Lecture 11 MATH 2401 - Harrell Copyright 2008 by Evans M. Harrell II. A very gradifying lectureScenes from our previous episode Tangent and normal vectors, Tangent and normal linesTips for the test – don’t lose points for trivial reasons! Show some work – There will be no partial credit for what isn’t shown. Make sure the grader sees the key facts/formulae somewhere Check your algebra/calculations. Common sense. An answer has to be the “right kind of animal.” Put the answer where indicated. Like…..intersecting surfaces. The intersection of two surfaces is usually a curve. How is it oriented – i.e., what is its tangent direction? Monday was fun, so let’s do some new (even more fiendishly intricate) geometric problemsIntersection of a plane and a cone Tangent to the curve at the intersection of 2 surfaces, such as a plane and a cone – the classic conic sections.But what about... The tangent to the curve where x = sin(π y z) and z = 2 x2 – 4 y2 + ¼ as it passes through (1, ¼, 2) ?The chain rule(s) (d/dt) f(r(t)) = ∇f(r(t)) • r′(t) Just like 1-D In components: df/dt = (∂f/∂x) dx/dt + (∂f/∂y) dy/dt + (∂ f/∂z) dz/dt + … Examples?The chain rule(s) Suppose the temperature in a plate is T(x,y) = 4 x2 – 2 x y – 4 y2, and that an object moves in a circle, r(t) = 2 cos(t) i + 2 sin(t) j . At what rate is the temperature changing?The chain rule(s) What about u(x,y), where x and y depend on s and t? For example, a change of variables.The chain rule(s) What about u(x,y), where x and y depend on s and t? (∂u/∂s) = (∂u/∂x)(∂x/∂s) + (∂u/∂y)(∂y/∂s) Remember: Add up all the possible routes for connecting u to the independent variable.Word problems Suppose that price of your widgets is P(t), you are selling at a rate of R(t) per month, and your expenses are F(t) + c(t) R(t) How rapidly is your profit changing, if P = 2, R = 3000, F = 2500, c = 1, P′(t) = .1, R′(t) = -20, c(t) = .05, and F′(t) = 5 ?Word problems Suppose that price of your widgets is P(t), you are selling at a rate of R(t) per month, and your expenses are F(t) + c(t) R(t) Profit = PR - F - c R d Profit/dt = R P′ + (P - c) R′ - F′ - c′R If f is differentiable on the line segment ab, there exists a position c so that f(b) - f(a) = ∇f(c)•(b - a) Why is this not as great as in 1D? A: the gradient doesn’t have to point in the direction (b - a), and |∇f(c)| isn’t the average slope |f(b) - f(a)|/|b - a|. Mean value theorem If f is differentiable on the line segment ab, there exists a position c so that f(b) - f(a) = ∇f(c)•(b - a) Example: (0,0) to (1,0), f(x,y) = x+y. Mean value theoremGradient determines f up to a constant Let U be open and connected, and f and g be differentiable on U. If ∇f = ∇g on U, then f(x) = g(x) + C. How would you prove this? Good enough to replace f by f-g and show that is ∇f = 0 on U, then f is a constant. Connect any two points by “polygonal path” and use M.V. Theorem.Which = 0.Gradient determines f up to a constant Let U be open and connected, and f and g be differentiable on U. If ∇f = ∇f on U, then f(x) = g(x) + C. How about an example? Arctan(y/x) vs.
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