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Chapter 3: Problem 19

Find the derivatives of the functions. $$p=\sqrt{3-t}$$

Short Answer

Expert verified
The derivative is \( \frac{dp}{dt} = -\frac{1}{2\sqrt{3-t}} \).

Step by step solution

01

Identify the Function Type

The function given is \( p = \sqrt{3-t} \), which is essentially a radical function. To make differentiation easier, we can express the square root as a fractional exponent: \( p = (3-t)^{1/2} \).
02

Differentiate Using the Power Rule

Using the power rule for differentiation, where \( \frac{d}{dx}[u^n]=nu^{n-1}\frac{du}{dx} \), we differentiate \( p = (3-t)^{1/2} \) with respect to \( t \). Here, \( u = (3-t) \) and \( n = 1/2 \). Thus, the derivative is: \[ \frac{dp}{dt} = \frac{1}{2}(3-t)^{-1/2} \times (-1) \]
03

Simplify the Expression

Multiply out the constant and simplify the expression: \[ \frac{dp}{dt} = -\frac{1}{2}(3-t)^{-1/2} \] Expressing \( (3-t)^{-1/2} \) as a fraction yields: \[ \frac{dp}{dt} = -\frac{1}{2\sqrt{3-t}} \]
04

Interpret and Finalize

The derivative of the function \( p = \sqrt{3-t} \) is \[ \frac{dp}{dt} = -\frac{1}{2\sqrt{3-t}} \]. This expression tells us how the function changes with respect to \( t \) and provides the rate of change of \( p \) at any point \( t \), as long as \( 3-t > 0 \).

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Key Concepts

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

Power Rule
The power rule is one of the fundamental rules in calculus used to differentiate functions that are expressed as powers of a variable. To use the power rule, you'll want a function of the form \( f(x) = x^n \). The derivative of this function is given by multiplying the exponent \( n \) with the function \( x \) raised to one less than the exponent: \( \frac{d}{dx}[x^n] = nx^{n-1} \).

This rule significantly simplifies differentiation, especially when dealing with polynomial expressions. For the function \( (3-t)^{1/2} \) from our exercise, we treat it as \( u^n \), where \( u = (3-t) \) and \( n = \frac{1}{2} \). By applying the power rule, we multiply \( \frac{1}{2} \) by \( (3-t)^{-1/2} \), and continue by differentiating \( u \) itself.

Understanding the power rule allows you to manage and calculate derivatives of various powers straightforwardly, making it an essential tool in solving calculus problems.
Fractional Exponents
Fractional exponents are a way to express roots in terms of powers. A square root, for instance, can be written as a power of one-half. In the exercise, \( \sqrt{3-t} \) is rewritten as \( (3-t)^{1/2} \). This transformation allows the power rule to be used more effectively.

When dealing with fractional exponents:
  • **Square roots** can be expressed as raising to the power of 1/2.
  • **Cube roots** are expressed as raising to the power of 1/3.
  • **n-th roots** in general are written as raising to the power of \( 1/n \).
This method of expressing roots simplifies the application of differentiation rules.

The manipulation of radical expressions into fractional exponents is a key concept that enhances your ability to apply different calculus rules effectively, making it easier to solve expressions involving roots.
Differentiation of Radical Functions
Differentiating radical functions often involves converting the radical expressions into functions with fractional exponents. This transformation enables the use of the power rule to find derivatives in a more straightforward manner.

Consider the function \( p = \sqrt{3-t} \). By expressing it as \( (3-t)^{1/2} \), we simplify the differentiation process. The power rule tells us how to proceed by giving a clear process: multiply the exponent by the base, now reduced by one in power, and then multiply by the derivative of the inside function, \( -(1) \).

The next step is simplification. After using the power rule, express the negative exponent back as a radical, ensuring the final result looks like \( \frac{1}{2\sqrt{3-t}} \).

Understanding differentiation of radical functions means converting them to a format that simplifies the process, delivering the rate of change in a more digestible form.

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Most popular questions from this chapter

Use a CAS to estimate the magnitude of the error in using the linearization in place of the function over a specified interval \(I\). Perform the following steps: a. Plot the function \(f\) over \(I\) b. Find the linearization \(L\) of the function at the point \(a\) c. Plot \(f\) and \(L\) together on a single graph. d. Plot the absolute error \(|f(x)-L(x)|\) over \(I\) and find its maximum value. e. From your graph in part (d), estimate as large a \(\delta>0\) as you can, satisfying \(|x-a|<\delta \quad \Rightarrow \quad|f(x)-L(x)|<\epsilon\) for \(\epsilon=0.5,0.1,\) and \(0.01 .\) Then check graphically to see if your \(\delta\) -estimate holds true. $$f(x)=x^{2 / 3}(x-2), \quad[-2,3], \quad a=2$$

By computing the first few derivatives and looking for a pattern, find \(d^{999} / d x^{999}(\cos x)\)

Give the position function \(s=f(t)\) of an object moving along the \(s\) -axis as a function of time \(t\). Graph \(f\) together with the velocity function \(v(t)=d s / d t=f^{\prime}(t)\) and the acceleration function \(a(t)=d^{2} s / d t^{2}=f^{\prime \prime}(t) .\) Comment on the object's behavior in relation to the signs and values of \(v\) and \(a .\) Include in your commentary such topics as the following: a. When is the object momentarily at rest? b. When does it move to the left (down) or to the right (up)? c. When does it change direction? d. When does it speed up and slow down? e. When is it moving fastest (highest speed)? Slowest? f. When is it farthest from the axis origin? $$s=t^{3}-6 t^{2}+7 t, \quad 0 \leq t \leq 4$$

Is there a value of \(b\) that will make $$g(x)=\left\\{\begin{array}{ll} x+b, & x < 0 \\ \cos x, & x \geq 0 \end{array}\right.$$ continuous at \(x=0 ?\) Differentiable at \(x=0 ?\) Give reasons for your answers.

Suppose that \(y=f(x)\) is differentiable at \(x=a\) and that \(g(x)=\) \(m(x-a)+c\) is a linear function in which \(m\) and \(c\) are constants. If the error \(E(x)=f(x)-g(x)\) were small enough near \(x=a\) we might think of using \(g\) as a linear approximation of \(f\) instead of the linearization \(L(x)=f(a)+f^{\prime}(a)(x-a) .\) Show that if we impose on \(g\) the conditions 1\. \(E(a)=0\) 2\. \(\lim _{x \rightarrow a} \frac{E(x)}{x-a}=0\) then \(g(x)=f(a)+f^{\prime}(a)(x-a) .\) Thus, the linearization \(L(x)\) gives the only linear approximation whose error is both zero at \(x=a\) and negligible in comparison with \(x-a\)
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