What is the slope(gradient) of a curve in 3dimensional geometry?
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the answer is pie over 2x plus 7
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I'll help :)\n\nSlope is calulated by taking the change in y over the change in x.  So (78-92)/(5-9)=  -14/-4 = 7/2\n\nThe slope is 7/2
184330
Gradient means slope!\n\nThe equation of a line with a slope of m and a y-intercetpt b is given by y=mX+b\n\nTherefore the equation for your question is y=-5x+6\n\nAlways remember the formula y=mx+b
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I'm pretty sure you are correct.  I am also working thru Calc III (at the moment, actually).  It helps to picture the gradient as perpendicular to the tangent plane.
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This is a Frequently Asked Question. Following is a modified previous answer:\n\n\nIf one has a function f as a map from the real line into itself, one also has the graph of the function as the set of all points (x,f(x)) as a curve in the Cartesian plane. If one wanted to know the area between the x-axis and this curve over some interval on the x-axis, one would use a definite integral. If one wanted to know the slope of a line tangent to the curve at some point (x,f(x)) on the curve, one would use the derivative of the function f at the point x. Both of these operators are very useful in physics, as most physical properties are defined as changes in some property. If p(x) is a graph of some physical property over the domain it is a function of, then the change in p with respect to x approaches the slope of the tangent line to p(x) as one considers smaller and smaller changes in x. Thus, the derivative is the ideal way of talking about continuous changes in physical properties.\n\nFor differentiation in one dimension, you can use the definition: the derivative of a function f at the (domain) point x is the limit as h approaches 0 of the slope of the secant line [f(x+h)-f(x)]/h. Visually, we take a secant line to the curve that passes through the points (x, f(x)) and (x+h, f(x+h)) and rotate it (see what happens as h gets smaller) until it only touches the curve once (it is now tangent to the curve). This line is unique at each point.\nIn more than one dimension, using the definition is cumbersome for anything other than proving when a derivative does or does not exist, so we use a useful theorem that says the derivative of a function f from R (the real line) into R^n (n-dimensional Euclidean space) is just the vector of partial derivatives of f. Ie., if f(x,y) = x^2 + y^2 (the graph is a circular paraboloid), then Df(x,y) = (df/dx, df/dy) = (2x, 2y). Partial derivatives are taken relative to one variable only; they end up treating the other variables as constants. The technical definition of the partial of f with respect to x at the point x0 is what you suspect it would be: the limit as h approaches zero of the slope of the secant line [f(x0+h, y) - f(x0,y)]/h.\nOnce one gets to these derivatives of multivariable functions though, one must be very careful, as the existence of all partial derivatives at a point does not mean the derivative exists at that point.\nOnce one gets a bit of experience taking derivatives using the definition, one can derive some simple derivative rules which allow you to take derivatives of elementary functions without explicitly doing all that limit mumbo-jumbo. A list of common differentiation rules can be found here: http://mathworld.wolfram.com/Derivative.html .\n\nIntegration is a more delicate matter. If you're doing a definite Riemann integral (finding the area under a curve), it is the limit of a Riemann sum as you increase the number of rectangles under the curve ( See http://mathworld.wolfram.com/RiemannSum.html ). The indefinite integral is just the function that will return the definite integral when you take the difference between two points.\nSometimes the rectangles will not converge to any finite value or it does not have a unique convergence and we say the function is not integrable on that interval (Take f(x) = 1/x on the interval [0,1] for example). A more technical definition of integration will help integrate more pathological functions, but you can look up this definition yourself in a real analysis text; it's not that different from summing rectangles.\nAgain, with integrals we have a very nice theorem that tells us that the indefinite integral of the derivative of a function f is f itself. This is the fundamental theorem of calculus, and it takes a lot of work out of finding an integral. For example, if f(x) = 2x, we know from our experience with taking derivatives that the indefinite integral of f(x) is just F(x) = x^2 + C where C is an unknown constant since the deriv
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First, look at where the curve intersects the line:\n\ncurve y=1/x\nline y=x/4\n\n1/x=x/4\nx=+/- 2\n\nNote that the curve is symmetrical about the origin for x&y positive and x&y negative, so the line intersects twice.\n\nThat's actually somewhat irrelevant, but it helped me visualize the line and the curve.\n\nTake the first derivative of the curve\ndy/dx=-1/x^2\nThis is the instantaneous slope of the curve at any point, which is the tangent by formal definition.\n\nA line perpendicular to the line y=.25x will have a slope of -4.\nSo set the derivative equal to -4 and solve for x\n-4=-1/x^2\nx=+/- 1/2\nsolve for y\nand find the points\n(1/2,2)\n(-1/2,-2)\nThis is where the two lines that are both tangent to the curve and perpendicular to the line intersect the line.\n\none line is y=-4x+4\nthe other is y=-4x -4\n\nj
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The slope is zero, it is a horizontal line.  The equation is y=-3.\n\ny=mx+b is the slope intercept form where m is the slope and b in the place where the line crosses the y axis.\n\nSlope is rise over run. The is the difference in y's over the distance in x's.
