Many applications of the derivative involve determining the rate of change at a given instant of a function with the independent variable time—which is why the term
instantaneous is used. Consider the height of a ball tossed upward with an initial velocity of 64 feet per second, given by
$\text{\hspace{0.17em}}s(t)=\mathrm{-16}{t}^{2}+64t+6,$ where
$\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ is measured in seconds and
$\text{\hspace{0.17em}}s\left(t\right)\text{\hspace{0.17em}}$ is measured in feet. We know the path is that of a parabola. The derivative will tell us how the height is changing at any given point in time. The height of the ball is shown in
[link] as a function of time. In physics, we call this the “
s -
t graph.”
Finding the instantaneous rate of change
Using the function above,
$\text{\hspace{0.17em}}s(t)=\mathrm{-16}{t}^{2}+64t+6,$ what is the instantaneous velocity of the ball at 1 second and 3 seconds into its flight?
The velocity at
$\text{\hspace{0.17em}}t=1\text{\hspace{0.17em}}$ and
$\text{\hspace{0.17em}}t=3\text{\hspace{0.17em}}$ is the instantaneous rate of change of distance per time, or velocity. Notice that the initial height is 6 feet. To find the instantaneous velocity, we find the
derivative and evaluate it at
$\text{\hspace{0.17em}}t=1\text{\hspace{0.17em}}$ and
$\text{\hspace{0.17em}}t=3:$
For any value of
$\text{\hspace{0.17em}}t$ ,
$\text{\hspace{0.17em}}{s}^{\prime}\left(t\right)\text{\hspace{0.17em}}$ tells us the velocity at that value of
$\text{\hspace{0.17em}}t.$
Evaluate
$\text{\hspace{0.17em}}t=1\text{\hspace{0.17em}}$ and
$\text{\hspace{0.17em}}t=3.$
The position of the ball is given by
$\text{\hspace{0.17em}}s(t)=\mathrm{-16}{t}^{2}+64t+6.\text{\hspace{0.17em}}$ What is its velocity 2 seconds into flight?
Using graphs to find instantaneous rates of change
We can estimate an instantaneous rate of change at
$\text{\hspace{0.17em}}x=a\text{\hspace{0.17em}}$ by observing the slope of the curve of the function
$\text{\hspace{0.17em}}f\left(x\right)\text{\hspace{0.17em}}$ at
$\text{\hspace{0.17em}}x=a.\text{\hspace{0.17em}}$ We do this by drawing a line tangent to the function at
$\text{\hspace{0.17em}}x=a\text{\hspace{0.17em}}$ and finding its slope.
Given a graph of a function
$\text{\hspace{0.17em}}f\left(x\right),\text{\hspace{0.17em}}$ find the instantaneous rate of change of the function at
$\text{\hspace{0.17em}}x=a.$
Locate
$\text{\hspace{0.17em}}x=a\text{\hspace{0.17em}}$ on the graph of the function
$\text{\hspace{0.17em}}f\left(x\right).$
Draw a tangent line, a line that goes through
$\text{\hspace{0.17em}}x=a\text{\hspace{0.17em}}$ at
$\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ and at no other point in that section of the curve. Extend the line far enough to calculate its slope as
$$\frac{\text{changein}y}{\text{changein}x}.$$
Estimating the derivative at a point on the graph of a function
From the graph of the function
$\text{\hspace{0.17em}}y=f\left(x\right)\text{\hspace{0.17em}}$ presented in
[link] , estimate each of the following:
$f(0)$
$f(2)$
$f\text{'}(0)$
$f\text{'}(2)$
To find the functional value,
$\text{\hspace{0.17em}}f\left(a\right),$ find the
y -coordinate at
$\text{\hspace{0.17em}}x=a.$
To find the
derivative at
$\text{\hspace{0.17em}}x=a,$$\text{\hspace{0.17em}}{f}^{\prime}\left(a\right),$ draw a tangent line at
$\text{\hspace{0.17em}}x=a,$ and estimate the slope of that tangent line. See
[link] .
