# 4.10 Antiderivatives  (Page 5/10)

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Solve the initial value problem $\frac{dy}{dx}=3{x}^{-2},y\left(1\right)=2.$

$y=-\frac{3}{x}+5$

Initial-value problems arise in many applications. Next we consider a problem in which a driver applies the brakes in a car. We are interested in how long it takes for the car to stop. Recall that the velocity function $v\left(t\right)$ is the derivative of a position function $s\left(t\right),$ and the acceleration $a\left(t\right)$ is the derivative of the velocity function. In earlier examples in the text, we could calculate the velocity from the position and then compute the acceleration from the velocity. In the next example we work the other way around. Given an acceleration function, we calculate the velocity function. We then use the velocity function to determine the position function.

## Decelerating car

A car is traveling at the rate of $88$ ft/sec $\left(60$ mph) when the brakes are applied. The car begins decelerating at a constant rate of $15$ ft/sec 2 .

1. How many seconds elapse before the car stops?
2. How far does the car travel during that time?
1. First we introduce variables for this problem. Let $t$ be the time (in seconds) after the brakes are first applied. Let $a\left(t\right)$ be the acceleration of the car (in feet per seconds squared) at time $t.$ Let $v\left(t\right)$ be the velocity of the car (in feet per second) at time $t.$ Let $s\left(t\right)$ be the car’s position (in feet) beyond the point where the brakes are applied at time $t.$
The car is traveling at a rate of $88\phantom{\rule{0.2em}{0ex}}\text{ft/sec}.$ Therefore, the initial velocity is $v\left(0\right)=88$ ft/sec. Since the car is decelerating, the acceleration is
$a\left(t\right)=-15{\phantom{\rule{0.2em}{0ex}}\text{ft/s}}^{2}.$

The acceleration is the derivative of the velocity,
${v}^{\prime }\left(t\right)=15.$

Therefore, we have an initial-value problem to solve:
${v}^{\prime }\left(t\right)=-15,v\left(0\right)=88.$

Integrating, we find that
$v\left(t\right)=-15t+C.$

Since $v\left(0\right)=88,C=88.$ Thus, the velocity function is
$v\left(t\right)=-15t+88.$

To find how long it takes for the car to stop, we need to find the time $t$ such that the velocity is zero. Solving $-15t+88=0,$ we obtain $t=\frac{88}{15}$ sec.
2. To find how far the car travels during this time, we need to find the position of the car after $\frac{88}{15}$ sec. We know the velocity $v\left(t\right)$ is the derivative of the position $s\left(t\right).$ Consider the initial position to be $s\left(0\right)=0.$ Therefore, we need to solve the initial-value problem
${s}^{\prime }\left(t\right)=-15t+88,s\left(0\right)=0.$

Integrating, we have
$s\left(t\right)=-\frac{15}{2}{t}^{2}+88t+C.$

Since $s\left(0\right)=0,$ the constant is $C=0.$ Therefore, the position function is
$s\left(t\right)=-\frac{15}{2}{t}^{2}+88t.$

After $t=\frac{88}{15}$ sec, the position is $s\left(\frac{88}{15}\right)\approx 258.133$ ft.

Suppose the car is traveling at the rate of $44$ ft/sec. How long does it take for the car to stop? How far will the car travel?

$2.93\phantom{\rule{0.2em}{0ex}}\text{sec},\phantom{\rule{0.2em}{0ex}}64.5\phantom{\rule{0.2em}{0ex}}\text{ft}$

## Key concepts

• If $F$ is an antiderivative of $f,$ then every antiderivative of $f$ is of the form $F\left(x\right)+C$ for some constant $C.$
• Solving the initial-value problem
$\frac{dy}{dx}=f\left(x\right),y\left({x}_{0}\right)={y}_{0}$

requires us first to find the set of antiderivatives of $f$ and then to look for the particular antiderivative that also satisfies the initial condition.

For the following exercises, show that $F\left(x\right)$ are antiderivatives of $f\left(x\right).$

$F\left(x\right)=5{x}^{3}+2{x}^{2}+3x+1,f\left(x\right)=15{x}^{2}+4x+3$

${F}^{\prime }\left(x\right)=15{x}^{2}+4x+3$

$F\left(x\right)={x}^{2}+4x+1,f\left(x\right)=2x+4$

$F\left(x\right)={x}^{2}{e}^{x},f\left(x\right)={e}^{x}\left({x}^{2}+2x\right)$

${F}^{\prime }\left(x\right)=2x{e}^{x}+{x}^{2}{e}^{x}$

$F\left(x\right)=\text{cos}\phantom{\rule{0.1em}{0ex}}x,f\left(x\right)=\text{−}\text{sin}\phantom{\rule{0.1em}{0ex}}x$

$F\left(x\right)={e}^{x},f\left(x\right)={e}^{x}$

${F}^{\prime }\left(x\right)={e}^{x}$

For the following exercises, find the antiderivative of the function.

