Question Statement

Find the instantaneous velocity at $t=3$ for the position function:

$f(t)=-4t^2+10t+6$

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Background and Explanation

This problem involves calculating instantaneous velocity using the limit definition of the derivative. Instantaneous velocity represents the rate of change of position at a specific moment in time, obtained by evaluating the derivative of the position function.

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Solution

Finding the General Velocity Function

The instantaneous velocity at time $t_1$ is given by the limit definition:

$V(t_1)={\lim }_{\Delta t\to 0}\frac{f(t_1+\Delta t)-f(t_1)}{\Delta t}$

Substitute the position function $f(t)=-4t^2+10t+6$:

$={\lim }_{\Delta t\to 0}\frac{\left[-4(t_1+\Delta t{)}^2+10(t_1+\Delta t)+6\right]-\left[-4t_1^2+10t_1+6\right]}{\Delta t}$

Expand $(t_1+\Delta t{)}^2=t_1^2+2t_1\Delta t+(\Delta t{)}^2$ and distribute:

$={\lim }_{\Delta t\to 0}\frac{-4(t_1^2+2t_1\Delta t+(\Delta t{)}^2)+10t_1+10\Delta t+6+4t_1^2-10t_1-6}{\Delta t}$

Simplify the numerator by distributing $-4$ and combining like terms:

$={\lim }_{\Delta t\to 0}\frac{-4t_1^2-8t_1\Delta t-4(\Delta t{)}^2+10t_1+10\Delta t+6+4t_1^2-10t_1-6}{\Delta t}$

Cancel terms ($-4t_1^2+4t_1^2=0$, $10t_1-10t_1=0$, $6-6=0$):

$={\lim }_{\Delta t\to 0}\frac{-8t_1\Delta t-4(\Delta t{)}^2+10\Delta t}{\Delta t}$

Factor out $\Delta t$ from the numerator:

$={\lim }_{\Delta t\to 0}\frac{\Delta t(-8t_1-4\Delta t+10)}{\Delta t}$

Cancel $\Delta t$:

$={\lim }_{\Delta t\to 0}(-8t_1-4\Delta t+10)$

Evaluate the limit as $\Delta t\to 0$:

$V(t_1)=-8t_1-4(0)+10=-8t_1+10$

Evaluating at $t=3$

Given $t_1=3$, substitute into the velocity function:

$\begin{array}{rl}V(3)& =-8(3)+10\\ & =-24+10\\ & =-14\end{array}$

Therefore, the instantaneous velocity at $t=3$ is $-14$ units per time.

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Key Formulas or Methods Used

• Limit definition of instantaneous velocity: $V(t_1)={\lim }_{\Delta t\to 0}\frac{f(t_1+\Delta t)-f(t_1)}{\Delta t}$
• Algebraic expansion: $(a+b{)}^2=a^2+2ab+b^2$
• Limit evaluation: Substituting $\Delta t=0$ after algebraic simplification

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Summary of Steps

1. Set up the limit definition of the derivative for instantaneous velocity
2. Substitute the position function $f(t)=-4t^2+10t+6$ into the limit expression
3. Expand $(t_1+\Delta t{)}^2$ and distribute coefficients
4. Simplify the numerator by combining like terms (cancel $t_1^2$, $t_1$, and constant terms)
5. Factor out $\Delta t$ and cancel with the denominator
6. Evaluate the limit as $\Delta t\to 0$ to obtain $V(t_1)=-8t_1+10$
7. Substitute $t=3$ to find $V(3)=-14$