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curvature - three dimensional spacein this part we want to briefly discuss the curvature of a smooth curve remind that for a smooth curve we require
optimization in this section we will learn optimization problems in optimization problems we will see for the largest value or the smallest
binormal vector - three dimensional spacenext is the binormal vector the binormal vector is illustrated to bebrarr t trarr t nrarr tsince the
unit normal vector - three dimensional spacethe unit normal vector is illustrated to ben t rarrt t trarr tthe unit normal is orthogonal or normal
demonstrates that f x 4 x5 x3 7 x - 2 has accurately one real rootsolutionfrom basic algebra principles we know that since f x is a 5th degree
the mean value theorem in this section we will discuss the mean value theorem before we going through the mean value theorem we have to cover the
utilizes the second derivative test to classify the critical points of the
tangent normal and binormal vectorsin this part we want to look at an application of derivatives for vector functions in fact there are a couple
smooth curve - three dimensional spacea smooth curve is a curve for which rarrr t is continuous and rarrr t ne 0 for any t except probably at the
determine or find out the domain of the subsequent functionrrarr t cos t ln 4- t radict1solutionthe first component is described for all ts the
domain of a vector functionthere is a vector function of a single variable in r2 and r3 have the formrrarr t f t gtrrarr t f t gt
vector functionswe very firstly saw vector functions back while we were looking at the equation of lines in that section we talked about them as we
level curves or contour curvesanother topic that we should look at is that of level curves or also known as contour curves the level curves of the
quadric surfacesearlier we have looked at lines and planes in three dimensions or r3 and when these are used fairly heavily at times in a calculus
scalar equation of planea little more helpful form of the equations is as follows begin with the first form of the vector equation and write a vector
write down the equation of the line which passes through the points 2 -1 3 and 1 4 -3 write all three forms of the equation of the linesolutionto
definition1 given any x1 amp x2 from an interval i with x1 lt x2 if f x1 lt f x2 then f x is increasing on i2
the shape of a graph part i in the earlier section we saw how to employ the derivative to finds out the absolute minimum amp maximum values of a
equations of lines in this part we need to take a view at the equation of a line in r3 as we saw in the earlier section the equation y mxb does
find out the absolute extrema for the given function and interval g t 2t 3 3t 2 -12t 4 on -4 2solution all we actually need to do here is
finding absolute extrema of fx on ab0 confirm that the function is continuous on the interval ab1 determine all critical points of fx
finding absolute extrema now its time to see our first major application of derivatives specified a continuous function fx on an interval ab we
fermats theorem if f x contain a relative extrema at x c amp f prime c exists then x c is a critical point of f x actually it will be a
three dimensional spacesin this section we will start taking a much more detailed look at 3-d space or r3 this is a major topic for mathematics
provide the vector for each of the followinga the vector from 2 -7 0 - 1 - 3 - 5 b the vector from 1-3-5 - 2 - 7 0c the position vector for - 90