amount of time, is called the derivative. ● The volume of a derivative is the amount of change that it represents.● The diameter of a circle is the distance from the center to any point on its edge. It is always twice as long as the radius. The diameter of a circle is usually denoted by "d" in equations. There are numerous situations involving related rates that can be encountered. The cone problem is connected to a secondary issue. A right cone has a diameter equal to its height, and sand falls out of the bottom at an increasingrate. A right cone with a radius of 1 centimeter is formed by placing it in an hourglass with anopening that can be adjusted. At what rate is the radius increasing when the radius is 1centimeter?

Usually, we are given the formula for finding the volume of a cone: one-third times pi times theradius squared multiplied by its height. So, we’d like to take the derivative. We are taking the derivative with respect to time of the velocity in order to find the instantaneousvelocity at time t. We have two variables that are changing, which is a product and fun. Thus, we will call this component f. And h will be our g. So 1/3 pi r squared. The instantaneous rate of change, or the derivative of h, which is dh/dt, can be calculated as 2/3 pi.Now we have 3. The integral of r dr over dt is f prime times g. Now, here's where the problem arises. Let us make a list of the things we require. dv dt. r. We are looking for a dh dt. We already have r. We need a dr dt, and we need an h. Let's consider the relationship between volume and time. Given that 3 cubic centimeters equals 1milliliter, we can conclude that in one minute, the cube will increase by 3 milliliters. The amount of sand in the upper chamber will increase. It means it will be a positive value. The radius is 1 centimeter. The height is not explicitly stated; however, we can infer that the heightis equal to the diameter because the formula for calculating volume of a cylinder states that "thevolume of a cylinder equals its radius times its height." Two radius values equal the diameter of a circle. To solve this problem, it is necessary to first determine at what rate the radius of a cone isincreasing when its height is 1 centimeter. Solving for the rate of change of radius, we get `dr=dt` .The radius is 2, so the height must be 2. A theoretical issue arises when we consider that we are not given information about dhdt. However, the expression that relates the height, h, to the diameter, d, of a circle is h = 2r. If we aregiven two such circles with the same diameter, then their heights must also be proportional to eachother. You take a derivative of a function. The derivative of dh dt is 2 dr dt's. So, instead of dh dt,substitute 2 dr dt's into the equation and then fill in the rest of the equation. The formula for calculating the volume of a circular cylinder is V = πr2h. The equationdv/dt=1/3πr2h is used to calculate the velocity of an object dropped from a height. Plugging inknown values for r, h, and v yields 1/3πr2h=2/3πrv+2/3πrh. Solving for v yields 2/3πrh=1−1/(3πr).To solve for the height of a cylinder, plug in the radius, r = 3; the diameter, d = 4; and the value ofpi, π = 3. Therefore V = πr2h: So the equation can be represented as 2/3 pi dr dt plus 4/3 pi dr dt, for a total of 6/3 times pi. Thisindicates that the rate of change in radius is 3 over 2 pi centimeters per minute. That was a good cone question. There is another way we could have done that. We could have gone at the beginning, said OK,and made a substitution of h for r in our formula for volume. So if v equals 1/3 pi r squared times h,then v equals 2 over 3 r cubed. And we could just go from there. The equation for the volume of a sphere is V = 4/3 πr3. If we substitute r for dr dt, we getV=(4/3)πr2r. Multiplying both sides by r2 gives us V=4/3 πr3. Given that dr dt is equal to 2 pi r squared, dv dt can be written as 3 over 2 pi.