Given y = f(x), the derivative dy/dx can be interpreted as the rate of change of y with respect to x in many areas such as physics, chemistry, biology, economics, and other sciences. Depending on your major or area of interest, please choose ONE example in Section 3.7 of the textbook (Calculus, Early Transcendentals, 6th edition, by James Stewart) and answer the following questions:
1) Specify the number of the example you chose (example 1, example 2, etc).
2) In this example, please analyze how the concept of the derivative is used.
3) In the context of your chosen area, please make up an example where a derivative needs to be computed to describe some rate of change.
If you do not have the textbook, you can look at the examples of Section 3.7 on the class website. Notice that the examples are classified under the associated areas (i.e., physics, chemistry, biology, and economics).
The deadline to submit your comments is a week from now (i.e., Thursday, 10/25). Late submission will be graded zero. Besides, there will be NO extension for this assignment since the Review Set for Test 3 will be posted next Thursday.
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Chemistry
Example 4
Example 4 in the book gives a practical example of how calculus relates to chemistry. It states that when there is a reaction between one or more substances there are two sides of a chemical equation. One side is made up of the reactants, the substances that are reacting, and the product, or end result of the reaction. As the reaction progresses over time the concentrations of the reactants decrease and the concentration of the product increases. The concentration of each is a function of time. Derivative is used to find the rate of reaction.
For example if you had two substances such as sodium, Na, and chloride, Cl, and were to combine them, the end result would be sodium chloride, NaCl, or salt. The chemical equation would look like, Na + Cl = NaCl. As the reaction between Na and Cl progresses and more and more NaCl is formed the amount of free Na and Cl decrease. Each concentration, Na, Cl, and NaCl, vary over the time of the reaction so it can be said that each concentration is a function of time. In order to find the rate of reaction over time the derivative is used. The equation would then be, rate of reaction = d [NaCl]/dt = - d [Na]/dt = - d [Cl]/dt. The concentration of NaCl increases as the reaction proceeds so it is positive and has a positive slope but the concentrations of Na and Cl decrease, and have a negative slope when graphed, so a negative sign is placed in front of the derivative to make it positive.
Karen Sanders
Physics
Example 3
In my example, we look at how a derivative is used in physics to calculate an (instantaneous) electric current. Electric currents exist when electric charges move; they are a measurement of the charge that passes through a surface at a given moment in time. Thus the function of an electric current is the derivative of the function of electric charge over time.
In physics, derivatives are often used to compute rates of change. For example, we have currents, like in my example above, which is the derivative of charge with respect to time. We also have velocity, which is the change in distance over time; acceleration is the change in velocity over time. As I just learned in physics class the other day, torque (a force exerted on a rotating object) is the derivative of angular momentum with respect to time.
Katie Moore
Economics
Example 8
In my example the definition of a derivative is used to determine the marginal cost of an item, which is the instantaneous rate of change of cost with respect to the number of items produced. My example takes a cost function, the total cost that a company incurs in producing x units of a certain commodity, C, and finds it's derivative with our definition. This new formula is the marginal function which determines the marginal cost at the production level of x.
With this derivative, one can determine the rate of change in additional costs to produce "nth" more units.
C'(n) = C(n+1)-C(n)
In conclusion, a company can use the derivative as a shorter, and precise equation to determine the cost of producing one more unit than they currently are.
Jonathan Kight
Physics
Example 1
In example 1 the derivate is being use to calculate Velocity from a position function and the the acceleration from the velocity function (second derivative). If you know the position function then you are able to find the instantaneous velocity at any time as well as the acceleration at anytime.
In physics these derivatives can be used to help determine how far and fast something will go before it stops. A practical application for this would be if one is designing cars. He would need to now how long it would take the car to stop given different starting velocities.
Peter Harris
Chemistry
Example 5
This example is about compressibility. If a substance is kept at a constant temperature, then its volume depends on its pressure. The rate of change of volume with respect to pressure is the derivative dV/dP. To define compressibility, just take the derivative, make it negative, and divide it by the volume.
An example where this can be applied is if you have a substance at constant temperature, and you also have the volume and temperature, you can find the compressibility, how fast the volume increases as the pressure increases.
Rachel Discavage
Example 1 the instantaneous rate of change of velocity with respect to time is acceleration: a(t)=v'(t)=s"(t). Here the velocity function is the derivative of the position function.
If s=f(t)=t^3-6t^2+9t, then v(t)=3t^2-12t+9. The velocity after 6s means the instantaneous velocity when t=4 v(t)=ds/dt=3(6)^2-12(6)+9=45m/s. The particle at rest when v(t)=0 is, 3t^2-12t+9=3(t^2-4t+3)=3(t-1)(t-3)=0 when t=1 or t=3. This means that the particle is at rest when t=1 and t=3. The particle moves in the positive direction in the time intervals t less than 1 and t greater than 3. It moves in the negative direction when 1 less than t less than 3.
The distance traveled in the first second is [f(1)-f(0)]=[9-0]=9 from t=1 to t=3 [f(3)-f(1)]=[0-9]=9 and from t=3 to t=5 [f(5]-f(3)]=[24-0]=24 the total distance is 9+9+24=42m. The acceleration is the derivative of the velocity function: a(t)=d^2s/dt^2=dv/dt=6t-12 so a(6)=6(6)-12=24m/s^2. The particle speeds up when the velocity and acceleratoin have the same sign(when 1 less than t less than 2 and t great than 3. The particle slows down when v and a have opposite signs, when and when 2 less than t less than 3.
Marcus Williams
Physics
Example 1
In this example, the concept of a derivative is used to link the three different functions of a moving object (position, velocity, and acceleration) with respect to time. The derivative of the position function gives you the graph of the velocity function, which can be used to find the instantaneous velocity of the object at any point in a specified time span. The derivative of the velocity function is the acceleration of the object over the given time period.
This relationship among the three functions of the object can be expressed as:
a(t) = v'(t) = s''(t)
From this, we can assume the following relationships are also true:
s'(t) = v(t)
v'(t) = a(t)
Example:
If the function s(t)=(t^2)-t+1 is given for the position of an object at time (t), and you are asked to find the acceleration of the object at t=2, you must take the second derivative of the position function to find the acceleration function:
s''(t)=a(t)=2t-1
Then, you plug the value of t into this equation:
a(2)=2(2)-1
a(2)=3
Thus, the acceleration of the object at t=2 is 3 m/s^2.
Levon Hoomes
1) I picked example number 1 about physics.
2) In physics if the position function S=f(t) is moving in a straight line then the change in seconds over the change in time is the average velocity over that time period. The derivative of that represents the instantaneous velocity which is the change of displacement with respect to time. The derivative of this ( the instantaneous change of velocity with respect to time) is its acceleration. This can be very useful in many aspects of life. Through the use of calculus and physics together you can explain and predict almost anything that relates to movement of an object.
Example: The position of an object is given by the equation f(x)9t^2-3t. What is the objects velocity at 2 seconds? In order to do this you would take the derivative => f'(x)=18t-3. Then simply plug in 2 into that so, f'(2)=18(2)-3. Which equals 33 m/s.
-Jasper W. Yonker
1. I used example 1
2. They give you the position function and ask you find specific velocities and also the general velocity for all times (t). In order to find velocity, you must take the derivative of the position function. The derivative also can tell you d by finding the critical points and finding where the particle is moving forward or increasing. In the next portion of the problem, it asks you to find the acceleration at time t which is simply the derivative of the velocity function or the second derivative of the position function.
3. A ball is thrown into the air and its position is found through the equation: s = f(x) = 2x^2 + 10x + 8
Find the velocity of the ball at time t=6 seconds.
Example 1
Physics:
IN this example we are given the position function relative to time of a particle. The example asks us to solve for many things but for the sake of derivatives i'm just going to focus on the velocity (the rate of distance relative to time)and acceleration (the rate of velocity relative tot time). For these problems knowing that velocity, v(t), is the derivative of the particles position, f(t), and that acceleration, a(t) is the derivative of velocity, v(t, is a very useful detail.
For parts a, b, and c all that is required is to find the derivative of f(t) to solve for v(t) and then plug i the t-values (2 and 4 seconds)to find your v-values. And use t=0 to find the particles velocity at rest. which should be zero.
For part g you must find the second derivative of f(t) and plug in the t-value (4seconds) to solve for the a(t)-value.
This information is very useful especially in physics since is is the science of motion, physicists are continually finding the speed(velocity) and acceleration of objects. Like if an engineer had to find how banked he could make a curved road for a race car driver to make it safely around the bend at top speeds without skidding and losing control. Velocity and acceleration need to be known for these situations and that is all possible through derivatives.
-Kelly Gustafson
Physics Example 1
In example 1 you are given the position function and asked to find the velocity and acceleration. you can find the velocity of the position function by taking the first derivative, and the acceleration can be found by taking the second derivative of the position function. If the position function is known then you can find the velocity and acceleration at any time.
