Space Curves. Recall the parametric equations of a curve in xyplane and compare them with parametric equations of a curve in space.


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1 Clculus 3 Li Vs Spce Curves Recll the prmetric equtions of curve in xyplne nd compre them with prmetric equtions of curve in spce. Prmetric curve in plne x = x(t) y = y(t) Prmetric curve in spce x = x(t) y = y(t) z = z(t) Given its prmetric equtions x = x(t), y = y(t), z = z(t), curve C cn be considered to be vector function, tht is function whose domin is in n intervl of rel numbers nd the rnge is set of vectors: r(t) = x(t), y(t), z(t). In this cse, the curve C is the grph of the vector function r(t). Any vlue t = t from the domin of r(t) corresponds to point (x, y, z ) on the curve C. The derivtive of vector function r = x(t), y(t), z(t) is the vector function r (t) = x (t), y (t), z (t) At point (x, y, z ) which corresponds to the vlue t of prmeter t, the vlue of the derivtive r (t ) = x (t ), y (t ), z (t ) represents the velocity vector of the tngent line t (x, y, z ). Note the nlogy with the twodimensionl scenrio. To find tngent line to the curve To find tngent line to the curve x = x(t), y = y(t) t t = t, use x = x(t), y = y(t), z = z(t) t t = t, use Point: (x(t ), y(t )), Point: (x(t ), y(t ), z(t )), Direction vector: x (t ), y (t ). Direction vector: x (t ), y (t ), z (t ). An eqution of the tngent line: An eqution of the tngent line: x = x(t ) + x (t )t x = x(t ) + x (t )t y = y(t ) + y (t )t y = y(t ) + y (t )t z = z(t ) + z (t )t 1
2 Recll tht the length L of prmetric curve x = x(t), y = y(t) with continuous derivtives on n intervl t b cn be obtined by integrting the length element ds from to b. L = The length element ds on sufficiently smll intervl cn be pproximted by the hypotenuse of tringle with sides dx nd dy nd so ds = dx + dy ds = dx + dy = (x (t)) + (y (t)) dt. Anlogously, the length element of spce curve stisfies ds = dx + dy + dz nd so ds = dx + dy + dz = (x (t)) + (y (t)) + (z (t)) dt. The expression (x (t)) + (y (t)) + (z (t)) represents the length of the derivtive vector r = x, y, z. Thus the length of spce curve on the intervl t b cn be found s L = ds = r (t) dt = (x (t)) + (y (t)) + (z (t)) dt. ds. Compre gin the two nd three dimensionl formuls. The length of the curve r = x(t), y(t) for t b is L = b (x (t)) + (y (t)) dt = r (t) dt The length of the curve r = x(t), y(t), z(t) for t b is L = b (x (t)) + (y (t)) + (z (t)) dt = r (t) dt Prctice Problems. 1. Describe the following curves. For those without prmetric representtion, find equtions of prmetric equtions. () The curve given by x = 1 + t, y = t, z = 1 + t. (b) The line segment from (1,, 4) to (3,, 1). (c) The curve given by x = cos t, y = sin t, z =. (d) The curve given by x = cos t, y = sin t, z = t. (e) The curve in the intersection of the cylinder x + y = 1 with the plne y + z =. (f) The tringle in the boundry of the prt of the plne 3x + y + z = 6 in the first octnt. (g) The boundry of the prt of the prboloid z = 4 x y in the first octnt.
