1 / 22

Problem of the Day

Problem of the Day. Let f and g be differentiable functions with the following properties: (i) g(x) > 0 for all x (ii) f (0) = 1. If h(x) = f(x)g(x) and h'(x) = f(x)g'(x), then f(x) =. A) f '(x) C) ex E) 1 B) g(x) D) 0.

netis
Download Presentation

Problem of the Day

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Problem of the Day Let f and g be differentiable functions with the following properties: (i) g(x) > 0 for all x (ii) f(0) = 1 If h(x) = f(x)g(x) and h'(x) = f(x)g'(x), then f(x) = A) f '(x) C) ex E) 1 B) g(x) D) 0

  2. Problem of the Day Let f and g be differentiable functions with the following properties: (i) g(x) > 0 for all x (ii) f(0) = 1 If h(x) = f(x)g(x) and h'(x) = f(x)g'(x), then f(x) = A) f '(x) C) ex E) 1 B) g(x) D) 0

  3. Tangent Line Approximations y = f(x) (a, f(a)) The equation of the tangent line to 
the curve y = f(x) at (a, f(a)) is y - f(a) = f '(a)(x - a) or y = f(a) + f '(a)(x - a) It is called the linear approximation of f at a 
and can be used to approximate values on the 
curve close to a.

  4. Example Find the linearization of the function f(x) = √x + 3 at 
a = 1 and use it to approximate the numbers √3.98 
and √4.05

  5. Example Find the linearization of the function f(x) = √x + 3 at 
a = 1 and use it to approximate the numbers √3.98 
and √4.05 f(x) = (x + 3)½ f '(x) = ½(x + 3)-½ 1) Find f '(x)

  6. Example Find the linearization of the function f(x) = √x + 3 at 
a = 1 and use it to approximate the numbers √3.98 
and √4.05 f(x) = (x + 3)½ f '(x) = ½(x + 3)-½ 1) Find f '(x) f(1) = √1 + 3 = 2 f '(1) = ½(1 + 3)-½ = ¼ 2) Find f(a) & f '(a)

  7. Example Find the linearization of the function f(x) = √x + 3 at 
a = 1 and use it to approximate the numbers √3.98 
and √4.05 f(x) = (x + 3)½ f '(x) = ½(x + 3)-½ 1) Find f '(x) f(1) = √1 + 3 = 2 f '(1) = ½(1 + 3)-½ = ¼ 2) Find f(a) & f '(a) 3) Put in linear approximation form f(x) = f(a) + f '(a)(x - a) f(x) = 2 + ¼(x - 1) = 7/4 + ¼x

  8. Example Find the linearization of the function f(x) = √x + 3 at 
a = 1 and use it to approximate the numbers √3.98 
and √4.05 3) Put in linear approximation form f(x) = 7/4 + ¼x 4) Approximate √3.98 (remember √3.98 is √.98 + 3, so what is x?) √x + 3 = √3.98 ≈7/4 + ¼(.98)≈1.995 Now find √4.05

  9. Example Find the linearization of the function f(x) = √x + 3 at 
a = 1 and use it to approximate the numbers √3.98 
and √4.05 3) Put in linear approximation form f(x) = 7/4 + ¼x 4) Approximate √4.05 (remember √4.05 is √1.05 + 3, so what is x?) √x + 3 = √4.05 ≈7/4 + ¼(1.05) ≈2.0125

  10. (c + Δx, f(c + Δx)) } (c, f(c)) f(c + Δx) { { f(c) c c + Δx Δx

  11. y = f(c) + f '(c)(x - c) (c + Δx, f(c + Δx)) } } Δy } (c, f(c)) f '(c)dx f(c + Δx) { { f(c) c c + Δx Δx

  12. Δy = the amount that the curve rises or falls dy = amount that the tangent line rises or falls as Δx changes, a change is produced in Δy as dx changes, a change is produced in dy

  13. Differentials y = f(c) + f '(c)(x - c) (c + Δx, f(c + Δx)) } } Δy } (c, f(c)) f '(c)dx f(c + Δx) { { f(c) c c + Δx lim f(c + Δx) - f(c) = dy Δx 0 Δx dx f ' (c) = dy dx f '(c) dx = dy Δx when Δx is small we get the best approximation

  14. Differentials Δy = f(x + Δx) - f(x) dy = f '(c)dx dy is the differential of y or Δy ≈ dy dx is the differential of x Δy = the amount that the curve rises or falls dy = amount that the tangent line rises or falls In many applications, the differential of y can be used 
as an approximation of the change in y.

  15. If y = x2 then dy/dx = 2x dy = 2xdx and Δy = (x + Δx)2 - x2 Find dy when x = 1 and dx = 0.01. Compare this value 
with Δy for x = 1 and Δx = 0.01. dy = f '(x)dx = f '(1)(0.01) = 0.02 Δy = f(x + Δx) - f(x) = f(1.01) - f(1) = 0.0201

  16. If y = x2 then dy/dx = 2x dy = 2xdx and Δy = (x + Δx)2 - x2 If you start at x = 1 and move along the 
tangent line so that the change in x is 2 
(i.e. dx and Δx), find dy and Δy

  17. If y = x2 then dy/dx = 2x dy = 2xdx Δy = (x + Δx)2 - x2 Δy = (x + Δx)2 - x2 Δy = (1 + 2)2 - 12 Δy = 9 - 1 Δy = 8 dy = 2xdx dy = 2(1)(2) = 4

  18. Calculating Differentials Differential Derivative Function y = x2 2x 2xdx y = 2 sin x 2 cos x 2 cos x dx y = 1/x -1/x2 -dx/x2

  19. Error propagation (Using differentials) Physicists and engineers use dy to approximate Δy. 
An example is estimating errors in measuring devices. { propagated error measurement error { { f(x + Δx) - f(x) = Δy { measured value exact value You can use differentials instead of calculating

  20. A sphere is measured and found to be 21 cm with a 
possible measurement error of 0.05 cm. What is the 
maximum error in using this value of the radius to 
compute the volume of the sphere? V = 4/3πr3 (if error in r is dr or Δr then error in V is dV or ΔV) dV = 4πr2dr = 4π(21)2(0.05) ≈277 Thus the maximum error in the calculated volume is 277 cm3

  21. A sphere is measured and found to be 21 cm with a 
possible measurement error of 0.05 cm. What is the 
maximum error in using this value of the radius to 
compute the volume of the sphere? The % error in volume is dV = 4πr2dr V 4/3πr3 dV = 4πdr V 4/3πr = 3(0.05) = .7% 21

More Related