Derivatives can be used to find lots of things in physics. In physics you can use derivatives to find out things like how fast a particular object is moving, and how far it will go.
william wigglesworth
1. I picked example number 1:Physics
2.A function that has a certain position, in physics, that is moving in a forward rate of change uses derivatives. For instance if you want to find the velocity of a specific function at a specific time you would take the first derivative. Now to find the acceleration of a given function at a specific time just take the derivative of the velocity which is the "double" derivative.
3. A football is thrown at a position function of F(x)=x^2+7x+7. What is the velocity after 1 second. To find this you must take the derivative of the function: F'(x)=2x+7. This gives you the velocity of the function. Then plug 1 into the function to find the velocity at the specific time: F'(1)= 2(1)+7=9. The velocity at 1 second for the function F(x)=x^2+7X+7 is 9m/s.
1.) In example four it uses derivatives as the rate of change of y, to solve a chemistry related problem.
2.) They set up an equation to find the rate of change of a concentration. The used the reactants yielding to products in an equation stated as A+B=C, then they took the derivative with respect to time because all these measurements are manipulated with time. They then set up the rate of change equal to the instantaneous rate of reaction. They solved the instantaneous rate of reaction and found that it will be positive, but the concentrations will decrease so they set the derivatives of A and B with respect to time as negative. Since the concentration of the product increases as the reactants decrease the equation of rate of reaction is now DC/dt=-DA/dt=-DB/dt. Therefore, for a reaction that has variables like aA+bB=cC+dD, it would look like this:
-1/a (DA/dt) = -1/b(DB/dt)= 1/c(DC/dt) =1/d(DD/dt).
3.) For example if you have two elements that combine like Hydrogen and Chloride. The end result would be hydrochloric acid. The chemical equation would look like: H+ + CL- =HCL, the Hydrogen Ion and the chloride Ion unite their concentration would be decreasing while the concentration of the new substance being created would be increasing all at the same rate. To be able to find that rate of change the derivative must be taken. If implemented into a graph the slope of the reactants in this case H+ and CL- would be decreasing while the slope of the product HCL would be increasing. Meaning the derivative of this equation needs to have a negative in front of the derivatives of the reactants to make them positive.
1)I chose example 1 from the physics lecture notes.
2)In physics, taking the derivative finds the velocity (second derivative finds acceleration) from speed. If you have the equation with the variable (X) being time, than you can take its derivative, and plug any number in for X (your time). This will give you your velocity at that given time frame.
3) Another time a derivative could be used to find the velocity of a falling object. The equation could be as fallows
s(x)= 2X^2+3X+1
taking the derivative you get
V(x)=4x+3
than you can plug in any time for X and find your velocity.
Michelle Sokoll
1. example 1
2. The derivative is used in this example to show that when you find the first derivative, it is the slope of the line which is equal to the velocity at that given point. If you were then to find the derivative of the velocity, that would give you the acceleration at that point.
3. In physics you can use it if you need to find the change in velocity or acceleration of a ball shot out of a cannon if you know the path of it's trajectory.
Trevor Hubbard
1) since I don't own the book, i'll choose example one from the physics lecture
2) in order to find the velocity, you need to take the derivative. once the derivative is found you can plug in your t (time) and find the velocity.
3) An example could be....finding the velocity of an object that is dropped from a building. In order to find the velocity, take the equation and find the derivative. Plug in the T to find the velocity.
Bishop Toups
1) I CHOSE THE EXAMPLE PERTAINING TO BIOLOGY.
> IN THE FIELD OF BIOLOGY YOU CAN USE DERIVATIVES TO FIND THE INCREASE OF A POPULATION OVER A SPECIFIC PERIOD OF TIME, YOU CAN ALSO FIND THE DIFFERENT DEPTH PERCEPTIONS OF AN ANIMALS LIFE SPAN OR FOOD CHAIN. TO FIND THE INCREASE OF A POPULATION OVER TIME YOU CAN JUST TAKE THE POPULATION OF THE FIRST GENERATION AND ALL OF THE POSSIBLE CONFLICTS WITHIN THAT GENERATION AND THEN TAKES ITS DERIVATIVE, I,E; d/dX(A^X) = a^XLN a. OR THE GROWTH RATE OF A POPULATION MAY IT BE BACTERIA A PARASITE OR MAMMAL dn/dt = d/dt(N0 2^t)= No2^tlN2. WHERE "N" IN THIS DERIVATIVE IS CONSIDERED THE POPULATION AND "t" BEING THE RATE OF CHANGE OF THAT POPULATION OVER TIME. ALMOST EVERYTHING IN BIOLOGY CONSISTS OF MATHEMATICS AND THE USE OF DERIVATIVES.IN THE FIELD OF ECOLOGY THE SCIENTISTS USE THE DERIVATIVES TO FIND THE GROWTH RATE OF SEEDS TO SEEDLINGS AND FLOWERS ALL THE TIME.
