What is the arc length of #f(x)= sqrt(x^3+5) # on #x in [0,2]#? How do you find the arc length of the curve #y=e^(x^2)# over the interval [0,1]? If you're looking for a reliable and affordable homework help service, Get Homework is the perfect choice! What is the arc length of #f(x) = 3xln(x^2) # on #x in [1,3] #? How do you find the length of the curve y = x5 6 + 1 10x3 between 1 x 2 ? As with arc length, we can conduct a similar development for functions of \(y\) to get a formula for the surface area of surfaces of revolution about the \(y-axis\). What is the arc length of #f(x)= sqrt(x-1) # on #x in [1,2] #? Here, we require \( f(x)\) to be differentiable, and furthermore we require its derivative, \( f(x),\) to be continuous. In this section, we use definite integrals to find the arc length of a curve. in the x,y plane pr in the cartesian plane. What is the arc length of the curve given by #f(x)=xe^(-x)# in the interval #x in [0,ln7]#? Find the length of the curve of the vector values function x=17t^3+15t^2-13t+10, y=19t^3+2t^2-9t+11, and z=6t^3+7t^2-7t+10, the upper limit is 2 and the lower limit is 5. example Note: the integral also works with respect to y, useful if we happen to know x=g(y): f(x) is just a horizontal line, so its derivative is f(x) = 0. Legal. What is the arc length of #f(x)=x^2/sqrt(7-x^2)# on #x in [0,1]#? In previous applications of integration, we required the function \( f(x)\) to be integrable, or at most continuous. I use the gradient function to calculate the derivatives., It produces a different (and in my opinion more accurate) estimate of the derivative than diff (that by definition also results in a vector that is one element shorter than the original), while the length of the gradient result is the same as the original. (This property comes up again in later chapters.). OK, now for the harder stuff. How do you find the arc length of the curve #y=ln(cosx)# over the What is the arc length of #f(x)=2x-1# on #x in [0,3]#? First, divide and multiply yi by xi: Now, as n approaches infinity (as wehead towards an infinite number of slices, and each slice gets smaller) we get: We now have an integral and we write dx to mean the x slices are approaching zero in width (likewise for dy): And dy/dx is the derivative of the function f(x), which can also be written f(x): And now suddenly we are in a much better place, we don't need to add up lots of slices, we can calculate an exact answer (if we can solve the differential and integral). The basic point here is a formula obtained by using the ideas of calculus: the length of the graph of y = f ( x) from x = a to x = b is arc length = a b 1 + ( d y d x) 2 d x Or, if the curve is parametrized in the form x = f ( t) y = g ( t) with the parameter t going from a to b, then arc length = a b ( d x d t) 2 + ( d y d t) 2 d t refers to the point of tangent, D refers to the degree of curve, We have \( f(x)=2x,\) so \( [f(x)]^2=4x^2.\) Then the arc length is given by, \[\begin{align*} \text{Arc Length} &=^b_a\sqrt{1+[f(x)]^2}\,dx \\[4pt] &=^3_1\sqrt{1+4x^2}\,dx. The graph of \(f(x)\) and the surface of rotation are shown in Figure \(\PageIndex{10}\). Note that the slant height of this frustum is just the length of the line segment used to generate it. f ( x). Derivative Calculator, What is the arc length of #f(x)=2/x^4-1/x^6# on #x in [3,6]#? We define the arc length function as, s(t) = t 0 r (u) du s ( t) = 0 t r ( u) d u. How to Find Length of Curve? What is the arclength of #f(x)=x^5-x^4+x # in the interval #[0,1]#? How do you find the length of a curve using integration? How do you find the arc length of the curve #y=(5sqrt7)/3x^(3/2)-9# over the interval [0,5]? What is the arc length of #f(x)=xsqrt(x^2-1) # on #x in [3,4] #? What is the arclength of #f(x)=[4x^22ln(x)] /8# in the interval #[1,e^3]#? Calculate the arc length of the graph of \( f(x)\) over the interval \( [1,3]\). What is the arclength of #f(x)=(x-3)e^x-xln(x/2)# on #x in [2,3]#? How do can you derive the equation for a circle's circumference using integration? How do you find the distance travelled from