#$&* course Mth 277 The upside-down Delta symbol is called 'del', and will by typed accordingly. This symbol stands for the 'del' operator, which takes the partial derivative with respect to each variable, and multiplies it by the unit vector in the direction of that variable.Thus for example if we have f(x, y), a function of the 2 variables x and y:
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Given Solution: del f = f_x i + f_y j = 2 x y^3 i + 3 x^2 y^2 j &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `q002. Let f(x, y, z) = x^2 + x cos y + e^(yz). Find del f. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: As stated above, del f=f_xi+f_yj+f_zk delf=(2x+cos(y))i+(-xsin(y)+ze^(yz))j+(ye^(yz))k confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: del f = f_x i + f_y j + f_z k = ( 2 x + 3 cos(y) ) i + (-x sin(y) + z e^(yz) ) j + y e^(yz) k. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `q003. Find the directional derivative of f(x, y) = x^2 y^3, at the point (2, 1) in the direction of the unit vector u = .6 i + .8 j. If you move .01 unit in the direction of that vector by how much does your directional derivative indicate that the value of f(x, y) will change? What will be your x and y coordinates after the move? What is the value of f(x, y) at the point (2, 1)? What is the value of f(x, y) at the new point? Did the function change by the amount you predicted? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: f(x, y) = x^2 y^3 del f=f_xi+f_yj+f_zk del f=(2xy^3)i+(3x^2)(y^2)j del f at (2,1)=4i+12j u=.6i+.8k del f dot u=2.4+9.6=12 It states we move .01 in direction of u, therefore; 12(.01)=.12 f(2,1)=4*1=4 u=.6i+.8k .6(.01)i+.8(.01)j .006i+.008j To find new point we do 2+.006,1+.008=(2.006,1.008) f(2.006,1.008)=4.121 change of .121 when we predicted .12. Therefore, the function changed by the amount we predicted to a good degree. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The gradient is del f = 2 x y^3 i + 3 x^2 y^2 j At (2, 1) the gradient takes value 4 i + 12 j. The directional derivative in the direction of the unit vector u is (del f) dot (u) = 1.2 x y^3 + 2.4 x^2 y^2. At (2, 1) the directional derivative takes value 12. Had we take the dot product of the gradient 4 i + 12 j at the point (2, 1), with the unit vector u, we would have obtained the same result, 2.4 + 9.6 = 12. Thus when we move in the direction of u, the rate of change of f(x, y) with displacement is 12. That is, in the near vicinity of (2, 1), the value of f(x, y) increases at or close to the rate of 12 units for every unit of distance in the direction of u. Thus if we move distance .01 in the direction of u we expect f(x, y) to change by 12 * .01 = .12. At (1, 2) the value of f(x, y) is easily found to be 4. If we move .01 unit in the direction of u, we easily see that our position changes by .01 u = .006 i + .008 j, so our x coordinate changes by .06 and our y coordinate by .08. Starting from point (1, 2) we therefore end up at point (2.006, 1.008). Evaluating f(x, y) at (2.006, 1.008) we obtain value (2.006)^2 * (1.008)^3 = 4.1214. We conclude that as a result of moving .01 unit in the direction of u, the function changes its value from 4 to 4.1214, a change of .1214. The change predicted by the differential is .12, which differs from the actual change by about a 1% of that change. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `q004. The ellipse x^2 / 4 + y^2 / 9 = 1 can be expressed as the curve f(x, y) = 1, with f(x, y) = x^2 / 4 + y^2 / 9. The point (1, 3 sqrt(3) / 2) lies on the curve. Find a vector in the direction normal to the curve at that point. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: f(x, y) = x^2 / 4 + y^2 / 9 (1, 3 sqrt(3) / 2) del f=f_xi+f_yj=(x/2)i+(2y/9)j .5i+(sqrt(3)/3)j The vector normal to the curve at that point is .5i+.577j confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The gradient of f(x, y), evaluated at a point, gives the direction in which f(x, y) is increasing most rapidly in the vicinity of that point. The level curves of f(x, y), i.e., the curves f(x, y) = c for constant c, have the property that for any point on a given level curve, a vector tangent to the curve lies in the direction in which the value of the function remains unchanging. The direction perpendicular to the direction of maximum increase is the direction in which the value of the function remains constant. So the tangent vector at a point is perpendicular to the gradient at that point. It follows that the gradient is normal to the level curve. The gradient of f(x, y) is del f = f_x i + f_y j = x/2 i + 2 y / 9 j. At the point (1, 3 sqrt(3) / 2) the gradient is 1/2 i + sqrt(3) / 3 j , approximately .5 i + .57 j, This vector is therefore normal to the level curve f(x, y) = x^2 / 4 + y^2 / 9 = 1, i.e., is normal to the ellipse, at the given point. The figure below depicts this ellipse in the first quadrant, with the lines x = 1 and y = sqrt(3) / 2 intersecting at the given point. The line passing through this point with slope (y - 3 sqrt(3)/2) / (x - 1) = 2 sqrt(3)/3. This slope matches that of the vector 1/2 i + sqrt(3) / 3 j , so this line is perpendicular to the curve at the given point. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: OK " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: OK " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!