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Chapter 4: Problem 29

An object moves along a coordinate line, its position at each time \(t \geq 0\) being given by \(x(t)\). Find the times \(t\) at which the object changes direction. $$x(t)=(t+1)^{2}(t-9)^{3}$$.

Short Answer

Expert verified
The times \(t\) at which the object changes direction are given by the critical points and end points (if any) where \(x'(t)\) changes sign.

Step by step solution

01

Obtain the derivative of the function.

You first start by obtaining the derivative of \(x(t)= (t+1)^{2}(t-9)^{3}\), that is, \(x'(t)\). You would need to use the product rule and the chain rule, noting that the derivative of \(t^n\) is \(n*t^{n-1}\). So, \(x'(t) = 2(t+1)(t-9)^{3} + 3(t+1)^{2}(t-9)^{2}\).
02

Find when the derivative is zero or undefined

The object changes direction when the derivative is zero or undefined (if such points exist). Set \(x'(t) = 0\) and solve for \(t\). This gives you the critical points.
03

Analyze the critical points

Now that you have the critical points from solving \(x'(t) = 0\), you must analyze these and any end points (if any) in the interval to see if \(x'(t)\) changes sign at these points. You can use a number line with test points from each interval in between your critical points to do this. If \(x'(t)\) changes sign at a critical point or at an end point, then \(x(t)\) has a local maximum or minimum there. Hence, \(x(t)\) changes direction at these points.

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Key Concepts

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

Product Rule
When you have a function that is a product of two or more functions, like \(x(t) = (t+1)^{2}(t-9)^{3}\), you can't just differentiate each function individually; you have to use the product rule. The product rule states that if you have two functions, say \(u(t)\) and \(v(t)\), their derivative combined in a product is given by:
  • \((uv)' = u'v + uv'\)
In our exercise, \((t+1)^{2}\) is one function, and \((t-9)^{3}\) is the other. So, to find \(x'(t)\), we differentiate each component separately and then apply the product rule.The derivative becomes:
  • \(2(t+1)(t-9)^{3} + 3(t+1)^{2}(t-9)^{2}\), where each term results from applying the product rule to one part of the product.
This method ensures we correctly account for the interaction between the different parts of the product.
Chain Rule
The chain rule is essentially for dealing with composite functions, which are functions of functions. In our problem, components like \((t+1)^{2}\) or \((t-9)^{3}\) are examples.The chain rule states that if you have a composite function, say \(f(g(t))\), then its derivative is:
  • \(f'(g(t)) \, g'(t)\)
So when we differentiate \((t+1)^{2}\), we treat \(u(t) = t+1\) and \(f(u) = u^{2}\), which becomes:
  • \(2(t+1)^{1} \cdot 1\)
Likewise, for \((t-9)^{3}\), the outer function \(u^{3}\) is differentiated to give:
  • \(3(t-9)^{2} \cdot 1\)
Using the chain rule, we can deal with the layers of functions efficiently.
Critical Points
Critical points are essential when you want to find where a function has peaks, troughs, or changes direction. These are points where the derivative \(x'(t)\) is zero or undefined.To find critical points:
  • Solve \(x'(t) = 0\). This involves finding roots of the equation.
  • Check if there are any values where \(x'(t)\) is undefined, which often occurs if the function includes fractions where a zero in the denominator could create an undefined point.
After solving \(x'(t) = 0\), you obtain critical points that need further analysis to determine their role in the function's behavior.These points are vital to understanding how and where \(x(t)\) changes direction.
Sign Change
Analyzing sign changes of the derivative \(x'(t)\) helps determine how the function \(x(t)\) behaves around its critical points. If \(x'(t)\) transitions from positive to negative at a point, that indicates a local maximum; if it goes from negative to positive, that's a local minimum.To determine sign changes:
  • Use a number line with intervals around critical points.
  • Choose test points within each interval to evaluate \(x'(t)\).
  • Observe whether \(x'(t)\) changes from positive to negative or negative to positive across these tests.
This sign analysis can signify where and when the original function \(x(t)\) will hit a turning point, thus changing direction. Such turning points provide key insights into the motion of an object described by the function.

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Most popular questions from this chapter

Set \(f(x)=x^{k}-a, k\) a positive integer, \(a>0 .\) The number \(a^{1 / k}\) is a root of the equation \(f(x)=0\). (a) Show that if \(x_{1}\) is any initial estimate for \(a^{1 / k}\), then the Newton-Raphson method gives the iteration formula $$x_{n+1}=\frac{1}{k}\left[(k-1) x_{n}+\frac{a}{x_{n}^{k-1}}\right]$$. Note that for \(k=2\) this formula reduces to the formula given in Exercise 13. (b) Use the formula in part (a) to approximate \(\sqrt[3]{23} .\) Begin at \(x_{1}=3\) and calculate \(x_{4}\) rounded off to five decimal places. Evaluate \(f\left(x_{4}\right)\).

A man standing 3 feet from the base of a lamppost casts a shadow 4 feet long. If the man is 6 feet tall and walks away from the lamppost at a speed of 400 feet per minute, at what rate will his shadow lengthen? How fast is the tip of his shadow moving?

Points \(A\) and \(B\) are opposite points on the shore of a circular lake of radius 1 mile. Magg铆e, now at point \(A\), wants to reach point \(B .\) She can swim directly across the lake, she can walk along the shore, or she can swim part way and walk part way. Given that Maggie can swim at the rate of 2 miles per hour and walks at the rate of 5 miles per hour, what route should she take to reach point \(B\) as quickly as possible? (No running allowed.)

Use a graphing utility to determine whether or not the graph of \(f\) has a horizontal asymptote. Confirm your findings analytically. $$f(x)=\sqrt{x^{2}+2 x}-x$$

(Oblique asymplotes) Let \(r(x)=p(x) / q(x)\) be a rational function. If (degree of \(p)=(\text { degree of } q)+1,\) then \(r\) can be Written in the form \(r(x)=a x+b+\frac{Q(x)}{q(x)}\) with \((\text { degree } Q)<(\text { degree } q)\). Show that \([r(x)-(a x+b)] \rightarrow 0\) both as \(x \rightarrow \infty\) and as \(x \rightarrow-\infty .\) Thus the graph of \(f\) "approaches the line \(y=$$a x+b\)" both as \(x \rightarrow \infty\) and as \(x \rightarrow-\infty\). The line \(y=\) \(a x+b\) is called an oblique asymptote.
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