Kenneth Kuttler

8.2. EXERCISES 193

Definition 8.1.10 Let r (t) give a point on R and regard t as time. This is calledthe position of the point. Then the velocity of the point is defined as v(t) = r′ (t). Theacceleration is defined as the derivative of the velocity. Thus the acceleration is a(t) ≡r′′ (t).

Example 8.1.11 Let the velocity v(t) of a point be given by t2 +1 and suppose the point isat 1 when t = 0. Find the position of the point.

Let the position of the point be r (t) . Then by definition of velocity, r′ (t) = t2 + 1 sor (t) = t3

3 +t+C. Now C must be determined. It is assumed that r (0) = 1. Therefore, C = 1

and so r (t) = t3

3 + t +1.

Example 8.1.12 The acceleration of an object is given by a(t) = t + 1. When t = 0, thevelocity is 1 and the position is 2. Determine the position.

It is given that r′′ (t) = t + 1,r′ (0) = 1. Therefore, r′ (t) = t2

2 + t + 1. Then r (t) =t3

6 + t2

2 + t +2.

8.2 Exercises1. Find the indicated antiderivatives.

(a)∫ x√

2x−3dx

(b)∫

x(3x2 +6

)5 dx

(c)∫

xsin(x2)

(d)∫

sin3 (2x)cos(2x)

(e)∫ 1√

1+4x2dx Hint: Remember the

sinh−1 function and its derivative.

2. Solve the initial value problems. There is an unknown function y and you are givenits derivative and its value at some point.

(a) dydx = x√

2x−3,y(2) = 1

(b) dydx = 5x

(3x2 +6

)5,y(0) = 3

(2x3),y(1) = 1

(d) y′ (x) = 1√1+3x2

.y(1) = 1

(e) y′ (x) = sec(x) ,y(0) = 3

(f) y′ (x) = xcsc(x2),y(1) = 1

3. An object moves on the x axis having velocity equal to 3t3

7+t4 . Find the position of the

object given that at t = 1, it is at the point 2. The position is 2+∫ t

13s2

7+s4 ds

4. An object moves on the x axis having velocity equal to t sin(2t2). Find the position

of the object given that at t = 1, it is at the point 1.

5. An object moves on the x axis having velocity equal to sec(t) . Find the position ofthe object given that at t = 1, it is at the point −2.

6. Find the indicated antiderivatives.

8.2. EXERCISES 193Definition 8.1.10 Le r(t) give a point on R and regard t as time. This is calledthe position of the point. Then the velocity of the point is defined as v(t) = r'(t). Theacceleration is defined as the derivative of the velocity. Thus the acceleration is a(t) =r’(t).Example 8.1.11 Let the velocity v(t) of a point be given by t? +1 and suppose the point isat 1 when t = 0. Find the position of the point.Let the position of the point be r(t). Then by definition of velocity, r’ (t) = 17 +1 sor(t)= o +t-+C.Now C must be determined. It is assumed that r (0) = 1. Therefore, C = 13and so r(t)= 5 +t+1.Example 8.1.12 The acceleration of an object is given by a(t) =t+1. When t =0, thevelocity is 1 and the position is 2. Determine the position.It is given that rv’ (t) =t+1,r’ (0) = 1. Therefore, 1’ (t) = ++ 1. Then r(t) =8.2. Exercises1. Find the indicated antiderivatives.(a) f Jas dx (d) f sin? (2x) cos (2x)5 I on(b) fx (3x +6) dx (e) f Jinan Hint: Remember the(c) fxsin (x?) dx sinh~! function and its derivative.2. Solve the initial value problems. There is an unknown function y and you are givenits derivative and its value at some point.@) F=BS.v2)=1 @ ¥@)= Fes y()=1(b) # =5x (3x7 +6)”, 9(0) =3 (e) y’ (x) =sec(x),y(0) =3(c) 4 = 3x? sin (2x3) y(1)=1 (f) y/ (x) =xesc (x*) ,y(1) =13. An object moves on the x axis having velocity equal to me Find the position of theobject given that at t = 1, it is at the point 2. The position is 2+ fi 7a ds4. An object moves on the x axis having velocity equal to ¢ sin (277) . Find the positionof the object given that at t = 1, it is at the point 1.5. An object moves on the x axis having velocity equal to sec (t). Find the position ofthe object given that at t = 1, it is at the point —2.6. Find the indicated antiderivatives.