t=0 to #t=pi# by an object whose motion is #x=3cos2t, y=3sin2t#? Because we have used a regular partition, the change in horizontal distance over each interval is given by \( x\). To help us find the length of each line segment, we look at the change in vertical distance as well as the change in horizontal distance over each interval. = 6.367 m (to nearest mm). We wish to find the surface area of the surface of revolution created by revolving the graph of \(y=f(x)\) around the \(x\)-axis as shown in the following figure. What is the arclength of #f(x)=x/(x-5) in [0,3]#? What is the arc length of #f(x)=(3/2)x^(2/3)# on #x in [1,8]#? The calculator takes the curve equation. Let \(f(x)\) be a nonnegative smooth function over the interval \([a,b]\). Consider the portion of the curve where \( 0y2\). To embed this widget in a post, install the Wolfram|Alpha Widget Shortcode Plugin and copy and paste the shortcode above into the HTML source. What is the arc length of #f(x)=cosx# on #x in [0,pi]#? What is the arclength of #f(x)=(x-2)/(x^2+3)# on #x in [-1,0]#? There is an unknown connection issue between Cloudflare and the origin web server. Here is a sketch of this situation . What is the arc length of #f(x)=xlnx # in the interval #[1,e^2]#? Arc Length Calculator - Symbolab Arc Length Calculator Find the arc length of functions between intervals step-by-step full pad Examples Related Symbolab blog posts My Notebook, the Symbolab way Math notebooks have been around for hundreds of years. What is the arclength of #f(x)=3x^2-x+4# on #x in [2,3]#? First we break the curve into small lengths and use the Distance Between 2 Points formula on each length to come up with an approximate answer: And let's use (delta) to mean the difference between values, so it becomes: S2 = (x2)2 + (y2)2 lines, Creative Commons Attribution-Noncommercial-ShareAlike 4.0 License. If the curve is parameterized by two functions x and y. Determine the length of a curve, \(x=g(y)\), between two points. The principle unit normal vector is the tangent vector of the vector function. By the Pythagorean theorem, the length of the line segment is, \[ x\sqrt{1+((y_i)/(x))^2}. What is the difference between chord length and arc length? Then, that expression is plugged into the arc length formula. integrals which come up are difficult or impossible to To help support the investigation, you can pull the corresponding error log from your web server and submit it our support team. We have \( f(x)=3x^{1/2},\) so \( [f(x)]^2=9x.\) Then, the arc length is, \[\begin{align*} \text{Arc Length} &=^b_a\sqrt{1+[f(x)]^2}dx \nonumber \\[4pt] &= ^1_0\sqrt{1+9x}dx. The arc length of a curve can be calculated using a definite integral. What is the arc length of #f(x)=sqrt(sinx) # in the interval #[0,pi]#? What is the arclength of #f(x)=x-sqrt(x+3)# on #x in [1,3]#? Surface area is the total area of the outer layer of an object. Set up (but do not evaluate) the integral to find the length of We begin by calculating the arc length of curves defined as functions of \( x\), then we examine the same process for curves defined as functions of \( y\). Figure \(\PageIndex{3}\) shows a representative line segment. Sn = (xn)2 + (yn)2. What is the arc length of #f(x)=x^2/12 + x^(-1)# on #x in [2,3]#? Determine the length of a curve, \(y=f(x)\), between two points. You can find the double integral in the x,y plane pr in the cartesian plane. The vector values curve is going to change in three dimensions changing the x-axis, y-axis, and z-axis and the limit of the parameter has an effect on the three-dimensional plane. Find the surface area of a solid of revolution. Let \(f(x)=\sqrt{x}\) over the interval \([1,4]\). Note: Set z (t) = 0 if the curve is only 2 dimensional. altitude $dy$ is (by the Pythagorean theorem) A piece of a cone like this is called a frustum of a cone. Then, for \( i=1,2,,n\), construct a line segment from the point \( (x_{i1},f(x_{i1}))\) to the point \( (x_i,f(x_i))\). What is the arclength of #f(x)=x+xsqrt(x+3)# on #x in [-3,0]#? To find the surface area of the band, we need to find the