Calc1231W3T1 - Total Differential Approximation
Saturday, 4 August 2018
11:14 AM
Created with Microsoft OneNote 2016.
![Untitled picture.png Machine generated alternative text:
Let t(x) be the tangent line to the function
at the point [4, 84]. The equation of this tangent is
Now evaluate
1
p(3.9) —
1(3.9)
80.44
The closeness between these values matches our expectation that the tangent line is close to the parabola for points near [4, 84]_ Below; the graph of p@) is plotted in red, and the tangent
t(x) is plotted in blue dots.
150
100
50
2
3
4
5
6
Untitled picture.png Machine generated alternative text:
Si(x) —
near the point A = [5, Si(5)]
[5, 1.550]
The differential approximation to Si near 5 is given by the formula
Si(x) 1.550 +
Since Si' (5)
, we have the linear approximation to Si@) near x = 5.
Evaluating integrals can be difficult. So let's instead use the total differential approximation to approximate the Sine Integral
t: 1.5
sin (t)
dt
t
• Use the linear approximation to evaluate (to 3 decimal places)
Si(4.9)
1.569
• Use the plot below (drag the slider to the appropriate value) to evaluate (to 3 decimal places)
1.57
Si(4.9)
sin
fix
1.325



](Calc1231W3T1%20-%20Total%20Differential%20Approximation_files/image001.png)
![Untitled picture.png Machine generated alternative text:
Si(x) —
near the point A = [5, Si(5)]
[5, 1.550]
The differential approximation to Si near 5 is given by the formula
Si(x) 1.550 +
Since Si' (5)
, we have the linear approximation to Si@) near x = 5.
Evaluating integrals can be difficult. So let's instead use the total differential approximation to approximate the Sine Integral
t: 1.5
sin (t)
dt
t
• Use the linear approximation to evaluate (to 3 decimal places)
Si(4.9)
1.569
• Use the plot below (drag the slider to the appropriate value) to evaluate (to 3 decimal places)
1.57
Si(4.9)
sin
fix
1.325
Untitled picture.png Machine generated alternative text:
The total differential approximation works in three-dimensional space, too. For example, consider the paraboloid
2
2
_4_
52
22
at the point [5, 1, 5/4].
The total differential approximation to z
F@, y) near [5, 1] is given by the formula
5/4+ D F (5, — 5) +
In fact; this gives the equation of the tangent plane to the surface at the point [5, 1, 5/4]_ Now
10/25
Dy
So we have the linear approximation to F(x, y) near [5, 1]
• Use this formula to approximate (to 3 decimal places)
F(5.1, 1.1)
• Compute directly from the definition (to 3 decimal places) the value
F(5.1, 1.1) =](Calc1231W3T1%20-%20Total%20Differential%20Approximation_files/image002.png)
![Untitled picture.png Machine generated alternative text:
The total differential approximation works in three-dimensional space, too. For example, consider the paraboloid
2
2
_4_
52
22
at the point [5, 1, 5/4].
The total differential approximation to z
F@, y) near [5, 1] is given by the formula
5/4+ D F (5, — 5) +
In fact; this gives the equation of the tangent plane to the surface at the point [5, 1, 5/4]_ Now
10/25
Dy
So we have the linear approximation to F(x, y) near [5, 1]
• Use this formula to approximate (to 3 decimal places)
F(5.1, 1.1)
• Compute directly from the definition (to 3 decimal places) the value
F(5.1, 1.1) =
Untitled picture.png Machine generated alternative text:
In the previous questions, it was possible to calculate the value of the function exactly. But it is also possible to make mathematical deductions without knowing the function (which might be
impossible or expensive to obtain), and instead only knowing the partial derivatives.
Assume that the approval rating of a Prime Minister is given by the function A(d, e), where d is defence spending (in billions) and e is education spending (in billions). The output of the
approval rating A(d, e) itself is a percentage between 0 and 100.
It is desirable to predict how changes to defence and education spending impact upon the PM's approval. With current spending at do and eo, the rate that approval (in percentage) changes
Dd (do, eo) = 8 so an increase in defence spending of 1 billion dollars will translate to an
with respect to defence spending (in billions) is measured by Newspoll to be the partial derivative
increase in approval of 8%
(do, eo) = 8.2_ Hence by the total differential approximation, for
Similarly the rate that approval changes with respect to education spending is measured to be the partial derivative
De
[d, e] in the neighbourhood of [do, eo]
A(d, e) A(dO, eo) + -H (do, — do) + % (do, — eo)
The current approval rating is A(d0, eo) = 54. If defence spending is decreased by 0.8 billion and education spending increased by 0.1 billion; then the approval rating approximately
changes to 48.42
Note: do not round your answer, approval rating is too important to be rounded.](Calc1231W3T1%20-%20Total%20Differential%20Approximation_files/image003.png)
![Untitled picture.png Machine generated alternative text:
In the previous questions, it was possible to calculate the value of the function exactly. But it is also possible to make mathematical deductions without knowing the function (which might be
impossible or expensive to obtain), and instead only knowing the partial derivatives.
Assume that the approval rating of a Prime Minister is given by the function A(d, e), where d is defence spending (in billions) and e is education spending (in billions). The output of the
approval rating A(d, e) itself is a percentage between 0 and 100.
It is desirable to predict how changes to defence and education spending impact upon the PM's approval. With current spending at do and eo, the rate that approval (in percentage) changes
Dd (do, eo) = 8 so an increase in defence spending of 1 billion dollars will translate to an
with respect to defence spending (in billions) is measured by Newspoll to be the partial derivative
increase in approval of 8%
(do, eo) = 8.2_ Hence by the total differential approximation, for
Similarly the rate that approval changes with respect to education spending is measured to be the partial derivative
De
[d, e] in the neighbourhood of [do, eo]
A(d, e) A(dO, eo) + -H (do, — do) + % (do, — eo)
The current approval rating is A(d0, eo) = 54. If defence spending is decreased by 0.8 billion and education spending increased by 0.1 billion; then the approval rating approximately
changes to 48.42
Note: do not round your answer, approval rating is too important to be rounded.
Untitled picture.png Machine generated alternative text:
How much toilet paper is left on the roll? The volume of a cylinder with length e and radius r is given by the formula:
TIr2
You've got some time on your hands, so you carefully measure your toilet roll and discover its dimensions to the nearest mm:
• length t = 94 mm,
• radius of the inner cardboard cylinder r = 29 mm, and
• radius of the outer cylinder R = 52 mm.
The volume of toilet paper on the roll is thus given by the formula:
V(t, r, R) = Tl(R2
According to your measurements, the volume of toilet paper is (to the nearest cubic mm):
O mm3
However, since your initial measurements were only accurate to the nearest mm, the true volume of toilet paper may be different
The total differential approximation can be used to estimate how the measured volume V(94, 29, 52) differs from the true value of V(C, r, R) as:
V(l, r, R) V(94, 29, 52) + DV (94, 29, - 94) + W (94,29, - 29) + 0494, 29,
DR
We calculate the partial derivatives (to the nearest integer):
- 52).
(94, 29, 52) -
(94, 29, 52) -
Dr
(94, 29, 52) -
DR
5853
-17128
30712
Since we measured t, r and R to the nearest mm,
It — 941 < 0.5, Ir
— 291 < 0.5 and IR
- 521 < 0.5.](Calc1231W3T1%20-%20Total%20Differential%20Approximation_files/image004.png)
