commit 7a0abad78d89f4e6d739e1efcf82380bd3fe8dd4
parent b1fcde25ed0ef5f27da96810d73a90830f5debd2
Author: Bradley Gannon <bradley@bradleygannon.com>
Date: Sun, 22 Mar 2026 16:56:33 -0400
Solve 1.{7,11,12,13}
Diffstat:
| M | main.typ | | | 215 | +++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ |
1 file changed, 215 insertions(+), 0 deletions(-)
diff --git a/main.typ b/main.typ
@@ -4,6 +4,9 @@
#let cross = math.times
#let definedAs = math.eq.triple
#let implies = math.arrow.r.double
+#let sepMag = text(font: "Script R", "r")
+#let sepVec = text(font: "Script R", weight: "bold", "r")
+#let sepUnit = math.hat(sepVec)
#set heading(numbering: (..nums) => {
let n = nums.pos()
@@ -155,6 +158,32 @@ Both sides are equal, which proves the *BAC-CAB* rule. $qed$
=== Problem 1.7
+The separation vector $sepVec$ is defined as
+
+#tag($ sepVec definedAs vec(r) - vec(r') "." $, "(Eq 1.23)")
+
+The field vector is $vec(r) = 4 unit(x) + 6 unit(y) + 8 unit(z)$, and the source vector is $vec(r')
+= 2 unit(x) + 8 unit(y) + 7 unit(z)$, so the separation vector is
+
+$
+sepVec = (4 - 2) unit(x) + (6 - 8) unit(y) + (8 - 7) unit(z) \
+#rect($sepVec = 2 unit(x) - 2 unit(y) + unit(z)$) "."
+$
+
+The magnitude of the separation vector is
+
+$
+sepMag = sqrt(sepMag_x^2 + sepMag_y^2 + sepMag_z^2) = sqrt(2^2 + (-2)^2 + 1^2) = sqrt(9) \
+#rect($sepMag = 3$) "."
+$
+
+The unit vector associated with the separation vector is
+
+$
+sepUnit = sepVec / sepMag = (2 unit(x) - 2 unit(y) - unit(z)) / 3 \
+#rect($sepUnit = 2/3 unit(x) - 2/3 unit(y) + 1/3 unit(z)$) "."
+$
+
=== Problem 1.8
=== Problem 1.9
@@ -163,6 +192,192 @@ Both sides are equal, which proves the *BAC-CAB* rule. $qed$
== Differential Calculus
+=== Problem 1.11
+
+#tag($
+gradient T definedAs (partial T) / (partial x) unit(x)
++ (partial T) / (partial y) unit(y)
++ (partial T) / (partial z) unit(z)
+$, "(Eq. 1.36)")
+
+==== Part (a)
+
+$
+gradient f = gradient (x^2 + y^3 + z^4) \
+gradient f = partial / (partial x) (x^2 + y^3 + z^4) unit(x)
++ partial / (partial y) (x^2 + y^3 + z^4) unit(y)
++ partial / (partial z) (x^2 + y^3 + z^4) unit(z) \
+#rect($ gradient f = 2 x unit(x) + 3 y^2 unit(y) + 4 z^3 unit(z) $)
+$
+
+==== