Chapter 5: Problem 4
In Exercises \(3-14,\) find the arc length of the graph of the function over the indicated interval. $$ y=2 x^{3 / 2}+3, \quad[0,9] $$
Short Answer
Expert verified
The arc length of the graph of the function over the interval [0, 9] is \(\frac{2}{3}*\frac{1}{9}[82^{3/2}-1^{3/2}]\)
Step by step solution
01
Find the derivative of the function
The function is \(y = 2x^{3/2} + 3\). To calculate derivative, power rule which states that the derivative of \(x^n\) is \(nx^{n-1}\), will be used. Therefore, the derivative of \(y\) with respect to \(x\) or \(y'\) is \(3x^{1/2}\).
02
Apply the Arc Length Formula
Now, the Arc Length formula can be used, but first, the derivative term in the formula has to be squared then added to 1 and the square root of the whole expression will be calculated. This gives us \(\sqrt{1 + (3x^{1/2})^2}\) which simplifies to \(\sqrt{1 + 9x}\). Now, using the definite integral from 0 to 9, we resolve \(\int_0^9 \sqrt{1 + 9x} dx\).
03
Use a suitable substitution
The integral seems hard to solve directly, so we use a substitution to make it simpler. Let \(u = 1 + 9x\). Then, as a differential, \(du = 9dx\). Now, \(dx = du/9\). The limits of integration change to u(0) = 1 and u(9) = 82. The integral becomes \[ \int_1^{82} \sqrt{u} (du/9) \]. This simplifies to \(\frac{1}{9}\int_1^{82}u^{1/2}du\).
04
Solve the integral
Using power rule in reverse for integration, we find the antiderivative of \(u^{1/2}\), which is \(\frac{2}{3}u^{3/2}\). When evaluated at 1 and 82, the result is \(\frac{2}{3}[82^{3/2}-1^{3/2}]\)
05
Multiply and Simplify
Don't forget to multiply the result by the constant \(\frac{1}{9}\) from the integral calculation. Therefore, the final arc length is \(\frac{2}{3}*\frac{1}{9}[82^{3/2}-1^{3/2}]\). Simplifying this yields a numerical value
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Derivative of Functions
Understanding the derivative of a function is foundational to calculus. When we speak of the derivative, we refer to the rate at which a function's value changes as its input changes. It's like measuring the speed of a car at a particular moment; just as speed tells us how fast the car is going, the derivative tells us how fast the function's value is increasing or decreasing at any given point.
The derivative is commonly denoted as \(y'\) or \(\frac{dy}{dx}\), and there are various rules to find the derivative, one of which is the power rule. The power rule is particularly useful for functions of the form \(x^n\), where \(n\) is any real number. According to the power rule, the derivative of \(x^n\) is \(nx^{n-1}\). For our exercise example, applying the power rule to \(y = 2x^{3/2} + 3\) gives us the derivative \(y' = 3x^{1/2}\), which is crucial for determining the arc length of the graph.
The derivative is commonly denoted as \(y'\) or \(\frac{dy}{dx}\), and there are various rules to find the derivative, one of which is the power rule. The power rule is particularly useful for functions of the form \(x^n\), where \(n\) is any real number. According to the power rule, the derivative of \(x^n\) is \(nx^{n-1}\). For our exercise example, applying the power rule to \(y = 2x^{3/2} + 3\) gives us the derivative \(y' = 3x^{1/2}\), which is crucial for determining the arc length of the graph.
Arc Length Formula
The arc length formula is instrumental in calculating the length of a curve represented by a function on a graph over a specific interval. The formula for the arc length of a function \(y=f(x)\) from \(a\) to \(b\) is given by \(L = \int_a^b \sqrt{1 + (f'(x))^2} dx\).
Here, \(f'(x)\) is the derivative of the function, which you've learned about in the previous section. To apply the arc length formula to our function \(y=2x^{3/2}+3\) over the interval \[0,9\], we first find the derivative, which is \(3x^{1/2}\), and substitute this into the formula. We then integrate the square root of \(1 + (3x^{1/2})^2\), resulting in the function \(\sqrt{1 + 9x}\) to be integrated with respect to \(x\) from 0 to 9.
Here, \(f'(x)\) is the derivative of the function, which you've learned about in the previous section. To apply the arc length formula to our function \(y=2x^{3/2}+3\) over the interval \[0,9\], we first find the derivative, which is \(3x^{1/2}\), and substitute this into the formula. We then integrate the square root of \(1 + (3x^{1/2})^2\), resulting in the function \(\sqrt{1 + 9x}\) to be integrated with respect to \(x\) from 0 to 9.
Definite Integral
The definite integral is a cornerstone concept in calculus, pivotal in many areas including computation of area under a curve and, as seen in our example, the arc length of a curve. It is represented symbolically as \(\int_a^b f(x) dx\), which essentially means summing up an infinite number of infinitesimally small areas under the function \(f(x)\) from \(x=a\) to \(x=b\).
The result of a definite integral is a number representing the total accumulation of the quantity for which the function stands. In the case of our arc length problem, after substituting the derived function into the arc length formula, we arrive at the definite integral \(\int_0^9 \sqrt{1 + 9x} dx\), which will give us the numerical length of the curve. Calculating this value might be complex, hence the need for the next step—substitution.
The result of a definite integral is a number representing the total accumulation of the quantity for which the function stands. In the case of our arc length problem, after substituting the derived function into the arc length formula, we arrive at the definite integral \(\int_0^9 \sqrt{1 + 9x} dx\), which will give us the numerical length of the curve. Calculating this value might be complex, hence the need for the next step—substitution.
Substitution in Integration
Substitution is a powerful technique in integration used to simplify an integral, making it easier to evaluate. It often involves setting a part of the integrand (the function being integrated) equal to a new variable. This method sometimes resembles a transformation, such as a change of variables, to make the integral more approachable.
In simpler terms, substitution is like changing the perspective to make a problem easier to solve. For our arc length problem, the substitution \(u=1+9x\) simplifies the integral \(\int_0^9 \sqrt{1 + 9x} dx\) by transforming it into the integral \(\frac{1}{9}\int_1^{82} u^{1/2} du\), by also adjusting the bounds to reflect the new variable \(u\). This new form of the integral is more straightforward to solve since it is now in a more familiar format.
In simpler terms, substitution is like changing the perspective to make a problem easier to solve. For our arc length problem, the substitution \(u=1+9x\) simplifies the integral \(\int_0^9 \sqrt{1 + 9x} dx\) by transforming it into the integral \(\frac{1}{9}\int_1^{82} u^{1/2} du\), by also adjusting the bounds to reflect the new variable \(u\). This new form of the integral is more straightforward to solve since it is now in a more familiar format.