Question

7. (14,5) A particular parametric curve is given by x=1-3, y = 21 +3 for -2 515 3. Sketch the curve using arrows to indicate the direction in which the curve is traced as increases. Then eliminate the parameter / to find a Cartesian equation of the curve. 8. (10,4) Find the equation of the plane through the point (-5, 4, 2) and with normal vector-31 +4j - k. Give your answer in both the vector equation of a plane and in the scalar equation of a plane. 9. (12,4) f(x,y)=-3x'y + 4y - y sin(xy), find the following partial derivatives: A. f, B. I

Answer 1

7). x= + 2 } y=attz for 24 ts3. y 1 sto al e no é 3 1 7 2 5 0 x = t - 3 y=at+3 Y-3=t Put the value of t in ean (1) we get x = (4-31-3 x+3= 2 y-3 Jit is in cartesian form

h =0 8) The equation of plane in vector form is : Cat-a).ño 04- a). ñ where ă = ~57 + uh tak r = normal vector -37 + 4 J1 A [04 - C-sî+ 47+0 R ).(-39+43-1) = o velou farem Now , in cartesian forum change M= alt yĝ+ ka [[277 y} t k 2) - (-577 454 27 )) (-37 49 7) = 0 (21+5)ỉ + (9-4)ġ tlx-) ) (-3 î+ vyl-F) - - -3(x+5) + 414-4)-(K-2)=0 cartesian form.

g) f(x,y) = -3x y + 4 yo y²sin(ay) я) у = 24 d (-3x²y + 4ys - y²sinny dy he ) (-35'y) - чау- ду'ѕinzy he he he - 37" + 4xy" - ( sinnya, y? t ya asipag -3x420y" - (ay sinay + y cosxy ne] -33* + гоу" - уліт у - у соту he he 7 huroshe - hxushe-groot, KE- 5) dy [an в) 499 = o't дуди ə ə-1 дудя ( Ауруу ну)) - 12 хуно - усотуху 2 -17°у -yeo37 y ) he с he he

- o С-13°у - уз олту) дудя ду - 18x3 - [ y3d.cosa y + coureyri - - уз = -17° - у* (-sinx чxx) + Зусот + xy² sinny +3y² cosxy 25 ду дz -) 273 +

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