1. Calculus III
Summer 2020
Lesson 2
Final Examination
Review
“A person who never made a
mistake never tried anything new.”
- Albert Einstein -
2. Lehman College, Department of Mathematics
Equations of Lines and Planes (1 of 2)
Example 0. Consider the following diagram:
Let 𝑃 be the point (1, 2, 3), and
suppose the vector 𝐫 is parallel
to 𝑃𝑅, where:
𝐫 = 2 𝐢 + 3 𝐣 + 4 𝐤
Let 𝑅 be the point (𝑥, 𝑦, 𝑧), then
the vector 𝑃𝑅 is given by:
𝑃𝑅 = (𝑥 − 1, 𝑦 − 2, 𝑧 − 3)
Parallel vectors simply mean
they are proportionate, so:
𝑥 − 1
2
=
𝑦 − 2
3
=
𝑧 − 3
4
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Equations of Lines and Planes (2 of 2)
Now, consider the plane perpendicular to the vector 𝐫
and passing through the point (1, 2, 3):
𝐫 = 2 𝐢 + 3 𝐣 + 4 𝐤
Let 𝑄 be the point (𝑥, 𝑦, 𝑧), then
the vector 𝑃𝑄 is given by:
𝑃𝑄 = (𝑥 − 1, 𝑦 − 2, 𝑧 − 3)
But 𝑃𝑄 is orthogonal to the
vector 𝐫, so their dot product
is zero:
3 𝑦 − 2 +2 𝑥 − 1 + 4 𝑧 − 3 = 0
𝐫 ⋅ 𝑃𝑄 = 0
2𝑥 + 3𝑦 + 4𝑥 − 20 = 0
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Tangent Planes and Normal Lines (1 of 6)
Example 1 (Ste., p. 916). Find an equation of the
tangent plane to the elliptic paraboloid given below:
at the point 1, 1, 3 .
Solution. Step 1. Write the equation of the surface as:
Step 2. Consider the function:
Step 3. Determine the first partial derivatives:
𝑧 = 2𝑥2 + 𝑦2
2𝑥2 + 𝑦2 − 𝑧 = 0
𝐹(𝑥, 𝑦, 𝑧) = 2𝑥2
+ 𝑦2
− 𝑧
𝐹𝑥(𝑥, 𝑦, 𝑧) = 𝐹𝑦(𝑥, 𝑦, 𝑧) =
𝐹𝑧(𝑥, 𝑦, 𝑧) =
4𝑥 2𝑦
−1
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Tangent Planes and Normal Lines (2 of 6)
Solution (cont’d).
Step 4. Evaluate the partial derivatives at the point:
Step 5. Find equation of the tangent plane at (1, 1, 3):
𝐹𝑥(1, 1, 3) = 4 1 = 4
𝐹𝑦(1, 1, 3) = 2 1 = 2
𝐹𝑧(1, 1, 3) = −1
4 𝑥 − 1 + 2 𝑦 − 1 − 1 𝑧 − 3 = 0
4𝑥 − 4 + 2𝑦 − 2 − 𝑧 + 3 = 0
4𝑥 + 2𝑦 − 𝑧 − 3 = 0
𝑧 = 4𝑥 + 2𝑦 − 3
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Tangent Planes and Normal Lines (3 of 6)
Solution (cont’d). From the previous slide:
Step 6. Find equation of the normal line at (1, 1, 3):
𝐹𝑥 1, 1, 3 = 4
𝐹𝑦 1, 1, 3 = 2
𝐹𝑧 1, 1, 3 = −1
𝑥 − 1
4
=
𝑦 − 1
2
=
𝑧 − 3
−1
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Tangent Planes and Normal Lines (4 of 6)
Example 2 (Lar., p. 947). Find an equation of the
tangent plane to the hyperboloid given below:
at the point 1, −1, 4 .
