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INSTRUCTORS Raffi Hovasapian John Zhu
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Lecture Comments (1)

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Post by Steve Denton on October 18, 2012

At 9:o5.That answer doesn't make any sense. How could the volume get 8pi*r^2 times bigger. It only gets 27 times bigger when it goes from r=1 to r=3. From r=13 to r=15 the V only gets 1.5 times bigger.???????
The ratio seems like it should be (r+2)^3 / r^3

Differentials

  • Used to approximate
    • Choose appropriate x and
    • Apply formula directly
    • Choose appropriate formula to relate all key variables
    • Differentiate
    • Apply appropriate substitutions and solve

Differentials

Approximate 3.13 using differentials
  • f(x) = x3
  • f(x + ∆x) ≈ f(x) + f′(x) ∆x
  • f′(x) = 3 x2
  • f(3 + 0.1) ≈ 33 + 3 (3)2 (0.1)
  • f(3.1) ≈ 27 + 2.7
3.13 ≈ 29.7
Approximate 3√{7.4} using differentials
  • f(x) = x[1/3]
  • f′(x) = [1/3] x−[2/3]
  • f(8 − .6) ≈ 3√{8} + [1/(3 (8)[2/3])] (−.6)
  • f(8 − .6) ≈ 2 − [.6/12]
  • f(7.4) ≈ 2 − [6/120]
  • f(7.4) ≈ 2 − [1/20]
  • f(7.4) ≈ 2 − .05
3√{7.4} ≈ 1.95
Approximate ln0.9 using Differentials
  • f(x) = lnx
  • f′(x) = [1/x]
  • f(1 − 0.1) ≈ ln1 + [1/1] (−0.1)
  • f(0.9) ≈ 0 − 0.1
ln0.9 ≈ −0.1
Approximate the volume of a cube with sides of length 4.2
  • Volume of a cube = l3
  • V(l) = l3
  • V′(l) = 3 l2
  • V(4 + .2) ≈ 43 + 3 (4)2 (.2)
  • V(4.2) ≈ 64 + 48 (.2)
  • V(4.2) ≈ 64 + 9.6
V(4.2) ≈ 73.6
The lengths of a cube are increased from 4 to 4.2. Use differentials to estimate the change in area.
  • V = l3
  • [d/dl] V = [d/dl] l3
  • [dV/dl] = 3 l2
  • dV = 3 l2 dl
  • dV ≈ 3 l2 ∆l
  • Notice that this turned into an approximation?
  • dy = f′(x)  dx
  • This formula can accurately calculate dy if f′(x) is unchanging, but if f′(x) is changing on that interval or change in x, it tends to become less accurate of an approximation the larger ∆x is used.
  • dV ≈ 3 (4)2 (4.2 − 4)
dV ≈ 9.6
Find the percent error when approximating 2.22 using using differentials with a ∆x of 0.2
  • f(x) = x2
  • f′(x) = 2x
  • f(2.2) ≈ 22 + 2(2)(0.2)
  • f(2.2) ≈ 4 + 0.8
  • f(2.2) ≈ 4.8
  • f(2.2) = 2.22 = 4.84
  • error = 100 % [(|4.84 − 4.8|)/4.84]
error ≈ 8.3%
Find the percent error when approximating 2.22 using differentials given 2.12 = 4.41
  • f(2.2) ≈ 2.12 + 2(2.1) (0.1)
  • f(2.2) ≈ 4.41 + .42
  • f(2.2) ≈ 4.83
  • f(2.2) = 2.22 = 4.84
  • error = 100 % [(|4.84 − 4.83|)/4.84]
error ≈ 0.21 %
A rectangle's height increases from 10 to 11 with a width of 2. Use differentials to estimate the change in area.
  • A = h  w
  • The width is constant, so we're only interested in the change of A with respect to h
  • [dA/dh] = w
  • dA = w  dh
  • dA ≈ w ∆h
  • dA ≈ 2 (11 − 10)
  • The approximation is not necessary here. f′(x) is a constant in this case.
dA = 2
A sphere's radius increases from 1 to 1.1. Use differentials to estimate the change in surface area.
  • A = 4 πr2
  • [dA/dr] = 8 πr
  • dA = 8 πr  dr
  • dA ≈ 8 π(1) (1.1 − 1)
dA ≈ 0.8 π
The height of a cone decreases from 2 to 1.8 with a constant radius of 3 at the base. Use differentials to estimate the change in volume
  • V = [1/3] πr2 h
  • [dV/dh] = [1/3] r2
  • In this case, [dV/dh] is a constant
  • dV = [1/3] πr2 ∆h
  • dV = [1/3] π32 (1.8 − 2)
  • dV = 3 π(−0.2)
dV = −0.6 π

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

Differentials

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Differentials 0:08
    • 1st Differential Formula
    • 2nd Differential Formula
  • Example 1 1:06
  • Example 2 3:21
  • Example 3 5:49
  • Example 4 7:19
  • Example 5 9:06