lateral surface area, \(S\), of the frustum (the area of just the slanted outside surface of the frustum, not including the areas of the top or bottom faces). From the source of tutorial.math.lamar.edu: Arc Length, Arc Length Formula(s). Accessibility StatementFor more information contact us atinfo@libretexts.orgor check out our status page at https://status.libretexts.org. So, applying the surface area formula, we have, \[\begin{align*} S &=(r_1+r_2)l \\ &=(f(x_{i1})+f(x_i))\sqrt{x^2+(yi)^2} \\ &=(f(x_{i1})+f(x_i))x\sqrt{1+(\dfrac{y_i}{x})^2} \end{align*}\], Now, as we did in the development of the arc length formula, we apply the Mean Value Theorem to select \(x^_i[x_{i1},x_i]\) such that \(f(x^_i)=(y_i)/x.\) This gives us, \[S=(f(x_{i1})+f(x_i))x\sqrt{1+(f(x^_i))^2} \nonumber \]. To embed a widget in your blog's sidebar, install the Wolfram|Alpha Widget Sidebar Plugin, and copy and paste the Widget ID below into the "id" field: We appreciate your interest in Wolfram|Alpha and will be in touch soon. The cross-sections of the small cone and the large cone are similar triangles, so we see that, \[ \dfrac{r_2}{r_1}=\dfrac{sl}{s} \nonumber \], \[\begin{align*} \dfrac{r_2}{r_1} &=\dfrac{sl}{s} \\ r_2s &=r_1(sl) \\ r_2s &=r_1sr_1l \\ r_1l &=r_1sr_2s \\ r_1l &=(r_1r_2)s \\ \dfrac{r_1l}{r_1r_2} =s \end{align*}\], Then the lateral surface area (SA) of the frustum is, \[\begin{align*} S &= \text{(Lateral SA of large cone)} \text{(Lateral SA of small cone)} \\[4pt] &=r_1sr_2(sl) \\[4pt] &=r_1(\dfrac{r_1l}{r_1r_2})r_2(\dfrac{r_1l}{r_1r_2l}) \\[4pt] &=\dfrac{r^2_1l}{r^1r^2}\dfrac{r_1r_2l}{r_1r_2}+r_2l \\[4pt] &=\dfrac{r^2_1l}{r_1r_2}\dfrac{r_1r2_l}{r_1r_2}+\dfrac{r_2l(r_1r_2)}{r_1r_2} \\[4pt] &=\dfrac{r^2_1}{lr_1r_2}\dfrac{r_1r_2l}{r_1r_2} + \dfrac{r_1r_2l}{r_1r_2}\dfrac{r^2_2l}{r_1r_3} \\[4pt] &=\dfrac{(r^2_1r^2_2)l}{r_1r_2}=\dfrac{(r_1r+2)(r1+r2)l}{r_1r_2} \\[4pt] &= (r_1+r_2)l. \label{eq20} \end{align*} \]. \[y\sqrt{1+\left(\dfrac{x_i}{y}\right)^2}. How do you find the lengths of the curve #y=int (sqrtt+1)^-2# from #[0,x^2]# for the interval #0<=x<=1#? Determine the length of a curve, \(y=f(x)\), between two points. Both \(x^_i\) and x^{**}_i\) are in the interval \([x_{i1},x_i]\), so it makes sense that as \(n\), both \(x^_i\) and \(x^{**}_i\) approach \(x\) Those of you who are interested in the details should consult an advanced calculus text. Notice that when each line segment is revolved around the axis, it produces a band. How do you find the distance travelled from t=0 to #t=2pi# by an object whose motion is #x=cos^2t, y=sin^2t#? Arc Length of 2D Parametric Curve. What is the arc length of #f(x)=sin(x+pi/12) # on #x in [0,(3pi)/8]#? Then, \[\begin{align*} \text{Surface Area} &=^d_c(2g(y)\sqrt{1+(g(y))^2})dy \\[4pt] &=^2_0(2(\dfrac{1}{3}y^3)\sqrt{1+y^4})dy \\[4pt] &=\dfrac{2}{3}^2_0(y^3\sqrt{1+y^4})dy. change in $x$ is $dx$ and a small change in $y$ is $dy$, then the Use a computer or calculator to approximate the value of the integral. These findings are summarized in the following theorem. We start by using line segments to approximate the curve, as we did earlier in this section. Taking a limit then gives us the definite integral formula. This almost looks like a Riemann sum, except we have functions evaluated at two different points, \(x^_i\) and \(x^{**}_{i}\), over the interval \([x_{i1},x_i]\). And the diagonal across a unit square really is the square root of 2, right? What is the formula for finding the length of an arc, using radians and degrees? How do you find the arc length of the curve #y=sqrt(x-3)# over the interval [3,10]? Well of course it is, but it's nice that we came up with the right answer! What is the arc length of the curve given by #f(x)=1+cosx# in the interval #x in [0,2pi]#? As a result, the web page can not be displayed. calculus: the length of the graph of $y=f(x)$ from $x=a$ to $x=b$ is The integral is evaluated, and that answer is, solving linear equations using substitution calculator, what do you call an alligator that sneaks up and bites you from behind. length of the hypotenuse of the right triangle with base $dx$ and Perform the calculations to get the value of the length of the line segment. Round the answer to three decimal places. $y={ 1 \over 4 }(e^{2x}+e^{-2x})$ from $x=0$ to $x=1$. The integrals generated by both the arc length and surface area formulas are often difficult to evaluate. What is the arc length of #f(x)=sqrt(4-x^2) # on #x in [-2,2]#? What is the arc length of #f(x)=secx*tanx # in the interval #[0,pi/4]#? What is the arclength of #f(x)=2-x^2 # in the interval #[0,1]#? 