Rate of change Word Problem

To find the "instantaneous" rate of change at x= a, you would need to find the derivative at x= a. If you are not familiar with derivatives, that will not be easy. The "vertical asymptotes" of a function are the values of x at which the function is undefined. The denominator of your function, C(x)= (3x^3- x^2+ 5x)/(x^3+ 10), is 0 when x= -10. That value of x is a "vertical asymptote".
  • #1
ArielM
2
0
Hi everyone, i am new to this website and i would like to ask one question that i don't quite get.
we have started a new topic in class about rates of change


Homework Statement


An Electrical current in a cicruit varies with time according to C=(3s^3-s^2+5s)/(S^3+10)
where currenct is "C" and time is "s" in seconds.

a) find the average rate of change from 0.75 seconds to 1.5 second
b)find the instantanious rate of change at 1.5 second.
c)Identify any vertical asymptotes.

Homework Equations





The Attempt at a Solution


I have managed to solved part a by substituting the values 0.75 and 1.5 to the equation.
i then took both answers and found the average rate of change by the formula y2-y1/x2-x1 (correct me if I am wrong)

as for finding the instanatnious rate of change - i am completley clueless :(
Thanks in advance !

Ariel Melichovich
 
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  • #2
Welcome to PF!

Hi Ariel! Welcome to PF! :smile:
ArielM said:
n Electrical current in a cicruit varies with time according to C=(3s^3-s^2+5s)/(S^3+10)
where currenct is "C" and time is "s" in seconds.

as for finding the instanatnious rate of change - i am completley clueless :(

Instantaneous rate of change is the derivative, dC/ds. :smile:
 
  • #3


tiny-tim said:
Hi Ariel! Welcome to PF! :smile:


Instantaneous rate of change is the derivative, dC/ds. :smile:

Thank you for your reply !
however, i am not quite sure what does the term 'd' mean. is that the derivative?

can the form (x,x+h) be apllied to this question where at the end i divide the equation by "h"?

Thanks again !
 
  • #4
If you do not know how to find a derivative, then you cannot do this problem. The derivative is the "instantaneous rate of change". But from what you say, you seem to have heard of the basics of the derivative: it is the limit, as h goes to 0, of the average rate of change from x to x+ h. However, for the rational function you have, that is going to be a difficult algebraic calculation.

Your calculation of the average rate of change is correct
 

1. What is a rate of change word problem?

A rate of change word problem involves finding the change in one variable over a specific period of time or under certain conditions. It requires identifying the independent and dependent variables, and calculating the rate of change between them.

2. How do I solve a rate of change word problem?

To solve a rate of change word problem, you first need to identify the independent and dependent variables. Then, you can use the given information to calculate the rate of change by dividing the change in the dependent variable by the change in the independent variable.

3. What is the formula for calculating rate of change?

The formula for calculating rate of change is: rate of change = (change in dependent variable) / (change in independent variable). This can also be written as: rate of change = (y2-y1) / (x2-x1), where (x1,y1) and (x2,y2) represent two points on the graph.

4. How is rate of change useful in real life?

Rate of change is useful in real life because it helps us understand how one variable affects another. For example, we can use it to calculate the average speed of a car, the growth rate of a population, or the change in stock prices over time.

5. What are some common misconceptions about rate of change?

One common misconception about rate of change is that it always has to be a constant rate. In reality, the rate of change can vary at different points on a graph. Another misconception is that the dependent variable always increases with the independent variable, when in fact it can also decrease or remain constant.

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