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2220hw3sol

# 2220hw3sol - Math 2220 Section 2.6 Problem Set 3 Solutions...

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Math 2220 Problem Set 3 Solutions Spring 2010 Section 2.6 : 4. Calculate the directional derivative of the function f ( x, y ) = 1 / ( x 2 + y 2 ) in the direction of the vector u = i j = (1 , 1) at the point (3 , 2). Solution. First normalize u to get a unit vector 1 2 (1 , 1). Next, we compute f = 1 ( x 2 + y 2 ) 2 ( 2 x, 2 y ) so f (3 , 2) = 1 13 2 ( 6 , 4). The directional derivative is then the dot product 1 13 2 ( 6 , 4) · 1 2 (1 , 1) = 10 13 2 2 . 6. Calculate the directional derivative of the function f ( x, y, z ) = xyz in the direction of the vector u = 2 k i 5 = 1 5 ( 1 , 0 , 2) at the point ( 1 , 0 , 2). Solution. The vector u is already a unit vector. We have f = ( yz, xz, xy ) so at ( 1 , 0 , 2) we have f = (0 , 2 , 0). The directional derivative is then (0 , 2 , 0) · ( - 1 5 , 0 , 2 5 ) = 0. 10. For the function f ( x, y ) = xy radicalbig x 2 + y 2 if ( x, y ) negationslash = (0 , 0) 0 if ( x, y ) = (0 , 0) (a) calculate f x (0 , 0) and f y (0 , 0) Solution. We have f ( x, 0) = 0 for all x (including x = 0) so f x ( x, 0) = 0 for all x , and in particular f x (0 , 0) = 0. Similarly, we have f (0 , y ) = 0 for all y so f y ( y, 0) = 0 for all y and in particular f y (0 , 0) = 0. (b) use Definition 6.1 (the definition of a directional derivative) to determine for which unit vectors v = ( v, w ) = v i + w j the directional derivative D v f (0 , 0) exists. Solution. The directional derivative in the direction v is the derivative with respect to t of the function f ( tv, tw ) = t 2 vw t 2 v 2 + t 2 w 2 . Since v is a unit vector we have v 2 + w 2 = 1 and so t 2 vw t 2 v 2 + t 2 w 2 = t 2 vw t 2 = t 2 vw | t | = | t | vw . If both v and w are nonzero then the function | t | vw is not a differentiable function of t at t = 0. Thus the directional derivative D v f (0 , 0) exists only when v or w is zero, so v = (1 , 0) or (0 , 1). In these two cases we have the directional derivatives f x (0 , 0) and f y (0 , 0) computed in part (a). 14. It is raining and rainwater is running off an ellipsoidal dome with equation 4 x 2 + y 2 + 4 z 2 = 16, where z 0. Given that gravity will cause the raindrops to slide down the dome as rapidly as possible, describe the curves whose paths the raindrops must follow. (Hint: You will need to solve a simple differential equation.) 1

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Math 2220 Problem Set 3 Solutions Spring 2010 Solution. Projecting to the xy -plane, the level curves are obtained by letting z be constant in the equation 4 x 2 + y 2 + 4 z 2 = 16, so the level curves are ellipses 4 x 2 + y 2 = c . We want to find the curves that meet these level curves orthogonally at all points. Thus we want curves y = g ( x ) such that dy dx equals the slope of the gradient vectors of the function f ( x, y ) = 4 x 2 + y 2 . We have f = (8 x, 2 y ), which is a vector of slope 2 y/ 8 x . Thus we want dy dx = 2 y/ 8 x . This can be rewritten as 4 dy y = dx x . Integrating this equation gives 4 ln y = ln x + C , so y 4 = e C x = Ax for arbitrary positive constants A .
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