1. Many real-world applications involve arc length. to. find the exact area of the surface obtained by rotating the curve about the x-axis calculator. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. Let \(g(y)=3y^3.\) Calculate the arc length of the graph of \(g(y)\) over the interval \([1,2]\). \end{align*}\], Using a computer to approximate the value of this integral, we get, \[ ^3_1\sqrt{1+4x^2}\,dx 8.26815. Because we have used a regular partition, the change in horizontal distance over each interval is given by \( x\). Embed this widget . We have \(g(y)=9y^2,\) so \([g(y)]^2=81y^4.\) Then the arc length is, \[\begin{align*} \text{Arc Length} &=^d_c\sqrt{1+[g(y)]^2}dy \\[4pt] &=^2_1\sqrt{1+81y^4}dy.\end{align*}\], Using a computer to approximate the value of this integral, we obtain, \[ ^2_1\sqrt{1+81y^4}dy21.0277.\nonumber \]. In some cases, we may have to use a computer or calculator to approximate the value of the integral. Let \(r_1\) and \(r_2\) be the radii of the wide end and the narrow end of the frustum, respectively, and let \(l\) be the slant height of the frustum as shown in the following figure. If we now follow the same development we did earlier, we get a formula for arc length of a function \(x=g(y)\). \[ \dfrac{}{6}(5\sqrt{5}3\sqrt{3})3.133 \nonumber \]. What is the arclength of #f(x)=sqrt((x-1)(x+2)-3x# on #x in [1,3]#? Both \(x^_i\) and x^{**}_i\) are in the interval \([x_{i1},x_i]\), so it makes sense that as \(n\), both \(x^_i\) and \(x^{**}_i\) approach \(x\) Those of you who are interested in the details should consult an advanced calculus text. Let \( f(x)=\sqrt{1x}\) over the interval \( [0,1/2]\). Calculate the arc length of the graph of \( f(x)\) over the interval \( [0,]\). \[ \dfrac{1}{6}(5\sqrt{5}1)1.697 \nonumber \]. The distance between the two-p. point. How do you find the arc length of the curve #sqrt(4-x^2)# from [-2,2]? http://mathinsight.org/length_curves_refresher, Keywords: More. What is the arc length of #f(x)= e^(3x)/x+x^2e^x # on #x in [1,2] #? These findings are summarized in the following theorem. How do you find the length of the curve for #y=x^(3/2) # for (0,6)? Accessibility StatementFor more information contact us atinfo@libretexts.orgor check out our status page at https://status.libretexts.org. What is the arclength of #f(x)=sqrt((x+3)(x/2-1))+5x# on #x in [6,7]#? These bands are actually pieces of cones (think of an ice cream cone with the pointy end cut off). For \(i=0,1,2,,n\), let \(P={x_i}\) be a regular partition of \([a,b]\). The length of the curve is also known to be the arc length of the function. Arc Length \( =^b_a\sqrt{1+[f(x)]^2}dx\), Arc Length \( =^d_c\sqrt{1+[g(y)]^2}dy\), Surface Area \( =^b_a(2f(x)\sqrt{1+(f(x))^2})dx\). The concepts we used to find the arc length of a curve can be extended to find the surface area of a surface of revolution. In some cases, we may have to use a computer or calculator to approximate the value of the integral. Before we look at why this might be important let's work a quick example. To find the length of a line segment with endpoints: Use the distance formula: d = [ (x - x) + (y - y)] Replace the values for the coordinates of the endpoints, (x, y) and (x, y). Solution: Step 1: Write the given data. Free Pre-Algebra, Algebra, Trigonometry, Calculus, Geometry, Statistics and Chemistry calculators step-by-step provides a good heuristic for remembering the formula, if a small How do you set up an integral from the length of the curve #y=1/x, 1<=x<=5#? We always struggled to serve you with the best online calculations, thus, there's a humble request to either disable the AD blocker or go with premium plans to use the AD-Free version for calculators. 