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Instantaneous Rate of Change and Average Rate of Change
I'm currently studying for my calculus AP exam and I'm currently stuck on a question.
Unless otherwise specified, the domain of a function f is assumed to be the set of all real numbers x for which f(x) is a real number.
$\begin{array}{|ccccc|}\hline x& -<!--\; -\; -->3& 0& 3& 6\\ f(x)& -<!--\; -\; -->5& 4& 1& 7\\ f^{\prime }(x)& -<!--\; -\; -->1& 2& -<!--\; -\; -->2& 4\\ \hline\end{array}$
The table above gives values of a twice-differentiable function f and its first derivative $f^{\prime }$ for selected values of x. Let g be the function defined by $g(x)=f(x^2-<!--\; -\; -->x)$.
Let h be the function with derivative given by $h^{\prime }(x)=4e^{\cos <!--\; \; -->x}$. At what value of x in the interval $-<!--\; -\; -->3\le <!--\; \le \; -->x\le <!--\; \le \; -->0$ does the instantaneous rate of change of h equal the average rate of change f over the interval $-<!--\; -\; -->3\le <!--\; \le \; -->x\le <!--\; \le \; -->0$?
I don't need someone to spoonfeed me the answer, but if someone would be able to tell me what theorem or what the first step or two I need to take to get started would be great. Thanks!

Find rate of change of implicit function.
There is a function like that : $\frac{y}{x}=a+b{x}^2+c{y}^4$. I want to find rate of change of $\frac{y}{x}$. If I find $\frac{dy}{dx}$ or $\frac{dx}{dy}$ it will give me rate of change of y (or x) respectively to x (or y). But I want to find rate of the change of the $\frac{dy}{dx}$ itself. How can I do it?

Instantaneous Rate of Change/Derivative
I am having a bit of trouble with a question on my Calc HW. Given $P(x)=3x^2+3,$ estimate the Instantaneous rate of change at x=5. This is what I have so far.
$f(x+h)=3(x+h{)}^2+3-<!--\; -\; -->(3x^2+3)$
$=(3x+3h)(x+h)+3-<!--\; -\; -->(3x^2+3)$
$=3x^2+3xh+3xh+3h^2+3-<!--\; -\; -->(3x^2+3)$
$=3x^2+6xh+3h^2+3-<!--\; -\; -->3x^2-<!--\; -\; -->3$
$=(6xh+3h^2/h)$
$=h(6x+3h)/h$
eliminate h and left with 6x+3h
I know this is incorrect, but don't know where I went wrong. I know for the last step, you have to plug in 5 to x and solve to get the Rate of change.

What is instantaneous rate of change mean in mathematics?
By definition a rate is a ratio of two different units. let us say those units refers to change in x and y of a function. But instant means no change and if there is no change there is nothing to ratio for. Surely instantaneous rate of change is senseless literally. But why do they keep stating it as instantaneous rate of change (Rate of change at a moment) instead of best linear approximation at a point?. Should it be fitting to include it in physics and not in math?

Maximum rate of change for a data
I came up with a question when studying calculus. Suppose that we have a data set, say $v=v(t)$. Assume that we know about their physical meaning and consequently we know that function V fits the data well. Now the maximum rate of change of the data is better approximated by:
a) maximum rate of change of V
b) average rate of change of V
c) we cannot say
I think (b) gives a more reasonable value. But is there a reasoning behind that? or other options?

