Constant Rate Of Change Worksheet With Answers

[Arleen Smelko]

"@context":" ","@type":"FAQPage","mainEntity":["@type":"Question","name":"What is the definition of constant rate in math?","acceptedAnswer":"@type":"Answer","text":"A constant rate in math is the absence of acceleration. In general, a function with a constant rate is one with a second derivative of 0. If you were to plot the function on standard graph paper, it would be a straight line, as the change in y (or rate) would be constant. ","@type":"Question","name":"How do you find the rate?","acceptedAnswer":"@type":"Answer","text":"To find the unit rate, divide the numerator and denominator of the given rate by the denominator of the given rate. So in this case, divide the numerator and denominator of 70/5 by 5, to get 14/1, or 14 students per class, which is the unit rate.","@type":"Question","name":"What does it mean to have a constant rate of change?","acceptedAnswer":"@type":"Answer","text":"When the value of x increases, the value of y remains constant. That is, there is no change in y value and the graph is a horizontal line. Example: Use the table to find the rate of change. Then graph it.","@type":"Question","name":"How do you figure out the rate of something?","acceptedAnswer":"@type":"Answer","text":"To find the number of hours she will work in 12 weeks, write a ratio equal to 60/3 that has a second term of 12. Tonya will work 240 hours in 12 weeks. You could also solve this problem by first finding the unit rate and multiplying it by 12."]

We know that air is being pumped into the balloon at a rate of 5 cm3/min. This is the rate at which the volume is increasing. Recall that rates of change are nothing more than derivatives and so we know that,

As in the previous section when we looked at implicit differentiation, we will typically not use the \(\left( t \right)\) part of things in the formulas, but since this is the first time through one of these we will do that to remind ourselves that they are really functions of \(t\).

Before working another example, we need to make a comment about the set up of the previous problem. When we labeled our sketch, we acknowledged that the hypotenuse is constant and so just called it 15 ft. A common mistake that students will sometimes make here is to also label the hypotenuse as a letter, say \(z\), in this case.

Plugging both of these values into the derivative give us same equation that we got in the example but required a little more effort to get to. It would have been easier to just label the hypotenuse 15 to start off with and not have to worry about remembering that \(z' = 0\).

So, in a general sense each problem was worked in pretty much the same manner. The only real difference between them was coming up with the relationship between the known and unknown quantities. This is often the hardest part of the problem. In many problems the best way to come up with the relationship is to sketch a diagram that shows the situation. This often seems like a silly step but can make all the difference in whether we can find the relationship or not.

So, as we can see if we take the relationship that relates \(r\) and \(h\) that we used in the first part and differentiate it we get a relationship between \(r'\) and \(h'\). At this point all we need to do here is use the result from the first part to get,

Also, this problem showed us that we will often have an equation that contains more variables that we have information about and so, in these cases, we will need to eliminate one (or more) of the variables. In this problem we eliminated the extra variable using the idea of similar triangles. This will not always be how we do this, but many of these problems do use similar triangles so make sure you can use that idea.

We will need \(w'\) to answer this part and we have the following equation from the similar triangle that relate the width to the height and we can quickly differentiate it to get a relationship between \(w'\) and \(h'\).

The tip of the shadow is then moving away from the pole at a rate of 3.6923 ft/sec. Notice as well that we never actually had to use the fact that \(x_p = 25\) for this problem. That will happen on rare occasions.

Below is a copy of the sketch in the problem statement with all the relevant quantities added in. The top of the shadow will be defined by the light rays going over the head of the person and so we again get yet another set of similar triangles.

We are all familiar with the fact that a car speeds up when we put our foot down on the accelerator. The rate of change of the velocity of a particle with respect to time is called its acceleration. If the velocity of the particle changes at a constant rate, then this rate is called the constant acceleration.

Since we are using metres and seconds as our basic units, we will measure acceleration in metres per second per second. This will be abbreviated as m/s\(^2\). It is also commonly abbreviated as ms\(^-2\).