148.72.209.19 length of a . \sqrt{\left({dx\over dt}\right)^2+\left({dy\over dt}\right)^2}\;dt$$, This formula comes from approximating the curve by straight The formula for calculating the length of a curve is given below: L = a b 1 + ( d y d x) 2 d x How to Find the Length of the Curve? What is the arc length of #f(x) = ln(x) # on #x in [1,3] #? We study some techniques for integration in Introduction to Techniques of Integration. We have just seen how to approximate the length of a curve with line segments. More. What is the arclength of #f(x)=x^3-e^x# on #x in [-1,0]#? Find the surface area of the surface generated by revolving the graph of \( f(x)\) around the \(x\)-axis. If an input is given then it can easily show the result for the given number. How do you find the arc length of the curve #y = 4x^(3/2) - 1# from [4,9]? We have just seen how to approximate the length of a curve with line segments. What is the arclength of #f(x)=1/sqrt((x+1)(2x-2))# on #x in [3,4]#? a = rate of radial acceleration. Round the answer to three decimal places. The arc length is first approximated using line segments, which generates a Riemann sum. How do you find the arc length of the curve #y=sqrt(cosx)# over the interval [-pi/2, pi/2]? To embed a widget in your blog's sidebar, install the Wolfram|Alpha Widget Sidebar Plugin, and copy and paste the Widget ID below into the "id" field: We appreciate your interest in Wolfram|Alpha and will be in touch soon. Taking the limit as \( n,\) we have, \[\begin{align*} \text{Arc Length} &=\lim_{n}\sum_{i=1}^n\sqrt{1+[f(x^_i)]^2}x \\[4pt] &=^b_a\sqrt{1+[f(x)]^2}dx.\end{align*}\]. How do you find the distance travelled from t=0 to #t=2pi# by an object whose motion is #x=cost, y=sint#? How do you find the lengths of the curve #y=(x-1)^(2/3)# for #1<=x<=9#? How do you find the length of the curve for #y=x^2# for (0, 3)? If we want to find the arc length of the graph of a function of \(y\), we can repeat the same process, except we partition the y-axis instead of the x-axis. \nonumber \]. How do you find the length of the curve #x^(2/3)+y^(2/3)=1# for the first quadrant? Disable your Adblocker and refresh your web page , Related Calculators: Arc Length Calculator. Then the arc length of the portion of the graph of \( f(x)\) from the point \( (a,f(a))\) to the point \( (b,f(b))\) is given by, \[\text{Arc Length}=^b_a\sqrt{1+[f(x)]^2}\,dx. Let \( g(y)=\sqrt{9y^2}\) over the interval \( y[0,2]\). When \( y=0, u=1\), and when \( y=2, u=17.\) Then, \[\begin{align*} \dfrac{2}{3}^2_0(y^3\sqrt{1+y^4})dy &=\dfrac{2}{3}^{17}_1\dfrac{1}{4}\sqrt{u}du \\[4pt] &=\dfrac{}{6}[\dfrac{2}{3}u^{3/2}]^{17}_1=\dfrac{}{9}[(17)^{3/2}1]24.118. What is the arc length of #f(x) = (x^2-x)^(3/2) # on #x in [2,3] #? Notice that we are revolving the curve around the \( y\)-axis, and the interval is in terms of \( y\), so we want to rewrite the function as a function of \( y\). Note that some (or all) \( y_i\) may be negative. with the parameter $t$ going from $a$ to $b$, then $$\hbox{ arc length The Length of Curve Calculator finds the arc length of the curve of the given interval. We need to take a quick look at another concept here. \nonumber \]. How do you find the length of the curve #y=x^5/6+1/(10x^3)# between #1<=x<=2# ? Legal. How do you find the length of the line #x=At+B, y=Ct+D, a<=t<=b#? What is the arc length of teh curve given by #f(x)=3x^6 + 4x# in the interval #x in [-2,184]#? \nonumber \]. a = time rate in centimetres per second. How do you find the length of the curve for #y= ln(1-x)# for (0, 1/2)? from. Find the surface area of the surface generated by revolving the graph of \( f(x)\) around the \(x\)-axis. Send feedback | Visit Wolfram|Alpha Length of Curve Calculator The above calculator is an online tool which shows output for the given input. L = /180 * r L = 70 / 180 * (8) L = 0.3889 * (8) L = 3.111 * Now, revolve these line segments around the \(x\)-axis to generate an approximation of the surface of revolution