MATH 51TA Section Notes for Thu 06 Nov 08Jason LoToday’s topics:• partial derivatives vs continuity• more on tangent plane• higher-order partial derivatives• (Newton’s method)• visualising multi-variable functions1 Partial derivatives vs ContinuityContinuing the discussion from last section (Tuesday), just to make sure thatwe are all on the same page regarding the relations between partial derivativesand continuity for a multi-variable function, let us write down what we know(and hopefully resolve lingering confusions once and for all). Given a functionf : X ⊂ Rn→ Rm, we know:• Theorem 3.10. If all partial derivatives ∂fi∂xjexist and are continuousin a neighbourhood of a in X, then f is differentiable at a• Theorem 3.9. If f is differentiable at a, then it is continuous at a.As a consequence, if all partial derivatives of f exist and are continuous in aneighbbourhood of a, then f would be continuous at a.For our function from last time,f(x, y) =(x3yx6+y2if (x, y) 6= (0, 0)0 if (x, y) = (0, 0).both partial derivatives fx, fyexist everywhere. Unfortunately, they are notcontinuous at the origin (you can check this), and so the function f(x, y) itselfis not continuous at the origin.Note that this is different from the single-variable scenario: for a singlevariable function f : X ⊂ R → R, as soon as the derivative f0(x) exists at apoint a ∈ X, f would be continuous at x = a.The function that I drew on the board and erased last time,g(x) =(x2if x 6= 01 if x = 01is not contiuous at the origin, and so its derivative cannot exist at the origin.We can see this directly:limh→0+g(0 + h) − g(0)h= limh→0+h2− 1h= −∞It was also asked last time: what do functions that are non-differentiable,but have well-defined partial derivatives, look like? Figure 2.50 on p.112 is thepicture of one such function. The function f above is another such function,but you would’ve had a hard time trying to visualise it using level curves!The slides give you an idea of what the function looks like. Note the ”non-smooth” behavior of the function near the origin. We can also tell from theslides that the partial derivatives are both equal to 0 at the origin.Matlab commands I used:syms x yf = (x3∗ y)/(x6+ y2)ezsurf(f,[-4,4,-4,4])2 More on Partial DerivativesExample 1. Use the tangent plane to approximate:f(0.9, 0.1), where f(x, y) = 3 + cos πxy,Solution. The partial derivatives of f exist and are continuous everywhere,and so f is differentiable on all of R2. So near any point on R2, we can approx-imate the function by the tangent plane at that point. Since (0.9, 0.1) is near(1, 0), we can approximate f(0.99, 0.01) using the tangent plane at (1, 0).∂f /∂x = πy cos πxy∂f /∂y = πx cos πxyand∂f /∂x(1, 0) = 0∂f /∂y(1, 0) = π cos 0 = πThe tangent plane at the point is thus given byz − f(1, 0) = 0 · (x − 1) + π · (y − 0) ⇔ z − 4 = πy(where z approximates f (x, y) near (1, 0)).So f(0.99, 0.01) ≈ π ·(0.01) + 4 = 4.031415.... (Compare this with the actualvalue f(0.99, 0.01) = 3.999....)2Example 2 (Ex 2.4.11). Determine all second-order partial derivatives off(x, y) = ey / x− ye−xWhat is the domain? Note that the two mixed partials agree on the domain off (on which f is continuous).3 Newon’s MethodEquation (6) on p.131.4 Visualising a Multi-Variable FunctionExample 3. Consider the function f(x, y) :=1x2+y2, whose domain is R2\{(0, 0}. We will try to visualise the behaviour of this function using level curves.So consider f(x, y) = c, where 1, 4, 9. Would it make sense if we put c = 0or c < 0?• When c = 1, we have x2+ y2= 1, so we get a circle of radius 1 centred atthe origin.• When c = 4, we have x2+ y2= (12)2, so we get a circle of radius 1/2centred at the origin.• When c = 9, we have x2+y2= (13)2, so we get a circl of radius 1/3 centredat the origin.So the graph of the function is like a spike reaching out to infinity at the origin,and extending towards 0 away from the origin.5 Extra
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