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Unformatted text preview: t the point is differentiable at , Theorem 13.3 applies. Again, we must ﬁrst normalize: 24 WATERLOO SOS E XAM -AID: MATH237 We have so that E XERCISE 25. Deﬁne the direction . by . Find the directional derivative of at the point in E XERCISE 26. Deﬁne by the point in the direction . Find the directional derivative of . at 14 Gradient Vector Revisited (7.2-7.3) Since there are inﬁnitely many possible directions from a point , there are in general inﬁnitely many . We may want to ﬁnd some special information about particular values for the directional derivative directional derivatives; in particular, we would like to know in which direction is the directional derivative maximized. The following theorem allows us to answer this in certain cases. T HEOREM 14.1. If is differentiable at and occurs when is the direction of . and , then the largest value of is , More generally, with the hypothesis of Theorem 14.1 we have so we can potentially determine directions in which the directional derivative is any proportion of the maximum value (between and ). E XAMPLE 29. Deﬁne by , and the direction in which it occurs. S OLUTION. Recall example 28. Notice is differentiable at . Find the largest rate of change of at the point and Then Theorem 14.1 applies: the largest rate of change of is and the direction in which it occurs is 25 . WATERLOO SOS E XAM -AID: MATH237 E XERCISE 27. Deﬁne by , and the direction in which it occurs. . Find the largest rate of change of at the point , E XERCISE 28. Deﬁne by . Find the largest rate of change of at the point and the direction in which it occurs. Then determine the direction in which the directional derivative is half its maximum value. The gradient vector will help with our graph sketching as well, by giving us information about the level curves. T HEOREM 14.2. If through . is differentiable at and , then is orthogonal to the level surface E XERCISE 29. Suppose we deﬁne ( if undeﬁned there) for integers . What is the relationship between such that the level curves of and intersect orthogonally everywhere except possibly Recall we deﬁned the tangent plane to a surface given by at the point required ? to be the plane However, many interesting surfaces are not immediately in the form . For example, we can look at the sphere . Rather than attempt to break up this function into multiple parts, Theorem 14.2 gives us a tool to ﬁnd the tangent plane very easily, as long as the conditions are satisﬁed. Suppose we have a surface in deﬁned by . For any in the tangent plane to the surface at , lies in the tangent plane so by Theorem 14.2 is orthogonal to . This gives where . Notice this reduces to our original formula when . E XERCISE 30. Find the equation of the tangent plane to the hyperboloid . at the point 15 Taylor Polynomials (8.1)
of at is given by D EFINITION 15.1. The second degree Taylor polynomial 26 Generally speaking, approximates than the linear approximation. for WATERLOO SOS E XAM -AID: MATH237 sufﬁciently close to , with better accuracy E XAMPLE 30. Calculate approximately (Use polynomial. Compare with the calculator value. S OLUTION. Compare with example 16. Let gives us and ) using the second degree Taylor . Differentiating Then We get the approximation The calculator gives us a value of about proximation, the error was about ). , for an error of about (when we used linear ap- E XERCISE 31. Calculate the following approximately (compare with the calculator value) using the second degree Taylor polynomial. Compare your results with those of exercise 9. (a) (b) . at . 16 Taylor’s Theorem (8.2) We have seen an example in which the error in approximating using the second degree Taylor polynomial is smaller tha...
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