find the domain and range of f(x)= 4x-7/x²-6x+8
find the range of f(x)=(x+1)(x+4)
-1, -4
Marcia
That's domain. The range is [-9/4,+infinity)
Jacob
If you're using calculus to find the range, you have to find the extrema through the first derivative test and then substitute the x-value for the extrema back into the original equation.
Jacob
Good morning,,, how are you
d/dx{1/y - lny + X^3.Y^5}
How to identify domain and range
hello
Akpevwe
He,,
Harrieta
hi
Dr
hello
velocity
I only talk to girls
Dr
women are smart then guys
Dr
Smarter
sorry
Dr
Dr
:(
Shun
was up
Dr
hello
is it chatting app?.. I do not see any calculus here. lol
Find the arc length of the graph of f(x) = In (sinx) on the interval [Π/4, Π/2].
Sand falling freely from a lorry form a conical shape whose height is always equal to one-third the radius of the base. a. How fast is the volume increasing when the radius of the base is (1m) and increasing at the rate of 1/4cm/sec Pls help me solve
show that lim f(x) + lim g(x)=m+l
list the basic elementary differentials
Differentiation and integration
yes
Damien
proper definition of derivative
the maximum rate of change of one variable with respect to another variable
terms of an AP is 1/v and the vth term is 1/u show that the sum of uv terms is 1/2(uv+1)
what is calculus?
calculus is math that studies the change in math, such as the rate and distance,
Tamarcus
what are the topics in calculus
Augustine
what is limit of a function?
what is x and how x=9.1 take?
what is f(x)
the function at x
Marc
also known as the y value so I could say y=2x or f(x)= 2x same thing just using functional notation your next question is what is dependent and independent variables. I am Dyslexic but know math and which is which confuses me. but one can vary the x value while y depends on which x you use. also
Marc
up domain and range
Marc
enjoy your work and good luck
Marc
I actually wanted to ask another questions on sets if u dont mind please?
Inembo
I have so many questions on set and I really love dis app I never believed u would reply
Inembo
Hmm go ahead and ask you got me curious too much conversation here
am sorry for disturbing I really want to know math that's why *I want to know the meaning of those symbols in sets* e.g n,U,A', etc pls I want to know it and how to solve its problems
Inembo
and how can i solve a question like dis *in a group of 40 students, 32 offer maths and 24 offer physics and 4 offer neither maths nor physics , how many offer both maths and physics*
Inembo
next questions what do dy mean by (A' n B^c)^c'
Inembo
The sets help you to define the function. The function is like a magic box where you put inside stuff(numbers or sets) and you get out the stuff but in different shapes (forms).
I dont understand what you wanna say by (A' n B^c)^c'
(A' n B (rise to the power of c)) all rise to the power of c
Inembo
Aaaahh
Ok so the set is formed by vectors and not numbers
A vector of length n
But you can make a set out of matrixes as well
I I don't even understand sets I wat to know d meaning of all d symbolsnon sets
Inembo
High-school?
yes
Inembo
am having big problem understanding sets more than other math topics
Inembo
So f:R->R means that the function takes real numbers and provides real numer. For ex. If f(x) =2x this means if you give to your function a real number like 2,it gives you also a real number 2times2=4
pls answer this question *in a group of 40 students, 32 offer maths and 24 offer physics and 4 offer neither maths nor physics , how many offer both maths and physics*
Inembo
If you have f:R^n->R^n you give to your function a vector of length n like (a1,a2,...an) where all a1,.. an are reals and gives you also a vector of length n... I don't know if i answering your question. Otherwise on YouTube you havr many videos where they explain it in a simple way
I would say 24
Offer both
Sorry 20
Actually you have 40 - 4 =36 who offer maths or physics or both.
I know its 20 but how to prove it
Inembo
You have 32+24=56who offer courses
56-36=20 who give both courses... I would say that
solution: In a question involving sets and Venn diagram, the sum of the members of set A + set B - the joint members of both set A and B + the members that are not in sets A or B = the total members of the set. In symbolic form n(A U B) = n(A) + n (B) - n (A and B) + n (A U B)'.
Mckenzie
In the case of sets A and B use the letters m and p to represent the sets and we have: n (M U P) = 40; n (M) = 24; n (P) = 32; n (M and P) = unknown; n (M U P)' = 4
Mckenzie
Now substitute the numerical values for the symbolic representation 40 = 24 + 32 - n(M and P) + 4 Now solve for the unknown using algebra: 40 = 24 + 32+ 4 - n(M and P) 40 = 60 - n(M and P) Add n(M and P), as well, subtract 40 from both sides of the equation to find the answer.
Mckenzie
40 - 40 + n(M and P) = 60 - 40 - n(M and P) + n(M and P) Solution: n(M and P) = 20
Mckenzie
thanks
Inembo
Simpler form: Add the sums of set M, set P and the complement of the union of sets M and P then subtract the number of students from the total.
Mckenzie
n(M and P) = (32 + 24 + 4) - 40 = 60 - 40 = 20
Mckenzie