Rates of change, compounding rates and exponentiationI have a very (apologies if stupidly) simple question about rates of change that has been bugging me for some time. I can't work out whether it relates to my misunderstanding what a rate of change is, to my misapplying the method for calculating a rate of change or something else. I'm hoping somebody on here can help.
For how I define a rate of change, take as an example a population of 1000 items (e.g. bacteria). I observe this population and after an hour I count the size of the population and see that it has increased by 10% (to 1100). I might hypothesise that the population is growing at the rate of 10% per hour, and if, an hour later, I see that it has grown by 10% again (to 1,210) then I might decide to conclude that it is growing at 10% per hour.
So, a rate of change of "proportion x per hour" means "after one hour the population will have changed by proportion x". If, after 1 hour, my population of bacteria was not 1,100, and if not 1,210 after 2 hours, that would mean that the rate of change was not 10% per hour.
First question: Is this a fair definition of a rate of change?
So far so good and it's easy to calculate the population after any given time using a compound interest-type formula.
But whenever I read about continuous change something odd seems to happen. Given that "grows at the rate of 10% per hour" means (i.e. is just another way of saying) "after 1 hour the original population will have increased by 10%", why do textbooks state that continuous change should be measured by the formula:
$P=P_0{e}^{rt}$
And then give the rate of change in a form where this seems to give the wrong answer (i.e. without adjusting it to account for the continuously compounded growth)? I've seen many texts and courses where 10% per day continuous growth is calculated as (for my above example, after 1 day):
$1000\ast <!--\; \ast \; -->e^{1\ast <!--\; \ast \; -->0.1}=1105.17$
This contradicts the definition of a rate of change expressed as "x per unit of time" stated above. If I was observing a population of 1000 bacteria and observed it grow to a population of 1105 after 1 hour I should surely conclude that it was growing at the rate of 10.5% per hour.
I can get the idea of a continuous rate just fine, and it's easy to produce a continuous rate of change that equates to a rate of 10% per day as defined above (that's just ln 1.1). But I struggle to see how a rate of change that means a population grows by 10.5% in an hours means it is growing at 10% per hour. That's like saying if I lend you money at 1% interest per month I'd be charging you 12% per year.
So what's wrong here? Have I got the wrong end of the stick with my definition of a rate of change, would most people interpret a population increase of 10.5% in an hour as a 10% per hour growth rate, or is something else amiss?

What is the Rate of change of $f(x^2)$ given rate of change of f (x)
Find the average rate of change of $f(x^2)$ on the interval [1,4] given that the average rate of change of $f(x)$ equals 9 on interval [1,16]?
This question has two different "answers" according to two different teachers
The first give answer of 9 by assuming $y=x^2$ and applying the formula of rate of change as following Let $y=x^2$ with I= [1,4] then $f(1)$=1,$f(4)$=16
$\frac{f(16)-<!--\; -\; -->f(1)}{(16-<!--\; -\; -->1)}$
The second give the answer of 45 as following
$\frac{f(16)-<!--\; -\; -->f(1)}{16-<!--\; -\; -->1}=9$
$\frac{f(16)-<!--\; -\; -->f(1)}{15}=9$
$f(16)-<!--\; -\; -->f(1)=9\ast <!--\; \ast \; -->15$
Now we find rate of change of $f(x^2)$ following $\frac{f(16)-<!--\; -\; -->f(1)}{4-<!--\; -\; -->1}=\frac{9\ast <!--\; \ast \; -->15}{3}=45$ what is the right answer?
Can we ensure the either answers geometrically? Thank you for helping

Rates of change
I’m having some trouble with part c) of the following questions,
a) What is the rate of change of the area A of a square with respect to its side x?
b) What is the rate of change of the area A of a circle with respect to its radius r?
c) Explain why one answer is the perimeter of the figure but the other answer is not.
So, knowing that if we have a square with side length x, then the area of the square as a function of its side is $A(x)=x^2$. The perimeter as a function of the side is $P(x)=4x$. And the rate of change of the area wrt its side is $\frac{dA}{dx}=2x$. With a circle, the area as a function of the radius is $A(r)=\mathrm{\pi <!--\; \pi \; -->}(r^2)$. And the rate of change of the area wrt its radius is $\frac{dA}{dr}=2\mathrm{\pi <!--\; \pi \; -->}(r)$. The circumference as a function of the radius is also $C(r)=2\mathrm{\pi <!--\; \pi \; -->}(r)$. Therefore it’s the circle that’s the figure with the rate of change of the area wrt its radius equal to its perimeter, and what I saw was that the square had a rate of change of area wrt its side equal to half the perimeter of the square, $\frac{dA}{dx}=2x=\frac{4x}{2}$
I inscribed a circle in a square with radius equal to half the square’s side length and went through the same work and then arrived at this, $A(r)=\mathrm{\pi <!--\; \pi \; -->}(\frac{x}{2}{)}^2=\frac{\mathrm{\pi <!--\; \pi \; -->}}{4}{x}^2$ and $C(\frac{x}{2})=2\mathrm{\pi <!--\; \pi \; -->}(\frac{x}{2}$ and that $\frac{dA}{dr}=\frac{\mathrm{\pi <!--\; \pi \; -->}}{2}x$
Somehow in this example, I don’t think it’s correct because the same fact about the rate of change of area wrt radius being equal to perimeter doesn’t hold. I appreciate any help in explaining this, thank you.