3 . For the following curves, find n eqution of the tngent line t the point where t =. Find the normliztion of direction vector t t =. () The curve given by x = cos t, y = sin t, z = t. (b) The curve in the intersection of the cylinder x + y = 1 with the plne y + z =. Use the prmetriztion of this curve from problem 1 (e). 3. For the following curves, find the length for t π. Use the clcultor to evlute the integrl in prt (b). () The curve given by x = cos t, y = sin t, z = t. (b) The curve in the intersection of the cylinder x + y = 1 with the plne y + z =. Use the prmetriztion of this curve from problem 1 (e). 4. Consider the curve C which is the intersection of the surfces x + y = 9 nd z = 1 y. () Find the prmetric equtions tht represent the curve C. (b) Find the eqution of the tngent line to the curve C t point (, 3, 8). (c) Find the length of the curve from (3,, 1) to (, 3, 8). You cn use the clcultor to evlute the integrl tht you re going to get. 5. Consider the curve C which is the intersection of the surfces y + z = 16 nd x = 8 y z. () Find the prmetric equtions tht represent the curve C. (b) Find the eqution of the tngent line to the curve C t point ( 8, 4, ). (c) Find the length of the curve from (4,, 4) to ( 8, 4, ). Use the clcultor to evlute the integrl tht you re going to get. 6. Find the length of the boundry of the prt of the prboloid z = 4 x y in the first octnt. Solutions. 1. () This curve is line pssing the point (1,, 1) in the direction of 1,,. (b) Any vector coliner with 3,, 1 1,, 4 =,, 5 cn be used s direction vector of the line pssing two points. You cn lso use ny of (1,,4) nd (3,, 1) for point on the line. For exmple, using (1,,4) we obtin prmetric equtions x = 1 + t, y = tz = 4 + 5t. In this cse, the initil point corresponds to t = nd the end point to t = 1. So, the line segment hs prmetriztion x = 1 + t, y = tz = 4 + 5t with t 1. (c) The xyequtions x = cos t, y = sin t, represent circle of rdius 1 in xyplne. Thus, the curve is on the cylinder determined by this circle. The zeqution z = represents the horizontl plne pssing on the zxis. So, this curve is the intersection of the cylinder with 3
4 the horizontl plne: it is circle of rdius 1 centered on the zxis in the horizontl plne pssing z =. (d) The xyequtions x = cos t, y = sin t, represent circle of rdius 1 in xyplne. Thus, the curve is on the cylinder determined by this circle. The eqution z = t hs n effect tht zvlues increse s tvlues increse. Thus, this curve is helix spirling up the cylinder s t increses. Use Mtlb to get precise grph. (e) The curve is the intersection of cylinder with n inclined plne. So, the curve is n ellipse. The xyequtions represent cylinder bsed t the circle of rdius 1 in xyplne. Thus, x nd y cn be prmetrized s x = cos t, y = sin t. To get the z eqution, solve the plne eqution y + z = for z, get z = y nd substitute tht y = sin t. Thus z = sin t. This gives us n eqution of the ellipse to be x = cos t, y = sin t, z = sin t. You cn use Mtlb to get precise grph. 1 (c) Circle in horizontl plne 1 (d) Helix 1 (e) Ellipse (f) The tringle in the boundry of the prt of the plne 3x+y+z = 6 in the first octnt consists of the three line segments, ech of which will hve different set of prmetric equtions. The intersection in xyplne z = cn be obtined by plugging z = in 3x + y + z = 6 nd using x, for exmple, s prmeter. Thus, we hve prt of the line 3x + y = 6 y = 3 3 x between its two intercepts (,) nd (,3) nd so x x = x, y = 3 3 x, z = or x = t, y = 3 3 t, z = with t. Alterntively, the prmetric equtions of this line cn be obtined s equtions of line pssing x nd y intercepts of the plne 3x+y +z = 6, (,, ) nd (,3, ). Using (,3,) s direction vector nd (,3,) s point on the line, we obtin the equtions x = t, y = 3 + 3t, z = with 1 t. Similrly, you cn find equtions of the remining two sides of the tringle. The intersection of xzplne y = cn be obtined by plugging y = in 3x + y + z = 6 nd using x, for exmple, s prmeter. Thus, we hve 3x + z = 6 z = 6 3x nd so x = x, y =, z = 6 3x or x = t, y =, z = 6 3t with t. The intersection of yzplne x = cn be obtined by plugging x = in 3x + y + z = 6 nd using y, for exmple, s prmeter. Thus, we hve y + z = 6 z = 6 y nd so x =, y = y, z = 6 y or x =, y = t, z = 6 t with t 3. (g) The boundry of the prt of the prboloid z = 4 x y in the first octnt consists of three curves, ech of which will hve different set of prmetric equtions. The prmetriztions 4
