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Thereβll be times where we might wish to find the slope of the tangent line to a polar curve given in the form π equals π of π.
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In this video, weβre going to learn how we can extend our understanding of finding the derivative of parametric equations to help us calculate the derivatives of polar curves and, hence, the slope of a polar curve.
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We begin by recalling that we can convert polar coordinates into Cartesian coordinates using the following equations, π₯ equals π cos π and π¦ equals π sin π.
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Weβre going to be considering a tangent line to a polar curve with the equation π is equal to π of π.
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So, letβs replace π in these equations with π of π so that π₯ is equal to π of π cos π and π¦ is equal to π of π times sin of π.
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Notice, we now have what looks a lot like a pair of parametric equations.
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We have equations for π₯ and π¦ in terms of a third parameter π.
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We recall that for two functions π₯ and π¦ in terms of π‘, their derivative dπ¦ by dπ₯ is dπ¦ by dπ‘ divided by dπ₯ by dπ‘.
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So, we replace π‘ with π, and we see that we can find the slope by calculating dπ¦ by dπ₯.
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Itβs dπ¦ by dπ divided by dπ₯ by dπ.
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But letβs have a look at our expressions for π₯ and π¦.
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Theyβre both actually the product of two functions in π, so we can use the product rule to find their derivatives.
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Remember, the product rule says that the derivative of the product of two differentiable functions π’ and π£ is π’ times dπ£ by dπ₯ plus three times dπ£ by dπ₯.
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This means dπ₯ by dπ is the derivative of π times cos π minus π times the derivative of cos π, which is sin π.
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So, thatβs π prime of π cos π minus π of π sin π.
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Now, letβs replace π of π with π and π prime of π with dπ dπ, and we see that dπ₯ by dπ is equal to dπ dπ cos π minus π sin π.
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By performing a similar process for dπ¦ by dπ, we see that itβs equal to dπ by dπ times sin π plus π cos π.
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And since we know that dπ¦ by dπ₯ is equal to dπ¦ by dπ divided by dπ₯ by dπ, we can say that dπ¦ by dπ₯ must be equal to dπ dπ sin π plus π cos π divided by dπ dπ cos π minus π sin π.
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Now, what this also means is that since we find horizontal tangents by finding the points where dπ¦ by dπ₯ is equal to zero, we can find horizontal tangents by letting dπ¦ by dπ equal to zero, provided that dπ₯ by dπ does not also equal zero.
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Similarly, we locate vertical tangents at the point where dπ₯ by dπ equals zero, assuming that π¦ by dπ₯ is not equal to zero at these points.
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Letβs now have a look at the application of these formulae.
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Find the slope of the tangent line to the curve π equals one over π at π equals π.
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Remember, when finding the slope, we need to evaluate the derivative of that curve at a specific point.
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The formula weβre interested in for a polar curve of the form π equals π of π is dπ¦ by dπ₯ equals dπ¦ by dπ divided by dπ₯ by dπ.
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And then, we have specific equations for dπ¦ by dπ and dπ₯ by dπ in terms of π and π.
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We can see that our equation is given as π equals one over π.
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So, weβll begin by working out the value of dπ dπ.
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Letβs start by writing one over π as π to the power of negative one.
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And of course, to differentiate a simple polynomial terms like this one, we multiply by the exponent and then reduce that exponent by one.
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So, thatβs negative one times π to the power of negative two, which is the same as negative one over π squared.
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Weβre also given that π is equal to π.
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So, all we need to do is substitute everything we know into our formula. dπ¦ by dπ is equal to negative one over π squared times sin π plus one over π times cos π.
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And then, we obtain dπ₯ by dπ to be equal to negative one over π squared cos π minus one over π sin π. dπ¦ by dπ₯ is the quotient of these too.
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And of course, we could simplify this somewhat by factoring one over π, for example.
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However, weβre going to evaluate this when π is equal to π, so we may as well go ahead and substitute that in.
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When we do, we find that the slope is equal to negative one over π squared times sin π plus one over π times cos π over negative one over π squared cos π minus one over π sin π.
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And in fact, sin of π is zero and cos of π is negative one, so this simplifies really nicely to negative one over π over one over π squared.
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Due to the nature of dividing fractions, this simplifies further to negative π squared over π, which is simply negative π.
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And we found the value of the derivative at π equals π.
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And therefore, the slope of the tangent line to the curve at this point is negative π.
