Solution: We are given $ D(t) = kt^2 + 5t $ and that $ D(3) = 48 $. Substituting $ t = 3 $: - Deep Underground Poetry
Solving for the Constant in the Quadratic Function: A Step-by-Step Guide Using $ D(t) = kt^2 + 5t $
Solving for the Constant in the Quadratic Function: A Step-by-Step Guide Using $ D(t) = kt^2 + 5t $
When working with mathematical modeling in science, engineering, or economics, analyzing functions of the form $ D(t) = kt^2 + 5t $ is common. These quadratic equations often represent real-world phenomena such as displacement, revenue, or cost over time. In this article, we’ll walk through how to determine the unknown constant $ k $ using the known value $ D(3) = 48 $, offering a clear, step-by-step solution that highlights key algebraic techniques useful for educators, students, and professionals.
Understanding the Context
Step 1: Understand the Function and Given Information
We are given a quadratic function:
$$
D(t) = kt^2 + 5t
$$
This function models a quantity $ D $ that depends on time $ t $, with the unknown constant $ k $ needing to be determined. We’re also told that at $ t = 3 $, $ D(3) = 48 $.
Substituting $ t = 3 $ into the function gives:
$$
D(3) = k(3)^2 + 5(3)
$$
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Key Insights
Step 2: Substitute and Simplify
Now compute each term:
$$
D(3) = k(9) + 15 = 9k + 15
$$
Since $ D(3) = 48 $, equate the expressions:
$$
9k + 15 = 48
$$
Step 3: Solve for $ k $
Subtract 15 from both sides:
$$
9k = 48 - 15 = 33
$$
Now divide both sides by 9:
$$
k = rac{33}{9} = rac{11}{3}
$$
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Step 4: Final Verification
To confirm, plug $ k = rac{11}{3} $ back into the original equation:
$$
D(t) = rac{11}{3}t^2 + 5t
$$
Now compute $ D(3) $:
$$
D(3) = rac{11}{3}(9) + 5(3) = 33 + 15 = 48
$$
The result matches the given value, verifying our solution.
Why This Technique Matters
This example illustrates a standard algebraic method for solving for unknown coefficients in quadratic functions—substitution followed by isolation of the unknown variable. Such skills are essential in fields ranging from physics (modeling motion) to finance (forecasting growth), where precise parameter estimation ensures accurate predictions.
Key Takeaways:
- Always substitute known values into the functional equation.
- Simplify expressions algebraically before isolating the unknown.
- Verify your solution by plugging it back into the original equation.
By mastering this step-by-step process, anyone can confidently solve similar problems involving quadratic models in real-world contexts.
