Calc1231W3T1 - Total Differential Approximation

Saturday, 4 August 2018

11:14 AM

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 




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) =
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.
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.
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. 

Untitled picture.png Machine generated alternative text:
Using the integer approximations above together with the triangle inequality 
R) - v(94, 29, 52)l (94, 29, 52)l - 94)l + (94, 29, 52)l - 29)l + I DV (94, 29, 52)l 
And so our measured volume V(94, 29, 52) differs from the true volume by 
O mm3 
To two decimal places; this represents an approximate error of no more than 
< 4.88%. 
Are you satisfied with this level of precision? (this part is not worth marks) 
@ No; we need more precise measurements. 
O Yes, this level of precision is sufficient for my needs 
O Better go buy some more toilet paper to account for this potentially significant error 
- 52)l-

 

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