Question Statement

Find the instantaneous velocity at time $t = 0$ for the position function:

$f (t) = t^{2} + \frac{1}{5 t + 1}$

Background and Explanation

Instantaneous velocity represents the rate of change of position at a specific moment in time, which mathematically corresponds to the derivative of the position function. When the derivative formula is not yet available, we use the limit definition of the derivative (the difference quotient) to calculate this rate of change from first principles.

Solution

Finding the General Velocity Function $V (t)$

The instantaneous velocity at time $t$ is defined as:

$V (t) = lim_{Δ t \to 0} \frac{f ( t + Δ t ) - f ( t )}{Δ t}$

Substituting $f (t) = t^{2} + \frac{1}{5 t + 1}$ into the difference quotient:

$V (t) = lim_{Δ t \to 0} \frac{[ ( t + Δ t ) ^{2} + \frac{1}{5 ( t + Δ t ) + 1} ] - [ t ^{2} + \frac{1}{5 t + 1} ]}{Δ t}$

Expanding $(t + Δ t)^{2} = t^{2} + 2 t Δ t + (Δ t)^{2}$ and simplifying the numerator:

$= lim_{Δ t \to 0} \frac{t ^{2} + 2 t Δ t + ( Δ t ) ^{2} + \frac{1}{5 t + 5Δ t + 1} - t ^{2} - \frac{1}{5 t + 1}}{Δ t}$

The $t^{2}$ terms cancel, leaving:

$= lim_{Δ t \to 0} \frac{( Δ t ) ^{2} + 2 t Δ t + [ \frac{1}{5 t + 5Δ t + 1} - \frac{1}{5 t + 1} ]}{Δ t}$

To combine the fractions, find a common denominator $(5 t + 5Δ t + 1) (5 t + 1)$ :

$\frac{1}{5 t + 5Δ t + 1} - \frac{1}{5 t + 1} = \frac{( 5 t + 1 ) - ( 5 t + 5Δ t + 1 )}{( 5 t + 5Δ t + 1 ) ( 5 t + 1 )} = \frac{- 5Δ t}{( 5 t + 5Δ t + 1 ) ( 5 t + 1 )}$

Substituting this back:

$= lim_{Δ t \to 0} \frac{( Δ t ) ^{2} + 2 t Δ t - \frac{5Δ t}{( 5 t + 5Δ t + 1 ) ( 5 t + 1 )}}{Δ t}$

Factor out $Δ t$ from the numerator:

$= lim_{Δ t \to 0} \frac{Δ t [ Δ t + 2 t - \frac{5}{( 5 t + 5Δ t + 1 ) ( 5 t + 1 )} ]}{Δ t}$

Cancel $Δ t$ from numerator and denominator:

$= lim_{Δ t \to 0} [Δ t + 2 t - \frac{5}{( 5 t + 5Δ t + 1 ) ( 5 t + 1 )}]$

Now evaluate the limit as $Δ t \to 0$ . The term $Δ t$ goes to $0$ , and in the denominator $(5 t + 5Δ t + 1) \to (5 t + 1)$ :

$= 0 + 2 t - \frac{5}{( 5 t + 1 ) ( 5 t + 1 )}$

Therefore, the velocity function is:

$V (t) = 2 t - \frac{5}{( 5 t + 1 ) ^{2}}$

Evaluating at $t = 0$

Substitute $t = 0$ into the velocity function:

$V (0) = 2 (0) - \frac{5}{( 5 ( 0 ) + 1 ) ^{2}} = 0 - \frac{5}{( 0 + 1 ) ^{2}} = - \frac{5}{1} = - 5$

Thus, the instantaneous velocity at $t = 0$ is $- 5$ units per time.

Key Formulas or Methods Used

Limit definition of instantaneous velocity: $V (t) = lim_{Δ t \to 0} \frac{f ( t + Δ t ) - f ( t )}{Δ t}$
Difference quotient: The fraction $\frac{f ( t + Δ t ) - f ( t )}{Δ t}$ representing average velocity over a small interval
Algebraic simplification: Expanding binomials $(t + Δ t)^{2}$ and combining rational expressions via common denominators
Limit evaluation: Taking $Δ t \to 0$ by direct substitution after algebraic cancellation

Summary of Steps

Set up the limit definition of the derivative for instantaneous velocity
Substitute the position function $f (t) = t^{2} + \frac{1}{5 t + 1}$ into the difference quotient
Expand $(t + Δ t)^{2}$ and group the rational terms separately
Combine the fractions by finding the common denominator $(5 t + 5Δ t + 1) (5 t + 1)$
Factor out $Δ t$ from the entire numerator to enable cancellation
Cancel $Δ t$ from numerator and denominator, then evaluate the limit as $Δ t \to 0$
Obtain the general velocity function: $V (t) = 2 t - \frac{5}{( 5 t + 1 ) ^{2}}$
Substitute $t = 0$ into the velocity function to get $V (0) = - 5$

Question Statement

Find the instantaneous velocity at time $t = 0$ for the position function:

$f (t) = t^{2} + \frac{1}{5 t + 1}$

Background and Explanation

Solution

Finding the General Velocity Function $V (t)$

The instantaneous velocity at time $t$ is defined as:

$V (t) = lim_{Δ t \to 0} \frac{f ( t + Δ t ) - f ( t )}{Δ t}$

Substituting $f (t) = t^{2} + \frac{1}{5 t + 1}$ into the difference quotient:

$V (t) = lim_{Δ t \to 0} \frac{[ ( t + Δ t ) ^{2} + \frac{1}{5 ( t + Δ t ) + 1} ] - [ t ^{2} + \frac{1}{5 t + 1} ]}{Δ t}$

Expanding $(t + Δ t)^{2} = t^{2} + 2 t Δ t + (Δ t)^{2}$ and simplifying the numerator:

$= lim_{Δ t \to 0} \frac{t ^{2} + 2 t Δ t + ( Δ t ) ^{2} + \frac{1}{5 t + 5Δ t + 1} - t ^{2} - \frac{1}{5 t + 1}}{Δ t}$

The $t^{2}$ terms cancel, leaving:

$= lim_{Δ t \to 0} \frac{( Δ t ) ^{2} + 2 t Δ t + [ \frac{1}{5 t + 5Δ t + 1} - \frac{1}{5 t + 1} ]}{Δ t}$

To combine the fractions, find a common denominator $(5 t + 5Δ t + 1) (5 t + 1)$ :

$\frac{1}{5 t + 5Δ t + 1} - \frac{1}{5 t + 1} = \frac{( 5 t + 1 ) - ( 5 t + 5Δ t + 1 )}{( 5 t + 5Δ t + 1 ) ( 5 t + 1 )} = \frac{- 5Δ t}{( 5 t + 5Δ t + 1 ) ( 5 t + 1 )}$

Substituting this back:

$= lim_{Δ t \to 0} \frac{( Δ t ) ^{2} + 2 t Δ t - \frac{5Δ t}{( 5 t + 5Δ t + 1 ) ( 5 t + 1 )}}{Δ t}$

Factor out $Δ t$ from the numerator:

$= lim_{Δ t \to 0} \frac{Δ t [ Δ t + 2 t - \frac{5}{( 5 t + 5Δ t + 1 ) ( 5 t + 1 )} ]}{Δ t}$

Cancel $Δ t$ from numerator and denominator:

$= lim_{Δ t \to 0} [Δ t + 2 t - \frac{5}{( 5 t + 5Δ t + 1 ) ( 5 t + 1 )}]$

Now evaluate the limit as $Δ t \to 0$ . The term $Δ t$ goes to $0$ , and in the denominator $(5 t + 5Δ t + 1) \to (5 t + 1)$ :

$= 0 + 2 t - \frac{5}{( 5 t + 1 ) ( 5 t + 1 )}$

Therefore, the velocity function is:

$V (t) = 2 t - \frac{5}{( 5 t + 1 ) ^{2}}$

Evaluating at $t = 0$

Substitute $t = 0$ into the velocity function:

$V (0) = 2 (0) - \frac{5}{( 5 ( 0 ) + 1 ) ^{2}} = 0 - \frac{5}{( 0 + 1 ) ^{2}} = - \frac{5}{1} = - 5$

Thus, the instantaneous velocity at $t = 0$ is $- 5$ units per time.

Key Formulas or Methods Used

Limit definition of instantaneous velocity: $V (t) = lim_{Δ t \to 0} \frac{f ( t + Δ t ) - f ( t )}{Δ t}$
Difference quotient: The fraction $\frac{f ( t + Δ t ) - f ( t )}{Δ t}$ representing average velocity over a small interval
Algebraic simplification: Expanding binomials $(t + Δ t)^{2}$ and combining rational expressions via common denominators
Limit evaluation: Taking $Δ t \to 0$ by direct substitution after algebraic cancellation

Summary of Steps

Set up the limit definition of the derivative for instantaneous velocity
Substitute the position function $f (t) = t^{2} + \frac{1}{5 t + 1}$ into the difference quotient
Expand $(t + Δ t)^{2}$ and group the rational terms separately
Combine the fractions by finding the common denominator $(5 t + 5Δ t + 1) (5 t + 1)$
Factor out $Δ t$ from the entire numerator to enable cancellation
Cancel $Δ t$ from numerator and denominator, then evaluate the limit as $Δ t \to 0$
Obtain the general velocity function: $V (t) = 2 t - \frac{5}{( 5 t + 1 ) ^{2}}$
Substitute $t = 0$ into the velocity function to get $V (0) = - 5$

2.3 Q-10

Question Statement

Background and Explanation

Solution

Finding the General Velocity Function $V (t)$

Evaluating at $t = 0$

Key Formulas or Methods Used

Summary of Steps

2.3 Q-10

Question Statement

Background and Explanation

Solution

Finding the General Velocity Function $V (t)$

Evaluating at $t = 0$

Key Formulas or Methods Used

Summary of Steps

Question Statement

Background and Explanation

Solution

Finding the General Velocity Function V(t)

Evaluating at t=0

Key Formulas or Methods Used

Summary of Steps

Question Statement

Background and Explanation

Solution

Finding the General Velocity Function V(t)

Evaluating at t=0

Key Formulas or Methods Used

Summary of Steps

Finding the General Velocity Function $V (t)$

Evaluating at $t = 0$

Finding the General Velocity Function $V (t)$

Evaluating at $t = 0$