Question Statement

The height $S$ above ground of a projectile at time $t$ is given by:

$S(t)=\frac{1}{2}g{t}^2+v_{0}t+s_0$

where $g$, $v_0$, and $s_0$ are constants. Find the instantaneous rate of change of $S$ with respect to $t$ at $t=4$.

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Background and Explanation

This problem requires finding the derivative of a position function to determine the instantaneous velocity (rate of change of height with respect to time). The derivative $\frac{dS}{dt}$ represents the velocity of the projectile at any given time $t$.

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Solution

To find the instantaneous rate of change, we differentiate the position function $S(t)$ with respect to time $t$ and then evaluate at $t=4$.

Step 1: Differentiate $S(t)$ with respect to $t$.

Using the power rule for differentiation on each term:

$\begin{array}{rl}\frac{dS}{dt}& =\frac{d}{dt}\left(\frac{1}{2}g{t}^2\right)+\frac{d}{dt}(v_{0}t)+\frac{d}{dt}(s_0)\\ & =\frac{1}{2}g(2t)+v_0(1)+0\\ & =gt+v_0\end{array}$

Note: The constant $s_0$ (initial height) disappears since the derivative of a constant is zero. The term $\frac{1}{2}g(2t)$ simplifies to $gt$ as the 2s cancel.

Step 2: Evaluate the derivative at $t=4$.

Substitute $t=4$ into the derivative:

$\begin{array}{rl}{\frac{dS}{dt}|}_{t=4}& =g(4)+v_0\\ & =4g+v_0\end{array}$

Thus, the instantaneous rate of change of $S$ with respect to $t$ at $t=4$ is:

$\frac{dS}{dt}=4g+v_0$

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Key Formulas or Methods Used

• Power Rule: $\frac{d}{dt}(t^n)=n{t}^{n-1}$
• Constant Multiple Rule: $\frac{d}{dt}[c\cdot f(t)]=c\cdot f^{\prime }(t)$
• Constant Rule: $\frac{d}{dt}(\text{constant})=0$
• Instantaneous Rate of Change: Value of the derivative $\frac{dS}{dt}$ at a specific point

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Summary of Steps

1. Differentiate the height function $S(t)$ term by term using basic differentiation rules.
2. Simplify the resulting expression to obtain $\frac{dS}{dt}=gt+v_0$.
3. Substitute $t=4$ into the derivative expression.
4. State the final result as $4g+v_0$.

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Question 5: Error Approximation Using Differentials

Question Statement

The side of a square is measured to be $10$ cm with a possible error of $\pm 0.3$ cm. Use differentials to find:

• The approximate maximum error in the area
• The approximate relative error
• The approximate percentage error

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Background and Explanation

When measurements have small uncertainties, differentials provide a linear approximation for how these errors propagate through calculations. For a function $A(x)$, the differential $dA\approx \Delta A$ estimates the maximum error in the output given a small error $dx$ in the input measurement.

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Solution

Step 1: Define the variables and the area function.

Let $x$ represent the side length of the square:

• Measured side: $x=10$ cm
• Possible error: $dx=\pm 0.3$ cm

The area of a square is: $A=x^2$

Step 2: Find the differential $dA$.

Differentiate $A$ with respect to $x$ to find the rate of change, then express as a differential: $\frac{dA}{dx}=2x$

Therefore: $dA=2xdx$

Step 3: Calculate the maximum error in the area.

Substitute $x=10$ and $dx=\pm 0.3$:

$\begin{array}{rl}dA& =2(10)(\pm 0.3)\\ & =20(\pm 0.3)\\ & =\pm 6{\text{ cm}}^2\end{array}$

Maximum error in Area: $\pm 6$ cm²

Step 4: Calculate the relative error.

First, find the actual area at $x=10$: $A=(10{)}^2=100{\text{ cm}}^2$

Relative error is the ratio of the error in the area to the total area:

$\begin{array}{rl}\text{Relative Error}& =\frac{dA}{A}\\ & =\frac{\pm 6}{100}\\ & =\pm 0.06\end{array}$

Step 5: Calculate the percentage error.

Multiply the relative error by $100\%$:

$\begin{array}{rl}\text{Percentage Error}& =\frac{dA}{A}\times 100\%\\ & =\pm 0.06\times 100\%\\ & =\pm 6\%\end{array}$

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Key Formulas or Methods Used

• Area of Square: $A=x^2$
• Differential Formula: $dA=A^{\prime }(x)dx=2xdx$
• Relative Error: $\frac{dA}{A}$ (or $\frac{\Delta A}{A}$)
• Percentage Error: $\frac{dA}{A}\times 100\%$

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Summary of Steps

1. Identify the side length $x=10$ cm and its possible error $dx=\pm 0.3$ cm.
2. Write the area formula $A=x^2$ and find its differential $dA=2xdx$.
3. Substitute values to find maximum area error: $dA=\pm 6$ cm².
4. Calculate the area $A=100$ cm² to use as the denominator for relative error.
5. Compute relative error: $\frac{\pm 6}{100}=\pm 0.06$.
6. Convert to percentage error by multiplying by 100: $\pm 6\%$.