5 cn be obtined by considering intersections with three coordinte plnes x =, y =, nd z = respectively. The intersection in xyplne z = is circle = 4 x y x +y = 4 which hs prmetric equtions x = cos t, y = sin t. Since we re considering just the prt with x nd y, we hve tht t π. Thus, this curve hs prmetric equtions x = cos t, y = sin t, z = with t π. The intersection in xzplne y = is prbol z = 4 x with x. Using x s prmeter produces prmetric equtions x = x, y =, z = 4 x or x = t, y =, z = 4 t with t. The intersection in yzplne x = is prbol z = 4 y with y. Using y s prmeter produces prmetric equtions x =, y = y, z = 4 y or x =, y = t, z = 4 t with t.. () To find point on the tngent, plug t = in the prmetric equtions of the curve. Get x = 1, y =, z =. To get direction vector, plug t = in the derivtive x = sin t, y = cos t, z = 1. Get, 1, 1. So, the eqution of the tngent is x = 1 + t, y = + 1t, z = + 1t x = 1, y = t, z = t. The direction vector, 1, 1 hs length so its normliztion is, 1, 1. (b) Use the prmetric equtions from problem 1 (d) x = cos t, y = sin t, z = sin t. To find point on the tngent, plug t = in the prmetric equtions. Get x = 1, y =, z =. To get direction vector, plug t = in the derivtive x = sin t, y = cos t, z = cos t. Get, 1, 1. So, the eqution of the tngent is x = 1 + t, y = + 1t, z = 1t x = 1, y = t, z = t. The direction vector, 1, 1 hs length so its normliztion is, 1, () L = π/ (x (t)) + (y (t)) + (z (t)) dt = π/ π/ π/ 1 + 1dt = dt = π. (b) L = π/ the clcultor, L (x (t)) + (y (t)) + (z (t)) dt = π/ ( sin t) + (cos t) + (1) dt = ( sin t) + (cos t) + ( cos t) dt. Using 4. () You cn prmetrize the cylinder x + y = 9 by x = 3 cos t nd y = 3 sin t. From the eqution z = 1 y, you obtin tht z = 1 (3 sin t) = 1 9 sin t. (b) To find direction vector, we need to plug tvlue tht corresponds to the point (, 3, 8) into the derivtive x = 3 sin t, y = 3 cos t, z = 18 sin t cos t. To find this tvlue, set x =, y = 3 nd z = 8 nd mke sure tht you find the tvlue tht stisfies ll three equtions. From the first eqution x = 3 cos t = t = ± π. From the second y = 3 sin t = 3 t = π. The vlue t = π stisfies the third eqution z = 1 9 sin π = 1 9 = 8. Thus, t = π. Plugging this vlue in the derivtives produces the direction vector 3,,. So, the tngent line is x = 3t, y = 3 + t, z = 8 + t x = 3t, y = 3, z = 8. 5
6 (c) From prt (b), we hve tht t = π corresponds to the point (, 3, 8). Thus, π is the upper bound. To find the lower bound, determine the tvlue tht corresponds to (3,, 1). Set x = 3, y = nd z = 1 nd mke sure tht you find the tvlue tht stisfies ll three equtions. From the first eqution x = 3 cos t = 3 cos t = 1 t =. From the second y = 3 sin t = sin t = t =. The vlue t = stisfies the third eqution z = 1 9 sin = 1 = 1. So, the bounds of integrtion re to π. The length is L = π/ ( 3 sin t) + (3 cos t) + ( 18 sin t cos t) dt = () You cn prmetrize the cylinder y + z = 16 by y = 4 cos t nd z = 4 sin t. From the eqution x = 8 y z, you obtin tht x = 8 (4 cos t) 4 sin t = 8 16 cos t 4 sin t. (b) To find direction vector, we need to plug tvlue tht corresponds to the point ( 8, 4, ) into the derivtive x = 3 cos t sin t 4 cos t, y = 4 sin t, z = 4 cos t. To find this tvlue, set x = 8, y = 4 nd z = nd mke sure tht you find the tvlue tht stisfies ll three equtions. From the second eqution y = 4 cos t = 4 cos t = 1 t = π. From the third z = 4 sin t = sin t = t = nd t = π. The vlue t = π grees with the tvlue we obtined using the yeqution. Plugging this vlue in the xeqution gives you x = 8 16 cos π 4 sin π = 8 16 = 8 which grees with the xcoordinte of ( 8, 4, ). Thus, t = π. Plugging this vlue in the derivtives produces the direction vector 4,, 4. So, the tngent line is x = 8 + 4t, y = 4 + t, z = 4t x = 8 + 4t, y = 4, z = 4t. (c) From prt (b), we hve tht t = π corresponds to the point ( 8, 4, ). Thus, π is the upper bound. To find the lower bound, determine the tvlue tht corresponds to (4,, 4). Set x = 4, y = nd z = 4 nd mke sure tht you find the tvlue tht stisfies ll three equtions. From the second eqution y = 4 cos t = cos t = t = ± π. From the third z = 4 sin t = 4 sin t = 1 t = π. So the vlue π obtined from the yeqution cn be discrded nd we obtin tht t = π. Plugging this vlue in the xeqution gives you x = 8 16 cos π 4 sin π = 8 4 = 4 which grees with the xcoordinte of (4,, 4). Thus, the lower bound is t = π. The length is L = π π/ (3 cos t sin t 4 cos t) + ( 4 sin t) + (4 cos t) dt = Recll tht we found prmetric equtions of the three curves in the intersection to be x = cos t, y = sin t, z = with t π, x = t, y =, z = 4 t with t, nd x =, y = t, z = 4 t with t. The three derivtive vectors nd length elements re x = sin t, y = cos t, z = x = 1, y =, z = t x =, y = 1, z = t ds = 4 sin t + 4 cos tdt = 4dt = dt ds = 1 + 4t dt, nd ds = 1 + 4t dt. The totl length cn be clculted s sum of the three integrls below. Using the clcultor for the second two produces π/ dt t dt t dt = π
Final Exam Solutions, MAC 3474 Calculus 3 Honors, Fall 2018
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