Casey mauldin
1) Physics example
2)In physics, the derivative is a very important aspect. The derivative of the position function represents the velocity, the rate of change of position with respect to time. The derivative of the velocity function is acceleration, therate of change of velocity with respect to time.
3) If a particle moves along the function f(t) = t^3 + 2t^2 + 5t, the velocity of the particle would be found using the function f'(x) = 3t^2 + 4t + 5. At a certain time, such as t=2, the instantanious velocity would be 25.
Barbra Giourgas
I chose example 1, physics. The derivative is used to find the velocity from the position function with respect to time. You can then find the rate of change of velocity with respect to time which is called acceleration by taking the second derivative.
If a car was leaving a certain place and arrives at a destination in a particular amount of time, you could use the derivatives to find instantaneous velocity, and acceleration.
Matthew Copello
Physics Example 3
In this example, we are looking at the average current of an electrical charge as it passes through a given surface(in this case a wire). We calculate this by dividing the change in the charge by the change in time. In order to find the instantaneous current of this charge we need to use our knowledge of limits combined with this knowledge about the derivative. As we let our time interval approach to zero we come to our instantaneous current, which we refer to as amps.
We use this information in almost all modern day technology, by discovering this rate at which the energy travels through a given source we can determine the amount of energy that will be supplied to our end unit.
Brian Thiele
Example 1
In this example calculus is used to analyze a position functio. With the use of derivatives, the acceleration and velocity of an object can be found at certain times. It is important to understand that the derivatives of a position function s=f(t) is equal to the instantaneous velocity v=ds/dt. Also, the derivative of the velocity is equal to the acceleration a=dv/dt. Once these equation are derived you can simply plug in a value of t(time). to find the acceleration or velocity at that time.
If the position of a particle is given by s=f(t)=t^3-6t^2+9t and we want to find the velocity and the acceleration of the object after a time of t=2seconds, you take the following steps. V9t)=ds/dt=3t^2-12t+9 is the derivative of the position and equal to the velocity of the object at time t. Plug 2 into this equation for t and you arrive at a velocity of -3m/s. To find the acceleration we take the derivative of the velocity, which is a(t)=dv/dt=6t-12. Then we plug in 2 for t and come to an acceleration of -12m/s^2.
Michael Lyons
Physics Example 1
Example 1 describes how to calculate velocity by finding the derivative of the position function. Once you find the velocity, a second derivative of that gives you the acceleration. Velocity and acceleration are huge components when determining physics.
One example of this application would be in the use of cops reviewing car crashes. You could measure where the tires started to screech as the position function and then determine, based on point of impact, how fast a certain car is going to determine whether the driver was speeding.
Trent Layton
1. I picked example number 1 about physics.
2. In this example the derivative is found by taking the slope of the line which we know is equal to the velocity at that given point. When you find the derivative of velocity that gives you the acceleration whihc is also crucial for physics.
3. In physics calculus, and the the derivative especially is used to find velocity and acceleration as describes in a practical convienetet fashion. An example of this is finding the acceleration as a ball is slingshotted into the air.
1) Example 6 Biology
2)In this example, the derivative is found as a way to find the rate of growth of a bacteria population. Since finding a rate of growth can not be completely accurate since there are different rates of death and birth at certain times, the use of finding an average rate of groth from a limit is not as accurate. That is why a curve is estimated by finding the derivative, like with the bacteria.
3) Most of biology uses mathematics and derivatives. When Biologists are trying to find correlations in genetic variation, they use derivatives to find the growth rate or occurance of certain genetic diseases or deformities. This type of calculus is used all of the time in biology research.
1. I Choose example 6 for my major Biology.
2. They used the concept of derivative to find the instantaneous rate of growth of a group of animals or plant populations. From looking at the graph of this calculation it shows that this function is discontinuos when a death or birth happens. In the other examples for biology they showed how derivatives were used to find flow of blood in a vessel and growth of bacteria in a certain time frame.
3. An example that I can think of to show a rate of change regarding biology would be the decay of a plant without water and nutrients. We did an experiment similar to this in my Bio lab and within a certain time period t=0 and t=2 hours we were able to calculate this rate of change.
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