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In this video, weβll learn how to use integration to find the arc length of a curve defined by parametric equations of the form π₯ equals π of π‘ and π¦ equals π of π‘.
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Weβll begin by recalling the formula for the arc length of a curve defined as π¦ is equal to some function of π₯.
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Then, weβll look at how we can generalize this formula for parametrically defined curves and consider a number of examples of the process.
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Given an equation π¦ in terms of π₯ and π₯-values greater than or equal to π and less than or equal to π, the arc length πΏ is given by the definite integral between π and π of the square root of one plus dπ¦ by dπ₯ squared with respect to π₯.
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We want to find a way to use this formula for curves defined parametrically.
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Remember, these are of the form π₯ equals π of π‘ and π¦ equals π of π‘.
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We also know that in this case, dπ¦ by dπ₯ is equal to dπ¦ by dπ‘ over dπ₯ by dπ‘, which can be written as π prime of π‘ over π prime of π‘.
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Now, using the fact that dπ₯ by dπ‘ is equal to π prime of π‘, weβre going to equivalently say that dπ₯ equals π prime of π‘ dπ‘.
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Redefining our limits so theyβre in terms of π‘ and we find that we can rewrite the arc length as the definite integral between πΌ and π½ of the square root of one plus dπ¦ by dπ‘ over dπ₯ by dπ‘ squared π prime of π‘ dπ‘.
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We separate the components of our fraction and rewrite π prime of π‘ as dπ₯ by dπ‘.
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Now, this looks really nasty.
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But weβre next going to add the fractions inside of our root by creating a common denominator of dπ₯ by dπ‘ squared.
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So inside of our root, we have dπ₯ by dπ‘ all squared over dπ₯ by dπ‘ all squared plus dπ¦ by dπ‘ all squared over dπ₯ by dπ‘ all squared.
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We can then factor out that denominator and notice we need to use the absolute value symbols as we need this to be positive to be able to do so.
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And we see that the arc length is equal to the definite integral between πΌ and π½ of one over the absolute value of dπ₯ by dπ‘ times the square root of dπ₯ by dπ‘ squared plus dπ¦ by dπ‘ squared times dπ₯ by dπ‘ dπ‘.
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Now, in fact, if we assume that the curve is traced out from left to right, we can lose the absolute value symbol.
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And then, we notice that one over dπ₯ by dπ‘ times dπ₯ by dπ‘ is simply one.
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And so, weβre left with the given formula for the arc length of a curve between the limits of π‘ equals πΌ and π‘ equals π½.
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Now, it is important to note that the Cartesian form of the arc length formula is only valid for π¦ equals π of π₯ when π prime is continuous on the closed interval π to π.
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So using this to make our parametric version has a similar consequence on what parametric equations we can use this formula on.
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In fact, if we start with the parametric equations π₯ equals π of π‘ and π¦ equals π of π‘, then π prime and π prime must be continuous on the closed interval from πΌ to π½.
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Itβs also worth remembering that with parametric equations the curve can loop over itself.
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This might lead to answers, which are longer than the actual arc length, in which case we would need to determine a range for π‘, for which the arc is traced out exactly once.
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So letβs have a look at how we can apply this formula.
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Express the length of the curve with parametric equations π₯ equals π‘ squared minus π‘ and π¦ equals π‘ to the fourth power, where π‘ is greater than or equal to one and less than or equal to four, as an integral.
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We recall that the formula for the arc length πΏ of a curve defined parametrically between the limits of π‘ equals πΌ and π‘ equals π½ is the definite integral between πΌ and π½ of the square root of dπ₯ by dπ‘ squared plus dπ¦ by dπ‘ squared with respect to π‘.
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Now, our curve is defined parametrically by π₯ equals π‘ squared minus π‘ and π¦ equals π‘ to the fourth power.
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And we want to find this arc length between the limits of π‘ equals one and π‘ equals four.
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So we let πΌ be equal to one, π½ be equal to four.
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And we see weβre going to need to differentiate π₯ and π¦ with respect to π‘.
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Now, to differentiate a polynomial term, we simply multiply the entire term by the exponent and then reduce the exponent by one.
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So the derivative of π‘ squared is two π‘.
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And when we differentiate negative π‘, we get negative one.
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dπ₯ by dπ‘ is, therefore, two π‘ minus one.
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And this satisfies the criteria that the derivative of this function is continuous.