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In our next example, weβll look at how we can use this formulae to locate horizontal and vertical tangent lines.
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Find the points at which π equals four cos π has a horizontal or vertical tangent line.
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Remember, the formula for the slope of the polar curve π is equal to π of π is dπ¦ by dπ₯ equals dπ¦ by dπ over dπ₯ by dπ, where dπ¦ by dπ is equal to dπ by dπ sin π plus π cos π and dπ₯ by dπ is equal to dπ dπ cos π minus π sin π.
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Horizontal tangents will occur when the derivative is equal to zero, in other words, where the numerator is equal to zero.
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So, we locate horizontal tangents by finding the points where dπ¦ by dπ equals zero, assuming that dπ₯ by dπ does not also equal zero.
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Vertical tangents occur where our denominator dπ₯ by dπ is equal to zero and the numerator dπ¦ by dπ is not equal to zero.
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In some sense, we can think of our gradient as positive or negative infinity here.
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But strictly speaking, we say that dπ¦ by dπ₯ is undefined, in other words, where the denominator dπ₯ by dπ is equal to zero, provided that d π¦ by dπ does not also equal to zero.
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So, weβre going to need to work out dπ¦ by dπ and dπ₯ by dπ.
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Weβre told that π is equal to four cos π.
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We quote the general result for the derivative of cos π as being negative sin π.
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And we see that dπ by dπ must be negative four sin π.
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This means dπ¦ by dπ is equal to negative four sin π times sin π plus π times cos π.
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And π is cos four cos π, so thatβs plus four cos π cos π.
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That simplifies to negative four sin squared π plus four cos squared π.
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And weβll set this equal to zero to find the location of any horizontal tangents.
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We begin to solve for π by dividing through by four and then adding sin squared π to both sides of the equation.
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Then, we divide through by cos squared π.
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And this is really useful because we know that sin squared π divided by cos squared π is equal to tan squared π.
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By finding the square root of both sides of this equation, remembering to take by the positive and negative square root of one, we obtain tan π to be equal to plus or minus one.
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Taking the inverse tan will give us the value of π.
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So, π is equal to the inverse tan of negative one or the inverse tan of one.
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That gives us values of π of negative π by four and π by four.
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Now, remember, weβre trying to find the points at which these occur, so we do need to substitute our values of π back into our original function for π.
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That gives us π is equal to four cos of negative π by four or four cos of π by four, which is two root two in both cases.
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And so, we see that we have horizontal tangent lines at the points two root two π by four and two root two negative π by four.
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Weβre now going to work out the vertical tangent lines.
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So, weβre going to repeat this process for dπ₯ by dπ, substituting π and dπ by dπ into our formula for dπ₯ by dπ.
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And we see that itβs equal to negative four sin π cos π minus four cos π sin π.
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We set this equal to zero, and we see that we can divide through by negative four.
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So, we obtain zero to be equal to sin π cos π plus cos π sin π, or two sin π cos π.
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Now, actually, we know that sin of two π is equal to two sin π cos π, so we can solve this by setting zero equal to sin of two π.
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So, two π is equal to the inverse tan of zero.
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Now, sin is periodic, and we need to consider that this is also the case for two π must be equal to π.
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And since π is greater than negative π and less than or equal to π, these are actually the only options we consider.
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Dividing through by two, we obtain π to be equal to zero and π by two.
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Remember, we substitute this back into our original equation for π.
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And we get that π equals four cos π or four cos π by two, which gives us values of π as four and zero.
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And we found the points at which π equals four cos π has horizontal and vertical tangent lines.
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The horizontal tangent lines are at the point two root two, π by four and two root two, negative π by four and the vertical tangent lines are at the points four, zero and zero, π by two.
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Weβll now consider how we can find the slopes of the tangent lines to a polar curve at the tips of the leaves.
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Find the slopes of the tangent lines to π equals two sin three π at the tips of the leaves.
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Remember the leaves in our polar curves are the repeating loops.
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And here, we recall that the graph of π equals sin of ππ β and in fact, multiples of sin ππ β has π loops when π is odd, and two π loops when π is an even integer.
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In our case, π is three; itβs odd.
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So, the graph of π equals two sin of three π is going to have three loops, or petals.
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So, how do we find the loops?
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Well, we begin by finding the two smallest non-negative values of π for which two sin of three π equals zero.
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These will describe the beginning and end point of each of our loops, of course, when π is equal to zero.