Solution. Step 1. Write the equation of the surface as:
Step 2. Consider the function:
Step 3. Determine the first partial derivatives:
𝑧2 − 2𝑥2 − 2𝑦2 = 12
𝑧2 − 2𝑥2 − 2𝑦2 − 12 = 0
𝐹(𝑥, 𝑦, 𝑧) = 𝑧2
− 2𝑥2
− 2𝑦2
− 12
𝐹𝑥(𝑥, 𝑦, 𝑧) = 𝐹𝑦(𝑥, 𝑦, 𝑧) =
𝐹𝑧(𝑥, 𝑦, 𝑧) =
−4𝑥 −4𝑦
2𝑧
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Tangent Planes and Normal Lines (6 of 6)
Solution (cont’d). From the previous slide:
Step 6. Find equation of the normal line at (1, −1, 4):
𝑥 − 1
−4
=
𝑦 + 1
4
=
𝑧 − 4
8
𝐹𝑥(1, −1, 4) = −4
𝐹𝑦(1, −1, 4) = 4
𝐹𝑧(1, −1, 4) = 8
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Relative Extrema (1 of 5)
Example 2 (Ste., p. 947). Find and classify the relative
extrema and saddle points of:
Solution. Step 1. Determine the first partial derivatives:
Step 2. Locate the critical points (set partials to zero):
The solutions yield the critical point:
Step 3. Determine the second partial derivatives:
𝑓 𝑥, 𝑦 = 𝑦2 − 𝑥2
𝑓𝑥(𝑥, 𝑦) = −2𝑥 𝑓𝑦(𝑥, 𝑦) = 2𝑦
−2𝑥 = 0 2𝑦 = 0and
(0, 0)
𝑓𝑥𝑥(𝑥, 𝑦) = −2 𝑓𝑥𝑦(𝑥, 𝑦) = 0 𝑓𝑦𝑦(𝑥, 𝑦) = 2
11. Lehman College, Department of Mathematics
Relative Extrema (2 of 5)
Solution. From the previous slide:
Step 4. Determine the quantity 𝑑:
The determinant used above is called the determinant
of the Hessian matrix of the function 𝑓.
Step 5. Evaluate 𝑑 for each critical point:
Since 𝑑 < 0, then the point (0, 0) is a:
𝑓𝑥𝑥(𝑥, 𝑦) = −2 𝑓𝑥𝑦(𝑥, 𝑦) = 0 𝑓𝑦𝑦(𝑥, 𝑦) = 2
𝑑 =
𝑓𝑥𝑥 𝑓𝑥𝑦
𝑓𝑦𝑥 𝑓𝑦𝑦
=
−2 0
0 2
= −4 − 0 = −4
saddle point
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Relative Extrema (3 of 5)
Example 3 (Ste., p. 947). Find and classify the relative
extrema and saddle points of:
Solution. Step 1. Determine the first partial derivatives:
Step 2. Locate the critical points (set partials to zero):
That is:
Substituting the first equation into the second :
𝑓 𝑥, 𝑦 = 𝑥4 + 𝑦4 − 4𝑥𝑦 + 1
𝑓𝑥(𝑥, 𝑦) = 4𝑥3 − 4𝑦 𝑓𝑦(𝑥, 𝑦) = 4𝑦3
− 4𝑥
4𝑥3
− 4𝑦 = 0 4𝑦3
− 4𝑥 = 0and
𝑥3 = 𝑦 and 𝑦3
= 𝑥
𝑥9
= 𝑥
𝑥9
− 𝑥 = 0 or 𝑥(𝑥8
− 1) = 0
or 𝑥(𝑥4
− 1)(𝑥4
+ 1) = 0
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Relative Extrema (4 of 5)
Solution. From the previous slide:
The solutions are: 𝑥 = 0, 𝑥 = 1 and 𝑥 = −1. Therefore,
the critical points are:
Step 3. Determine the second partial derivatives:
Step 4. Determine the quantity 𝑑 from the Hessian:
𝑑 =
𝑓𝑥𝑥 𝑓𝑥𝑦
𝑓𝑦𝑥 𝑓𝑦𝑦
= 12𝑥2
−4
−4 12𝑦2 = 144𝑥2 𝑦2 − 16
𝑥(𝑥 − 1)(𝑥 + 1)(𝑥2 + 1)(𝑥4 + 1) = 0
or𝑥(𝑥4
− 1)(𝑥4
+ 1) = 0
(0, 0), (1, 1), (−1, −1)
𝑓𝑥𝑥(𝑥, 𝑦) =12𝑥2 𝑓𝑥𝑦(𝑥, 𝑦) = −4 𝑓𝑦𝑦(𝑥, 𝑦) = 12𝑦2
14. Lehman College, Department of Mathematics
Relative Extrema (5 of 5)
Solution. From the previous slides:
Step 5. Evaluate 𝑑 for each critical point:
For the point (0, 0):
For the point 1, 1 :
We also evaluate:
For the point −1, −1 :
We also evaluate:
saddle point
𝑑 = 144𝑥2 𝑦2 − 16
𝑑 = −16 < 0
𝑑 = 144 − 16 > 0
𝑓𝑥𝑥 = 12 > 0
𝑓𝑥𝑥(𝑥, 𝑦) = 12𝑥2and
local minimum
𝑑 = 144 − 16 > 0
𝑓𝑥𝑥 = 12 > 0 local minimum
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Iterated Integrals and Volume (1 of 6)
Example 2 (Ste., p. 991). Find the volume of the solid
that lies under the paraboloid 𝑧 = 𝑥2
+ 𝑦2
and above
the region 𝐷 in the 𝑥𝑦-plane bounded by the line 𝑦 = 2𝑥
and the parabola 𝑦 = 𝑥2
.