$f(0)\text{\hspace{0.17em}}$ is the
y -coordinate at
$\text{\hspace{0.17em}}x=0.\text{\hspace{0.17em}}$ The point has coordinates
$\text{\hspace{0.17em}}\left(0,1\right),$ thus
$\text{\hspace{0.17em}}f(0)=1.$
$f(2)\text{\hspace{0.17em}}$ is the
y -coordinate at
$\text{\hspace{0.17em}}x=2.\text{\hspace{0.17em}}$ The point has coordinates
$\text{\hspace{0.17em}}\left(2,1\right),$ thus
$\text{\hspace{0.17em}}f(2)=1.$
${f}^{\prime}(0)\text{\hspace{0.17em}}$ is found by estimating the slope of the tangent line to the curve at
$\text{\hspace{0.17em}}x=0.\text{\hspace{0.17em}}$ The tangent line to the curve at
$\text{\hspace{0.17em}}x=0\text{\hspace{0.17em}}$ appears horizontal. Horizontal lines have a slope of 0, thus
$\text{\hspace{0.17em}}{f}^{\prime}(0)=0.$
${f}^{\prime}(2)\text{\hspace{0.17em}}$ is found by estimating the slope of the tangent line to the curve at
$\text{\hspace{0.17em}}x=2.\text{\hspace{0.17em}}$ Observe the path of the tangent line to the curve at
$\text{\hspace{0.17em}}x=2.\text{\hspace{0.17em}}$ As the
$\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ value moves one unit to the right, the
$\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ value moves up four units to another point on the line. Thus, the slope is 4, so
$\text{\hspace{0.17em}}{f}^{\prime}(2)=4.$
The center is at (3,4) a focus is at (3,-1) and the lenght of the major axis is 26 what will be the answer?
Rima
I done know
Joe
What kind of answer is that😑?
Rima
I had just woken up when i got this message
Joe
Can you please help me. Tomorrow is the deadline of my assignment then I don't know how to solve that
Rima
i have a question.
Abdul
how do you find the real and complex roots of a polynomial?
Abdul
@abdul with delta maybe which is b(square)-4ac=result then the 1st root -b-radical delta over 2a and the 2nd root -b+radical delta over 2a. I am not sure if this was your question but check it up
Nare
This is the actual question: Find all roots(real and complex) of the polynomial f(x)=6x^3 + x^2 - 4x + 1
Abdul
@Nare please let me know if you can solve it.
Abdul
I have a question
juweeriya
hello guys I'm new here? will you happy with me
mustapha
The average annual population increase of a pack of wolves is 25.
Period =2π
if there is a coefficient (b), just divide the coefficient by 2π to get the new period
Am
if not then how would I find it from a graph
Imani
by looking at the graph, find the distance between two consecutive maximum points (the highest points of the wave). so if the top of one wave is at point A (1,2) and the next top of the wave is at point B (6,2), then the period is 5, the difference of the x-coordinates.
Am
you could also do it with two consecutive minimum points or x-intercepts
the range is twice of the natural number which is the domain
Morolake
A cell phone company offers two plans for minutes. Plan A: $15 per month and $2 for every 300 texts. Plan B: $25 per month and $0.50 for every 100 texts. How many texts would you need to send per month for plan B to save you money?
Hey I am new to precalculus, and wanted clarification please on what sine is as I am floored by the terms in this app? I don't mean to sound stupid but I have only completed up to college algebra.
I don't know if you are looking for a deeper answer or not, but the sine of an angle in a right triangle is the length of the opposite side to the angle in question divided by the length of the hypotenuse of said triangle.
Marco
can you give me sir tips to quickly understand precalculus. Im new too in that topic.
Thanks
Jenica
if you remember sine, cosine, and tangent from geometry, all the relationships are the same but they use x y and r instead (x is adjacent, y is opposite, and r is hypotenuse).
Natalie
it is better to use unit circle than triangle .triangle is only used for acute angles but you can begin with. Download any application named"unit circle" you find in it all you need. unit circle is a circle centred at origine (0;0) with radius r= 1.
SLIMANE
What is domain
johnphilip
the standard equation
of the ellipse that has vertices (0,-4)&(0,4) and foci (0, -15)&(0,15)
it's standard equation is x^2 + y^2/16 =1
tell my why is it only x^2? why is there no a^2?
This term is plural for a focus, it is used for conic sections. For more detail or other math questions. I recommend researching on "Khan academy" or watching "The Organic Chemistry Tutor" YouTube channel.