Find the points (x, y) at which the curve has: (a) a horizontal tangent: (b) a vertical tangent.

step 1 In this question, we are given the parameter equations. XT is equal to sun 2T and YT is equal to

Find the points (x, y) at which the curve has: (a) a...

Key Concepts

Parametric Equations

Parametric equations use one or more independent parameters to express the coordinates of the points on a curve. This approach is particularly useful for describing curves that cannot be easily represented by a single function y=f(x) or x=f(y). It allows the separate description of x and y variables as functions of an independent parameter that can simplify analysis and computations, such as finding tangents and plotting the curve.

Derivatives in Parametric Form

When a curve is given by parametric equations, the slope of the tangent line (dy/dx) can be determined by taking the derivative of both x and y with respect to the parameter and then dividing them: dy/dx = (dy/dt)/(dx/dt), provided dx/dt is not zero. This method is essential for understanding the behavior of the curve and determining the nature of its tangents.

Horizontal and Vertical Tangents

Horizontal tangents occur where the derivative dy/dx equals zero, typically when dy/dt = 0 and dx/dt ? 0, indicating a flat slope at that point on the curve. Conversely, vertical tangents occur where the derivative is undefined, which usually happens when dx/dt = 0 and dy/dt ? 0, meaning the tangent line is vertical. Identifying these conditions is important for a detailed understanding of the curve’s geometry.

Elimination of the Parameter

Eliminating the parameter involves rewriting the parametric equations to obtain a direct relation between x and y. Although not always necessary for finding tangents, this process can provide insights into the overall shape and structure of the curve and facilitate analysis and visualization.

Sketching Parametric Curves

Sketching a curve defined parametrically involves plotting key points, determining the behavior at the endpoints or critical points, and analyzing symmetry or periodicity. It requires understanding how x and y vary with the parameter and using the computed tangents to capture the geometry correctly. This holistic visualization is crucial in many applications of calculus and analytical geometry.

Transcript

00:01 In this question, we are given the parameter equations.

00:04 Xt is equal to sun 2t and yt is equal to sine 2t.

00:15 Now we know that a horizontal tangent occurs at d .dx is equal to zero or when dydx is equal to zero rather.

00:27 And a vertical tangent only occurs where dydx is equal to zero, is undefined, undefined.

00:40 So here to find, d .x, which is our gradient expression, we have to first find dxd2, and dxdt here is 2, 2, kosoff, 2t, and our d -y -d -t here is kosoff t.

01:02 Our d -y -d -x is given by d.

01:08 D .t over dx dt.

01:13 Plugging in what we have here we have we have we have cost t is dydd t over two course of two t and now we know that at horizontal target d y dx is equal to zero and d y dx is equal to zero and for this expression to be equal to zero cost t has to be equal to zero so we are looking for the values of t for which cost t is equal to zero now we know that and this has to be well cost of two t or the denominator is not equal to zero so those values are t will be equal to pi over two or three pi over two or three pi over 2.

02:11 These are the values of t for which cost t is equal to 0.

02:18 Now at t is equal to pi over 2 to find the points, we plug in pi over 2 for x into x.

02:28 So pi over 2, here we have sign of pi and we get 0.

02:36 At y pi over 2, we will have sign of pi over 2.

02:43 We get one.

02:45 So the point is 0 1 where we have a horizontal tangent and at t is equal to at t is equal to 3 pi over 2 our x will be given my our x at 3 pi over 2 is equal to sign of 3 pi so here sign of 3 pi gives us 0 and y at 3 pi over 2 we have sign of 3 pi over 2 and we get a minus 1 so the point at which we have another horizontal tangent is 0 and minus 1 like that so moving on for vertical tangents at vertical tangent you know that d x is undefined for do idx to be undefined for to find the denominator here should give us a zero.

03:50 So 2 cost of 2 t should give us a 0, dividing both sides by 2.

03:57 We have cause of 2t 0 is equal to a 0 and therefore 2t is equal to arc cause of 0.

04:11 When we arc course of 0 will give us pi over 2 and 3.

04:18 Pi over and these are plus in o minus plus or minus these values so dividing by two t is equal to plus or minus pi over four or plus or minus three pi over four so these are the values of t for which we will have du idx undefined such that at those points we have the horizontal tangent, the vertical tangent.

04:51 So plugging in these points at t is equal to plus or minus pi over four, we are going to get x as plus or minus one.

05:05 Our y will be equal to plus or minus one over root of two...

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