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Chapter 3: Problem 14

In Exercises \(13-16,\) find \(d s / d t\) $$ s=t^{2}-\sec t+1 $$

Short Answer

Expert verified
\( \frac{ds}{dt} = 2t - \sec t \tan t \)

Step by step solution

01

Understand the Derivative

We need to find the derivative of the function \( s(t) = t^2 - \sec t + 1 \) with respect to \( t \). This means we are looking for \( \frac{ds}{dt} \).
02

Differentiate Each Term

Differentiate each term of the function separately. The first term is \( t^2 \), which differentiates to \( 2t \). The second term is \( -\sec t \), which differentiates to \( -\sec t \tan t \). The third term is a constant, \( 1 \), and its derivative is \( 0 \).
03

Combine the Derivatives

Combine the derivatives from Step 2 to get \( \frac{ds}{dt} = 2t - \sec t \tan t + 0 \).
04

Simplify the Expression

The simplified form of the expression for \( \frac{ds}{dt} \) is \( 2t - \sec t \tan t \).

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Key Concepts

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

Derivative Calculation
Calculating derivatives is a fundamental skill in calculus, involving finding the rate at which a function changes with respect to a variable. In our exercise, we focus on differentiating the function \( s(t) = t^2 - \sec t + 1 \). Here are the steps involved:
  • Identify the terms that need differentiation: \( t^2 \), \(-\sec t\), and the constant \( 1 \).

  • Apply the power rule for \( t^2 \): the derivative is \( 2t \), as the power rule states that the derivative of \( t^n \) is \( nt^{n-1} \).

  • Use the derivative for trigonometric functions: the derivative of \(-\sec t\) is \(-\sec t \tan t\), which is derived from known formulas in calculus.

  • Remember, the derivative of a constant like \( 1 \) is always \( 0 \).
Combining these derivatives, the overall derivative \( \frac{ds}{dt} \) is \( 2t - \sec t \tan t \). This calculation gives insight into how the original function \( s(t) \) changes as \( t \) changes.
Trigonometric Functions
Trigonometric functions such as \( \sec t \) and \( \tan t \) play essential roles in calculus, especially in differentiation. In this exercise, we specifically dealt with the trigonometric function \(-\sec t\):
  • The secant function, \( \sec t \), is the reciprocal of the cosine function, \( \sec t = \frac{1}{\cos t} \).
  • The tangent function, \( \tan t \), is defined as \( \tan t = \frac{\sin t}{\cos t} \).

  • For differentiation, the derivative of \( \sec t \) with respect to \( t \) is \( \sec t \tan t \). Therefore, \(-\sec t\) differentiates to \(-\sec t \tan t\).
These functions are vital in understanding more complex motion dynamics in calculus. They frequently appear in various applications including physics and engineering, where wave-like motion is modeled.
Mathematical Expressions
Understanding mathematical expressions and simplifying them is crucial in calculus. The expression we dealt with, \( s(t) = t^2 - \sec t + 1 \), is a combination of different types of terms:
  • Polynomial terms like \( t^2 \) which are straightforward to differentiate using simple rules like the power rule.

  • Trigonometric terms such as \(-\sec t\) require specific knowledge of trigonometric derivatives.

  • Constant numbers like \( 1 \) simplify the equation by having a derivative of zero.
Simplifying the expression after differentiation is key to achieving a clear result, \( 2t - \sec t \tan t \) in this case. As you work through calculus problems, ensure each step is clear and each term is addressed properly for accurate results.

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Most popular questions from this chapter

Quadratic approximations Let \(Q(x)=b_{0}+b_{1}(x-a)+b_{2}(x-a)^{2}\) be a quadratic approximation to \(f(x)\) at \(x=a\) with the properties: i. \(Q(a)=f(a)\) ii. \(Q^{\prime}(a)=f^{\prime}(a)\) iii. \(Q^{\prime \prime}(a)=f^{\prime \prime}(a)\) Determine the coefficients \(b_{0}, b_{1},\) and \(b_{2}\) b. Find the quadratic approximation to \(f(x)=1 /(1-x)\) at \(x=0\) . c. Graph \(f(x)=1 /(1-x)\) and its quadratic approximation at \(x=0 .\) Then zoom in on the two graphs at the point \((0,1)\) . Comment on what you see. d. Find the quadratic approximation to \(g(x)=1 / x\) at \(x=1\) . Graph \(g\) and its quadratic approximation together. Comment on what you see. e. Find the quadratic approximation to \(h(x)=\sqrt{1+x}\) at \(x=0 .\) Graph \(h\) and its quadratic approximation together. Comment on what you see. f. What are the linearizations of \(f, g,\) and \(h\) at the respective points in parts (b), (d), and (e)?

Falling meteorite The velocity of a heavy meteorite entering Earth's atmosphere is inversely proportional to \(\sqrt{s}\) when it is \(s\) \(\mathrm{km}\) from Earth's center. Show that the meteorite's acceleration is inversely proportional to \(s^{2} .\)

Use a CAS to perform the following steps in Exercises \(77-84\) . a. Plot the equation with the implicit plotter of a CAS. Check to see that the given point \(P\) satisfies the equation. b. Using implicit differentiation, find a formula for the derivative \(d y / d x\) and evaluate it at the given point \(P .\) c. Use the slope found in part (b) to find an equation for the tangent line to the curve at \(P\) . Then plot the implicit curve and tangent line together on a single graph. $$ x+\tan \left(\frac{y}{x}\right)=2, \quad P\left(1, \frac{\pi}{4}\right) $$

The diameter of a sphere is measured as \(100 \pm 1 \mathrm{cm}\) and the volume is calculated from this measurement. Estimate the percent- age error in the volume calculation.

Use a CAS to perform the following steps in Exercises \(77-84\) . a. Plot the equation with the implicit plotter of a CAS. Check to see that the given point \(P\) satisfies the equation. b. Using implicit differentiation, find a formula for the derivative \(d y / d x\) and evaluate it at the given point \(P .\) c. Use the slope found in part (b) to find an equation for the tangent line to the curve at \(P\) . Then plot the implicit curve and tangent line together on a single graph. $$ x \sqrt{1+2 y}+y=x^{2}, \quad P(1,0) $$
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