What do instantaneous rates of change really represent?The derivative of $f(x)$ is the value of the limit of the average rate of change of y with respect to x as the change in x approaches 0. This is the value, in other words, that the average rate of change approaches but NEVER hits.
This means that it is NOT the infintesimal rate of change of y with respect to $dy/dx$ merely approaches the derivative's value. If the rate of change did actually achieve 0 change in x, you'd get 0/0 which is an indeterminant form.
So if the derivative is the literal rate of change at an exact instant -- a rate of change with an interval of 0, what does that actually tell you? Can a specific moment in time really have a rate of change? Is that rate of change ever even maintained, even at a specific instant?
I know that a point by itself can't have a rate of change, you need a continuum of points around it to determine one (hence a limit). What does an instantaneous rate of change tell you?

Rate of change of surface area of a sphere given the rate of change of the radius
Air is pumped into a spherical balloon such that the radius of the balloon increases at the rate of ${\displaystyle \frac{1}{20}}\mathrm{\pi <!--\; \pi \; -->}$ cm/s when the radius is 8.5cm. Find the rate of change of the surface area of the balloon at this instant.
How do I do this? I still don't really understand how rate of change works here.

Rate of Change with DerivativesWe just started with rate of change while using derivatives and I am stuck on a question, hope you can help.
A particle moves on a vertical line so that its altitude at time t is $y=t^3-12·t+3$, where $t\ge 0$. Find the distance that the particle travels in the time interval $0\le t\le 3$

Rate of change $\frac{dV}{dt}$
I have the following task: "Poiscuille's Law: $V=\frac{P}{4Lv}(R^2-<!--\; -\; -->r^2)$. Assume that r is a constant as well as P,L,v. Find the rate of change $\frac{dV}{dt}$ in terms of R and $\frac{dR}{dt}$ when L=1mm, p=100, $v=0.05$"
I cannot understand how can how can diff that, what is the t here? I can write $V=500(R^2-<!--\; -\; -->r^2)$. But what after? And what is "find the rate of change in terms of...? Something like:$\frac{dV}{dt}=\frac{dV}{dR}\frac{dR}{dt}$? I am right?

Find highest rate of change of any function
Is there any way to find where the rate of change of function is maximum/highest?
Suppose we have a sine wave and i want to know where the rate of change of the function is maximum highest?
How can i find that?

What is the unit of this "Rate of Change"?
the function $f(x)=0.9x$
So, the rate of change of y with respect to x is 0.9.
My question is, what is the unit of this Rate of Change?
0.9 per x?

Rate of change vs. average rate of change of a function
What is the difference between the both? I am trying to find a separate definitions for both in a textbook, but it only defines average rate of change.
I know that average rate of change is:
$\frac{changeiny}{changeinx}$
But what is just rate of change then?

Rate of change of cross-section of cylinder
I got this task on my calculus class and I got stuck at process of figuring it out
A clay cylinder is being compressed so that its height is changing at the rate of 4 millimeters per second, and its diameter is increasing at the rate of 2 millimeters per second. Find the rate of change of the area of the horizontal cross-section of the cylinder when its height is 1 centimeter.
What I know:
Volume is unknown and constant
Rate of change of diameter is
${\displaystyle \frac{dD}{dt}}=0,2\text{cm}/\text{sec}$
Rate of change of height is
${\displaystyle \frac{dh}{dt}}=-<!--\; -\; -->0,4\text{cm}/\text{sec}$
Trying to find ${\displaystyle \frac{dD}{dt}}$ by using formula
$V=\mathrm{\pi <!--\; \pi \; -->}{r}^{2}h$