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dπ¦ by dπ‘ is the first derivative of π‘ to the fourth power.
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Thatβs four π‘ cubed, which is also a continuous function.
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Now, we noticed that weβre going to have to square these in our formula for the arc length.
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So letβs work out dπ₯ by dπ‘ squared and dπ¦ by dπ‘ squared before substituting into the formula.
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By distributing the parentheses, we find that two π‘ minus one all squared is four π‘ squared minus four π‘ plus one.
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And four π‘ cubed all squared is 16π‘ to the sixth power.
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Our final step is to substitute into the formula for the arc length.
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And we find that πΏ is equal to the definite integral between one and four of the square root of four π‘ squared minus four π‘ plus one plus 16π‘ to the sixth power dπ‘.
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We might choose to rewrite the expression inside our root in descending powers of π‘.
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And when we do, we find that the arc length of the curve defined by our parametric equations for π‘ is greater than or equal to one and less than or equal to four is the integral shown.
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In our next example, weβre going to look at how to actually evaluate one of these expressions.
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Find the length of the curve with parametric equations π₯ equals three cos π‘ minus cos three π‘ and π¦ equals three sin π‘ minus sin three π‘, where π‘ is greater than or equal to zero and less than or equal to π.
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We recall that the formula we used to find the arc length of a curve defined parametrically for values of π‘ from πΌ to π½ is the definite integral between πΌ and π½ of the square root of dπ₯ by dπ‘ squared plus dπ¦ by dπ‘ squared with respect to π‘.
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In this case, π₯ is equal to three cos π‘ minus cos three π‘ and π¦ is equal to three sin π‘ minus sin three π‘.
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And weβre interested in the length of the curve between π‘ is greater than or equal to zero and less than or equal to π.
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So weβll let πΌ be equal to zero and π½ be equal to π.
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Weβre also going to need to work out dπ₯ by dπ‘ and dπ¦ by dπ‘.
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And so, since weβre working with trigonometric expressions, we recall the derivative of cos of ππ‘ and sin of ππ‘.
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They are negative π sin of ππ‘ and π cos ππ‘, respectively, for real constant values of π.
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This means dπ₯ by dπ‘ is negative three sin π‘ minus negative three sin three π‘.
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And of course, that becomes plus three sin three π‘.
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Similarly, dπ¦ by dπ‘ is three cos π‘ minus three cos three π‘.
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Before we substitute into the formula, weβre actually going to square these and find their sum.
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Negative three sin π‘ plus three sin three π‘ all squared is nine sin squared π‘ minus 18 sin π‘ sin three π‘ plus nine sin square three π‘.
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Then, three cos π‘ minus three cos three π‘ all squared is nine cos squared π‘ minus 18 cos π‘ cos three π‘ plus nine cos squared three π‘.
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At this stage, we recall the trigonometric identity sin squared π‘ plus cos squared π‘ equals one.
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And we see that we have nine sin squared π‘ plus nine cos squared π‘.
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Well, that must be equal to nine.
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Similarly, we have nine sin squared three π‘ plus nine cos squared three π‘, which is also equal to nine.
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And we also have negative 18 times sin π‘ sin three π‘ plus cos π‘ cos three π‘.
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All Iβve done here is factored the negative 18 out.
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Next, weβre going to use the trigonometric identity cos of π΄ minus π΅ is equal to cos π΄ cos π΅ plus sin π΄ sin π΅.
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And this means that sin π‘ sin three π‘ plus cos π‘ cos three π‘ must be equal to cos of three π‘ minus π‘, which is, of course, simply cos of two π‘.
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So this becomes 18 minus 18 cos of two π‘.
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And so, we find that the arc length is equal to the definite integral between zero and π of the square root of 18 minus 18 cos of two π‘ dπ‘.
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Letβs clear some space and evaluate this integral.
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Now, in fact, the integral of the square root of 18 minus 18 cos of two π‘ still isnβt particularly nice to calculate.
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And so, we go back to the fact that cos of two π‘ is equal to two cos squared π‘ minus one.
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We replace cos of two π‘ with this expression and then distribute the parentheses.
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And our integrand is now equal to the square root of 36 minus 36 cos squared π‘.
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We take out the common factor of 36 and then rearrange the identity sin squared π‘ plus cos squared π‘ equals one.
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So that one minus cos squared π‘ is equal to sin squared π‘.