For example, if the velocity of a particle moving in a straight line changes uniformly (at a constant rate of change) from 2 m/s to 5 m/s over one second, then its constant acceleration is 3 m/s\(^2\).

Let \(t\) be the time in seconds from the beginning of the motion of a particle. If the particle has a velocity of 4 m/s initially (at \(t=0\)) and has a constant acceleration of 2 m/s\(^2\), find the velocity of the particle:

Over the first three seconds, the particle's speed is decreasing (the particle is slowing down). At three seconds, the particle is momentarily at rest. After three seconds, the velocity is still decreasing, but the speed is increasing (the particle is going faster and faster).

Throughout this section, we have been considering motion in a straight line with constant acceleration. This situation is very common; for example, a body moving under the influence of gravity travels with a constant acceleration.

Note. Each of the five equations involve four of the five variables \(u, v, x, a, t\). If the values of three of the variables are known, then the remaining values can be found by using two of the equations.

For the graph on the right, the displacement can be found by considering the two triangles between the graph and the \(t\)-axis. One of the triangles has positive signed area and the other has negative signed area.

Motion due to gravity is a good context in which to demonstrate the use of the constant-acceleration formulas. As discussed earlier, our two directions in vertical motion are up and down, and a decision has to be made as to which of the two directions is positive. Acceleration due to gravity is a constant, with magnitude denoted by \(g\). In the following example, we take the upwards direction to be positive and take \(g = 10\) m/s\(^2\).

A man dives from a springboard where his centre of gravity is initially 12 metres above the water, and his initial velocity is 4.9 m/s upwards. Regard the diver as a particle at his centre of gravity, and assume that the diver's motion is vertical.

A constant rate infusion (CRI) is a medication continuously administered to a patient and is used to maintain consistent plasma levels of that medication. CRIs are commonly administered to patients to achieve appropriate levels of pain management, blood pressure management, sedation, anaesthesia, electrolyte supplementation, insulin, and liquid nutrition via a feeding tube. Delivering a CRI will avoid peak and trough levels of pain management and allow titration to suit the individual patient. When using a CRI to manage blood pressure, medication can easily be increased or decreased to obtain optimal effect and discontinued as needed. While CRI management requires 24 hour monitoring and specialised knowledge by the veterinary staff, the ability to maintain medications at therapeutic levels at all times make CRIs worth the time and knowledge. The veterinary nurse needs to not only understand the effects of the drugs being administered, but also how to calculate and create a variety of CRIs. This article will cover different types of CRI calculations and management.

A constant rate infusion (CRI) is prepared to give a patient a continuous dose of drug in intravenous (IV) fluids. This method is advantageous for administering continuous pain management or for drugs with a short half life, as the drug is maintained at effective plasma concentrations for the duration of the CRI (Creedon et al, 2012). CRIs can also be used to deliver liquid nutrition via feeding tubes for those patients unable to tolerate bolus feedings. CRIs can be utilized for many different patient needs such as pain management, blood pressure management, electrolyte supplementation, sedation, anaesthesia, and insulin administration. The drug can be administered without dilution, calculated to be mixed with concurrent IV fluids already being delivered at a pre-set rate, or prepared separate from IV fluids and titrated to different rates depending on patient needs (Silverstein and Hopper, 2009).

Example 6.2.2 A plane is flying directly away from you at 500 mph at an altitude of3 miles. How fast is the plane's distance from you increasing at themoment when the plane is flying over a point on the ground 4 milesfrom you?

Example 6.2.4 Water is poured into a conical container at the rate of 10cm$^3$/sec. The cone points directly down, and it has a height of30 cm and a base radius of 10 cm; see figure 6.2.2.How fast is the water level rising when the water is 4 cm deep (at itsdeepest point)?

We have seen that sometimes there are apparently more than twovariables that change with time, but in reality there are just two, asthe others can be expressed in terms of just two. But sometimes therereally are several variables that change with time; as long as youknow the rates of change of all but one of them you can find the rateof change of the remaining one. As in the case when there are just twovariables, take the derivative of both sides of the equation relating all ofthe variables, and then substitute all of the known values and solve forthe unknown rate.

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