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On a graph, a line can have a slope. The slope is defined as how many spaces the line moves up or down divided by how many spaces the line moves left or right from any two given points. You may see it written as\n\nrise\n----\nrun\n\nOR\n\nchange in Y\n-----------\nchange in X\n\nOR\n\n{for the line through points (X1, Y1) and (X2, Y2)}\n\nY1-Y2\n-----\nX1-X2\n\n\nA line from upper right to lower left will have a positive slope.\nFrom upper left to lower right will have a negative slope.\nStraight up and down will have an undefined slope.\nLeft to right will have a slope of 0.\n\nHope this helped! Good luck!
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You draw it under the curve\n\nThe tangent line will only touch the curve at one point for this type of smooth upward bending curve.(At least for the part of the graph shown).\n\nYour graph plots total displacement verses time. If the graph was a straight line then this would indicate a constant "velocity" but since the curve bends upwards you are dealing with a positive acceleration.\n\nPractically, from your graph, suppose you wanted to know the approximate slope of the tangent line at t=5. Use the two closest data points: t=4, d=8.4 and t=6, d=16.5. Then the slope is about (16.5-8.4)/(6-4) = 4.05 at t=5. it will be slightly greater than this. You can use a straight edge to make a visual adjustment and then you can find the slope by dividing the displacement value at t=5 by (5 minus the value of t where the straight edge crosses the time axis).\n\nThe other way is to use the data to deduce, a closely as possible, the formula of the function being graphed. For example if the displacement was directly proportional to the square of the time, d = k t^2 for a determined value of k, then the slope of the tangent line at any time, t, would simply be 2 k t which is the time derivatiive of the function.
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although Question is little vague, i understand that u want to find slope of a line ???\n\nsay coordinatyes of A and B are (x1,y1) and (x2,y2)\n\nThe Slope of line m= (y1-y2)/(x1-x2)
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equation of slope is (y2-y1)/(x2-x1)\n\n(8,-2) means x1=8 and y1=-2\n(3,5) means x2=3 and y2=5\n\nso slope= (5-(-2))/(3-8)\n=7/-5\n\nso slope= -7/5
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Basically, its a straight line going through (1,3), with gradient, m=-3. we have y=mx+c. putting the point (1,3) into here, \n\n3=(-3)(1)+c, so c=6, which is the y-intercept. so the line equation is,\n\ny=-3x + 6, which is a straight line with y-intercept at y=6. the line will be sloping downwards since the gradient (m) is negative. if you plot a few values of x and y which satisfy the equation you can see the line. to check axis intercepts, put y=0 for the x-intercept and x=0 for the y-intercept.
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You need to differentiate the equation y=2x^2. \n\ndy/dx = 4x therfore the gradient at x=3 = 4*x = 12
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There are two methods of solving this question.  One method is applying using the calculus, and other method is using Coordinate geometry.  Since u have not mentioned a particular method, easier way is applying the Calculus.\n\nLet (x1, y1) be a point on the curve.\nDifferentiating the given equation to find the slope of the tagent at (x1, y1), we get\ndy/dx = 6x -5\nAt (x1,y1) dy/dx = 6x1-5\n\nGive condition is the slope of the normal at this pt is = 2\nSo the slope of the tangent = -1/2\n\nTherefore, 6x1 - 5 = -1/2 which leads to x1 = 3/4\nSubstituting this value in the given equation we get,\ny1 = 3.(3/4)^2 - 5.(3/4)+1 \nThat is y1 = -17/16\n\nAs asked the abscissa is 3/4
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slope = (-1-2)/(-1-0) = -3/-1 = 3
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Slope is defined as increase in y over increase in x,\n\nIncrease in y is 5-2=3\nIncrease in x is -3/4-5/4=-0.5\nSo slope is 3/(-0.5)=-3/2
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Slope = 1/4\n\nSlope is the "change in y" divided by the "change in x" (AKA "rise" over "run").\n\n(x1, y1) = (1, 3)\n(x2, y2) = (5, 4)\n\nThus:\n\nm (slope) = "delta y"/"delta x"\nm = (y2 - y1)/(x2 - x1)\nm = (4 - 3)/(5 - 1)\nm = 1/4