BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805
BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

Solutions

Chapter 2.6, Problem 14E

(a)

To determine

To find: The velocity of the rock after one second.

Expert Solution

Answer to Problem 14E

The velocity of the rock after one second is 6.28 m/s.

Explanation of Solution

Given:

The height (in meters) of the rock after t seconds is H=10t1.86t2.

The position function of the rock is, f(t)=10t1.86t2.

Formula used:

The derivative v(a) is the velocity of the particle at time t=a.

v(a)=limh0f(a+h)f(a)h (1)

Zero productivity rule: “If p×q=0 then p=0 or q=0 or both  p=0 and q=0.”

Calculation:

Obtain the velocity of the rock after one second.

Take the position function H=f(t) and substitute the time a=1 in equation (1),

v(1)=limh0f(1+h)f(1)h=limh0(10(1+h)1.86(1+h)2)(10(1)1.86(1)2)h=limh0(10+10h1.86(h2+2h+1))(101.86)h=limh0(10+10h1.86h23.72h1.86)(8.14)h

Simplify further and obtain the velocity as follows.

v(1)=limh01.86h2+(10h3.72h)+(101.868.14)h=limh01.86h2+6.28h+(1010)h=limh0h(1.86h+6.28)h

Cancel the common term h from both the numerator and the denominator,

v(1)=limh0(1.86h+6.28)=(1.86(0)+6.28)=6.28

Thus, the velocity of the rock at time t=1 s is 6.28 m/s.

(b)

To determine

To find: The velocity of the rock when t=a.

Expert Solution

Answer to Problem 14E

The velocity of the rock when t=a is 103.72a m/s.

Explanation of Solution

Calculation:

Obtain the velocity of the rock when t=a.

Take the position function H=f(t) and substitute the time t=a in (1),

v(a)=limh0f(a+h)f(a)h=limh0(10(a+h)1.86(a+h)2)(10(a)1.86(a)2)h=limh0(10a+10h1.86(h2+2ah+a2))(10a1.86a2)h=limh0(10a+10h1.86h23.72ah1.86a2)(10a1.86a2)h

Simplify further and obtain the velocity as follows.

v(a)=limh010a+10h1.86h23.72ah1.86a210a+1.86a2h=limh0(10a10a)+10h1.86h23.72ah+(1.86a2+1.86a2)h=limh010h1.86h23.72ahh=limh0h(101.86h3.72a)h

Cancel the common term h of both the numerator and denominator,

v(a)=limh0(101.86h3.72a)=(101.86(0)3.72a)=103.72a

Thus, the velocity of the rock at time t=a is v(a)=(103.72a) m/s_.

(c)

To determine

To find: Obtain the time t when the rock hit the surface.

Expert Solution

Answer to Problem 14E

The rock will hit the surface when time t=5.4 s.

Explanation of Solution

The rock will hit the surface when height of the rock is zero.

H(t)=010t1.86t2=0t(101.86t)=0

Use the zero productivity rule to take t=0 or 101.86t=0.

Let 101.86t=0 and simplify further and obtain the time as follows.

101.86t=01.86t=10

Isolate t as shown below.

t=101.865.4

Thus, the rock will hit the surface when time t=5.4 s_.

(d)

To determine

To find: The velocity of the rock to hit the surface.

Expert Solution

Answer to Problem 14E

The velocity of the rock to hit the surface is 10 m/s_.

Explanation of Solution

The rock will hit the surface when time t=5.4 s.

Substitute a=5.4 s in part (b),

v(5.4)=103.72(5.4)=1020.088=10.08810

Thus, the velocity of the rock to hit the surface is 10 m/s.

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