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Find the values of the derivatives. $$ \left.\frac{d s}{d t}\right|_{t=-1} \text { if } s=1-3 t^{2} $$

Short Answer

Expert verified
The value of the derivative at \( t = -1 \) is 6.

Step by step solution

01

Understand the Expression

The function given is \( s = 1 - 3t^2 \), and we are required to find the derivative of \( s \) with respect to \( t \) evaluated at \( t = -1 \). This means we need to differentiate \( s \) and then substitute \( t = -1 \) into that derivative.
02

Differentiate the Function

We need to calculate \( \frac{ds}{dt} \). Start by differentiating \( s = 1 - 3t^2 \) with respect to \( t \). The derivative of a constant is 0, and the derivative of \( -3t^2 \) is \(-6t\). Therefore, \( \frac{ds}{dt} = 0 - 6t = -6t \).
03

Evaluate the Derivative at \( t = -1 \)

Substitute \( t = -1 \) into the derivative \( \frac{ds}{dt} = -6t \). This gives us \( -6(-1) = 6 \). Therefore, \( \left. \frac{ds}{dt} \right|_{t=-1} = 6 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Differentiate Function
Differentiation is a fundamental concept in calculus that involves finding the rate at which a function changes at any given point. This process is essential for understanding how variables interact with each other in a dynamic way.
In this exercise, you need to find the derivative of the function \( s = 1 - 3t^2 \) with respect to \( t \). Differentiating a function means calculating its derivative, which essentially represents the slope of the tangent line at any point on the curve.
Here's a simple breakdown:
  • Identify the function: Here, the function is \( s = 1 - 3t^2 \).
  • Apply derivation rules: The function is composed of a constant \( 1 \) and a polynomial term \( -3t^2 \). The derivative of a constant is always 0, while the power rule can be used for terms like \( -3t^2 \).
Applying the power rule, \( \frac{d}{dt}(t^n) = nt^{n-1} \), the derivative of \( -3t^2 \) is \( -6t \). Therefore, the derivative of \( s \) is \( \frac{ds}{dt} = 0 - 6t = -6t \). This derivative tells us how \( s \) changes with small changes in \( t \).
Evaluate Derivative
Once we have the derivative of a function, evaluating it at a specific point gives us the instantaneous rate of change at that particular value of \( t \). This step helps in predicting the behavior of the function around that point.
After differentiating \( s = 1 - 3t^2 \) to obtain \( \frac{ds}{dt} = -6t \), the next task is to substitute \( t = -1 \) to find the slope of \( s \) at this point.
Here's how you do it:
  • Substitute the value: Replace \( t \) in the derivative \( -6t \) with \( -1 \).
  • Perform the arithmetic: Calculate \( -6(-1) \), which results in 6.
Thus, the evaluated derivative \( \left. \frac{ds}{dt} \right|_{t=-1} \) equals 6. This means, at \( t = -1 \), the rate of change of \( s \) with respect to \( t \) is 6, indicating that the function is increasing at this moment.
Calculus Problem-Solving
Solving calculus problems often involves a set of methodical steps. These steps allow you to break down complex problems into manageable segments, which is especially useful when dealing with derivatives.
To efficiently solve problems like this one, follow these guidelines:
  • Analyze the problem: Start by understanding what is being asked. In this case, it's finding a derivative and evaluating it at a particular value.
  • Apply differentiation rules: Differentiate using known rules like the power rule, product rule, or chain rule as required. Here, the power rule was used to find the derivative.
  • Substitute and solve: After finding the derivative, plug in the given values to solve for the specific instance required.
By sticking to these steps, one can navigate the intricacies of calculus with ease, ensuring each problem is approached logically and systematically. This methodical approach not only aids in solving the immediate problem but also strengthens overall understanding of calculus.

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