as shown in the following figure. As we have done many times before, we are going to partition the interval \([a,b]\) and approximate the surface area by calculating the surface area of simpler shapes. Definitely well worth it, great app teaches me how to do math equations better than my teacher does and for that I'm greatful, I don't use the app to cheat I use it to check my answers and if I did something wrong I could get tough how to. This page titled 6.4: Arc Length of a Curve and Surface Area is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Gilbert Strang & Edwin Jed Herman (OpenStax) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request. Polar Equation r =. Bundle: Calculus, 7th + Enhanced WebAssign Homework and eBook Printed Access Card for Multi Term Math and Science (7th Edition) Edit edition Solutions for Chapter 10.4 Problem 51E: Use a calculator to find the length of the curve correct to four decimal places. What is the arc length of #f(x)=sqrt(18-x^2) # on #x in [0,3]#? This makes sense intuitively. So the arc length between 2 and 3 is 1. Laplace Transform Calculator Derivative of Function Calculator Online Calculator Linear Algebra 2. For objects such as cubes or bricks, the surface area of the object is the sum of the areas of all of its faces. You just stick to the given steps, then find exact length of curve calculator measures the precise result. For objects such as cubes or bricks, the surface area of the object is the sum of the areas of all of its faces. For \(i=0,1,2,,n\), let \(P={x_i}\) be a regular partition of \([a,b]\). We start by using line segments to approximate the length of the curve. where \(r\) is the radius of the base of the cone and \(s\) is the slant height (Figure \(\PageIndex{7}\)). How do you find the length of the curve for #y= 1/8(4x^22ln(x))# for [2, 6]? The integrals generated by both the arc length and surface area formulas are often difficult to evaluate. Thus, \[ \begin{align*} \text{Arc Length} &=^1_0\sqrt{1+9x}dx \\[4pt] =\dfrac{1}{9}^1_0\sqrt{1+9x}9dx \\[4pt] &= \dfrac{1}{9}^{10}_1\sqrt{u}du \\[4pt] &=\dfrac{1}{9}\dfrac{2}{3}u^{3/2}^{10}_1 =\dfrac{2}{27}[10\sqrt{10}1] \\[4pt] &2.268units. You can find formula for each property of horizontal curves. This is important to know! arc length of the curve of the given interval. How do you find the arc length of #y=ln(cos(x))# on the interval #[pi/6,pi/4]#? Find the surface area of the surface generated by revolving the graph of \( f(x)\) around the \( y\)-axis. The curve length can be of various types like Explicit Reach support from expert teachers. What is the arclength of #f(x)=1/e^(3x)# on #x in [1,2]#? Let \( f(x)=y=\dfrac[3]{3x}\). How do you find the arc length of the curve # f(x)=e^x# from [0,20]? Notice that we are revolving the curve around the \( y\)-axis, and the interval is in terms of \( y\), so we want to rewrite the function as a function of \( y\). We can then approximate the curve by a series of straight lines connecting the points. $\begingroup$ @theonlygusti - That "derivative of volume = area" (or for 2D, "derivative of area = perimeter") trick only works for highly regular shapes. Here, we require \( f(x)\) to be differentiable, and furthermore we require its derivative, \( f(x),\) to be continuous. We have \( g(y)=(1/3)y^3\), so \( g(y)=y^2\) and \( (g(y))^2=y^4\). }=\int_a^b\; Note that the slant height of this frustum is just the length of the line segment used to generate it. What is the arclength of #f(x)=sqrt(4-x^2) # in the interval #[-2,2]#? How do you find the length of the curve defined by #f(x) = x^2# on the x-interval (0, 3)? by cleaning up a bit, = cos2( 3)sin( 3) Let us first look at the curve r = cos3( 3), which looks like this: Note that goes from 0 to 3 to complete the loop once. How do you evaluate the following line integral #(x^2)zds#, where c is the line segment from the point (0, 6, -1) to the point (4,1,5)? How do you find the distance travelled from t=0 to t=1 by a particle whose motion is given by #x=4(1-t)^(3/2), y=2t^(3/2)#? 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