Solution.
Step 1. Sketch the region 𝐷 in
the 𝑥𝑦-plane:
Step 2. Write the region 𝐷 in
set notation:
𝐷 = 𝑥, 𝑦 | 0 ≤ 𝑥 ≤ 2, 𝑥2 ≤ 𝑦 ≤ 2𝑥
16. Lehman College, Department of Mathematics
Iterated Integrals and Volume (2 of 6)
Solution (cont’d). From the previous slide:
Step 3. Therefore, the volume 𝑉 under 𝑥2
+ 𝑦2
and
above the region 𝐷 is given by the integral:
𝐷 = 𝑥, 𝑦 | 0 ≤ 𝑥 ≤ 2, 𝑥2 ≤ 𝑦 ≤ 2𝑥
𝑉 =
0
2
𝑥2
2𝑥
𝑥2 + 𝑦2 𝑑𝑦 𝑑𝑥
=
0
2
𝑥2 𝑦 +
𝑦3
3 𝑥2
2𝑥
𝑑𝑥
=
0
2
𝑥2
2𝑥 +
2𝑥 3
3
− 𝑥2
𝑥2
+
𝑥2 3
3
𝑑𝑥
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Iterated Integrals and Volume (4 of 6)
Example 8 (Prob. Set 3). Use rectangular coordinates
to set up an iterated integral that represents the volume
of the solid bounded by the surfaces 𝑧 = 𝑥2
+ 𝑦2
+ 3,
𝑧 = 0 and 𝑥2
+ 𝑦2
= 1.
Solution.
Step 1. Sketch the region 𝐷 in
the plane 𝑧 = 0 (𝑥𝑦-plane):
Step 2. Describe the region 𝐷
in set notation:
𝐷 = 𝑥, 𝑦 | − 1 ≤ 𝑥 ≤ 1, − 1 − 𝑥2 ≤ 𝑦 ≤ 1 − 𝑥2
𝐷 in the unit circle 𝑥2 + 𝑦2 = 1
19. Lehman College, Department of Mathematics
Iterated Integrals and Volume (5 of 6)
Solution (cont’d). From the previous slide:
Step 3. Therefore, the volume 𝑉 under 𝑥2
+ 𝑦2
+ 3 and
above the region 𝐷 is given by the integral:
(b) Evaluate the iterated integral in (a) by converting to
polar coordinates:
𝑉 =
−1
1
− 1−𝑥2
1−𝑥2
𝑥2 + 𝑦2 + 3 𝑑𝑦 𝑑𝑥
𝐷 = 𝑥, 𝑦 | − 1 ≤ 𝑥 ≤ 1, − 1 − 𝑥2 ≤ 𝑦 ≤ 1 − 𝑥2
𝑉 =
0
2𝜋
0
1
𝑟2
+ 3 𝑟𝑑𝑟 𝑑𝜃
21. Lehman College, Department of Mathematics
Lagrange Multipliers (1 of 2)
Example 2 (Ste. p. 960). Find the extreme values of the
function 𝑓(𝑥, 𝑦) = 𝑥2
+ 2𝑦2
on the circle 𝑥2
+ 𝑦2
= 1.
Solution. Let 𝑔(𝑥, 𝑦) = 𝑥2 + 𝑦2, then the method of
Lagrange multipliers requires that we solve the system:
That is:
Factoring, we obtain:
Which means:
If 𝑥 = 0, then:
If 𝜆 = 1, then:
Evaluate 𝑓 at the points:
𝑓𝑥 = 𝜆 𝑔 𝑥 𝑓𝑦 = 𝜆 𝑔 𝑦 𝑔(𝑥, 𝑦) = 1
2𝑥 = 𝜆(2𝑥) 4𝑦 = 𝜆(2𝑦) 𝑥2
+ 𝑦2
= 1
𝑥 𝜆 − 1 = 0 𝑦 𝜆 − 2 = 0
𝑥 = 0 or 𝜆 = 1
𝑦 = ±1
𝑦 = 0 and 𝑥 = ±1
(0, −1),(0, 1), (1, 0), (−1, 0)
22. Lehman College, Department of Mathematics
Lagrange Multipliers (2 of 2)
Solution (cont’d). Since 𝑓(𝑥, 𝑦) = 𝑥2
+ 2𝑦2
, then:
It follows that the maximum value of 𝑓(𝑥, 𝑦) = 𝑥2
+ 2𝑦2
on the circle 𝑥2
+ 𝑦2
= 1 is 2 and it occurs at the points
0, 1 and 0, −1 .
The minimum value of the function is 1, which occurs at
the points 1, 0 and −1, 0 .
𝑓 0, −1 =
𝑓 0, 1 =
𝑓 1, 0 =
𝑓 −1, 0 =
2
2
1
1