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By dividing through by two, we see that sin of three π equals zero.
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And then, we take the inverse sin of zero, and we see that three π is equal to zero or, alternatively, π.
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This means that π must be equal to zero or π by three.
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One leaf is, therefore, described by letting π go from zero to π by three.
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That might look a little something like this.
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This means the tip of its petal will occur when π is exactly halfway between π equals zero and π equals π by three.
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Thatβs π equals π by six.
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Since we have three loops, we continue by adding a third of a full turn.
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Thatβs two π by three.
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So, we find that π is equal to π by six, π by six plus two π over three, which is five π by six, and then we also subtract two π by three from π by six.
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And that gives us negative π by two.
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Itβs important to substitute these values into our original equation for π, and we find that π is equal to two throughout.
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We can say that the tips of the leaves are at two, π by six; two, five π by six; and two, negative π by two.
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Now, we can alternatively redefine our final point as negative two, π by two.
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And this is because a negative value of π takes us through the origin and out the other side at exactly the same distance.
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So, these are actually the exact same points.
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Remember, we need to find the slopes of the tangent lines to our curve at these points.
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And we recall that we can find the derivative and, therefore, the slope by using the given formula.
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The derivative of sin of three π is three cos of three π.
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So, dπ by dπ is six cos of three π.
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We can substitute this into our formula for dπ¦ by dπ₯.
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And we see that itβs equal to six cos of three π sin π plus two sin of three π cos π over six cos three π cos π minus two sin of three π sin π.
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Weβll evaluate this at the tip of our first leaf by substituting π equals π by six.
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And that gives us a slope of negative root three.
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We then substitute in π is equal to five π by six, and we find the slope here is root three.
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And finally, we substitute in π equals π by two, which gives us a slope of zero.
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And itβs important to realise that have we substituted the π equals negative π by two, we would have got the same answer.
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So, we find the slope at the first tip is a negative root three, and thatβs at the point two, π by six.
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The second leaf tip is at two, five π by six.
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And the slope there is root three.
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And our third is at negative two, π over two or two, negative π over two.
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And the slope there is zero.
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We can also use the techniques outlined in this video to find an equation for the tangent or normal to a polar curve at a point.
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Letβs see what that might look like.
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Find the equation of the tangent line to the polar curve with equation π equals cos two π at π is equal to π by four.
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Remember, the equation for a straight line is given by π¦ minus π¦ one equals π times π₯ minus π₯ one.
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Here, π is the slope, whereas π₯ one, π¦ one are the point where the tangent line hits the curve.
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So, weβre going to need to begin by working out the slope of our tangent line.
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The formula we can use is dπ¦ by dπ₯ equals dπ by dπ sin π plus π cos π over dπ dπ cos π minus π sin π.
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Well, our π is equal to cos two π, so dπ by dπ must be equal to negative two sin of two π.
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We can substitute this into our equation for dπ¦ by dπ₯, and we get negative two sin two π sin π plus cos two π cos π all over negative two sin two π cos π minus cos two π sin π.
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Remember, weβre looking to find the slope when π is equal to π by four.
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So, letβs evaluate this when π is equal to π by four.
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When we replace π with π by four, we find dπ¦ by dπ₯ to be equal to one.
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So, we know the slope of the curve, but what about π₯ one, π¦ one?
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Well, we use the fact that π₯ is equal to π cos π and π¦ is equal to π sin π.
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And since π is equal to cos two π, we find that π₯ is equal to cos two π cos π and π¦ is equal to cos two π sin π.
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Then, when π is equal to π by four, we find that π₯ is equal to cos of two times π by four times cos of π by four, which is simply equal to zero.
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Similarly, π¦ is equal to cos of two times π by four times sin of π by four, which is also zero.
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So, the equation of the tangent line to our curve at π is equal to π by four is π¦ minus zero equals one times π₯ minus zero, which is simply π¦ equals π₯.
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In this video, weβve seen that we can use the formula dπ¦ by dπ₯ equals dπ¦ by dπ over dπ₯ by dπ to find the derivative of a polar curve given in the form π equals π of π, where dπ¦ by dπ and dπ₯ by dπ are as shown.
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By setting dπ¦ by dπ equals zero, we can establish the existence of horizontal tangents.
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And by setting the dπ₯ by dπ equal to zero, we find the vertical tangents.
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We can also combine this formula with the equation of a straight line to help us find the equation of the tangent and normal to a curve.