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What is the physical interpretation of gradient of a scalar field and directional derivative?
The elevation of points on a hill is a scalar 'field'. It can have a different value at every point, but each one is a scalar value. Imagine a lumpy bumpy irregular hill, and pick a point to talk about, say, somewhere on the side of the hill. At that point, the directional derivative of the elevation is the rate at which the elevation changes leaving the point in that direction. It has different values in different directions: If you're looking up the hill, then the d.d. is positive in that direction; if you're looking down the hill, the d.d. is negative in that direction. If you're looking along the side of the hill, the d.d. could be zero, because the elevation doesn't change in that particular direction. The directional derivative is a vector. The direction is whatever direction you're talking about, and the magnitude is the rate of change in that direction. The gradient is the vector that's simply the greatest positive directional derivative at that point. Its direction is the direction of the steepest rise, and its magnitude is the rate of rise in that direction. If your hill is, say, a perfect cone, and you're on the side, then the gradient is the vector from you straight toward the top, with magnitude equal to the slope of the side of the cone. Any other vector is a directional derivative, with a smaller slope, and it isn't the gradient.
What is the height of the hill if the elevation of a hill at a place due east of it is 39 degrees at a place B due south of A the elevation is 27 degrees and the distance from A to B is 500m?
It is not clear from the question where the point A is. This answer assumes that A is the point to the East of the hill. Suppose the height of the hill is h metres. Suppose O is the point, on ground level, under the apex of the hill. Then tan(39) = h/OA so that OA = h/tan(39) Also tan(27) = h/OB so that OB = h/tan(27) Now, triangle OAB has a right angle at A, so by Pythagoras, OB2 = OA2 + AB2 Therefore h2/tan2(27) = h2/tan2(39) + 5002 that is h2/0.2596 = h2/0.6558 + 250000 so that 2.3269h2 = 250000 h2 = 107441 h = 327.8 metres.
What does it mean for a line to have a negative slope?
If a line has a negative slope it is going 'down hill' and if it has a positive slope it is going 'up hill'
what- A paved road runs down a hill as shown, with endpoints (0, 8) and (8, 0). What is the slope of the hill?
-1
What is the speed of a biker that bikes up a 40 m hill for 10 seconds?
A 40m hill IN 10sec? 144 km/h or 89mph.