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So our integrand is six times the square root of sin squared π‘, which is, of course, simply six sin π‘.
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When we integrate six sin π‘, we get negative six cos π‘.
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So the arc length is equal to negative six cos π‘ evaluated between those limits.
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Thatβs negative six cos of π minus negative six cos of zero, which is equal to 12.
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And so, we found that the arc length of the curve that weβre interested in is 12 units.
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As you might expect, not only does this process work for curves defined by trigonometric equations, but also those defined by exponential and logarithmic ones.
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Find the length of the curve with parametric equations π₯ equals π to the power of π‘ minus π‘ and π¦ equals four π to the power of π‘ over two, where π‘ is greater than or equal to zero and less than or equal to two.
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We know that the formula we used to find the arc length of curves defined parametrically from values of π‘ greater than or equal to πΌ and less than or equal to π½ is the definite integral between πΌ and π½ of the square root of dπ₯ by dπ‘ squared plus dπ¦ by dπ‘ squared with respect to π‘.
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Now, in this case, weβre interested in the length of the curve, where π‘ is greater than or equal to zero and less than or equal to π‘.
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So weβll let πΌ to be equal to zero and π½ be equal to two.
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Then, our parametric equations are π₯ equals π to the power of π‘ minus π‘ and π¦ equals four π to the power of π‘ over two.
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Itβs quite clear that weβre going to need to work out dπ₯ by dπ‘ and dπ¦ by dπ‘.
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And so, we firstly recall that the derivative of π to the power of π‘ is π to the power of π‘.
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The derivative of negative π‘ is one.
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So dπ₯ by dπ‘ is π to the power of π‘ minus one.
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Now, weβre going to use the chain rule to differentiate π¦ with respect to π‘.
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We let π’ be equal to π‘ over two.
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So that dπ’ by dπ‘ is equal to one-half.
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Then, dπ¦ by dπ‘ is dπ¦ by dπ’ times dπ’ by dπ‘.
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Now, π¦ is equal to four π to the power of π’.
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So dπ¦ by dπ‘ is four π to the power of π’ times a half, which is two π to the power of π’.
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But of course, we want dπ¦ by dπ‘ in terms of π‘.
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So we replace π’ with π‘ over two.
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And we find that dπ¦ by dπ‘ equals two π to the power of π‘ over two.
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Now, in fact, for our arc length formula, we need dπ₯ by dπ‘ squared and dπ¦ by dπ‘ squared.
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So weβre going to square each of our expressions.
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When we do, we find that π to the power of π‘ minus one squared is π to the power of two π‘ minus two π to the power of π‘ plus one.
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And dπ¦ by dπ‘ squared is simply four π to the power of π‘.
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Now, we substitute everything we know into our formula for the arc length.
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And we get the definite integral between zero and two of the square root of π to the power of two π‘ minus two π to the power of π‘ plus one plus four π to the power of π‘ with respect to π‘.
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We notice that negative two π to the power of π‘ plus four π to the power of π‘ is two π to the power of π‘.
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And thatβs great because we see we can factor π to the power of two π‘ plus two π to the power of π‘ plus one a little like we would with a quadratic.
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We get π to the power of π‘ plus one times π to the power of π‘ plus one or π to the power of π‘ plus one squared.
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And of course, the square root of π to the power of π‘ plus one squared is just π to the power of π‘ plus one.
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When we integrate π to the power of π‘, we get π to the power of π‘.
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And the integral of one is π‘.
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This means the arc length is π squared plus two minus π to the power of of zero plus zero.
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And, of course, π to the power of zero is one.
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So this simplifies to π to the power of two plus one.
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And the length of the curve with our parametric equations for values of π‘ from zero to two is π to the power of two plus one units.
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In this video, weβve learned that for a curve defined parametrically by π₯ equals π of π‘ and π¦ equals π of π‘, the arc length of the curve for values of π‘ greater than or equal to πΌ and less than or equal to π½ is given by the definite integral between πΌ and π½ of the square root of dπ₯ by dπ‘ squared plus dπ¦ by dπ‘ squared with respect to π‘.
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In this case, π prime and π prime must be continuous functions on the closed interval πΌ to π½.
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We saw that itβs also worth remembering that with parametric equations, the curve can loop over itself.
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And in which case, we do need to determine a range for π‘, for which the arc is traced out exactly once.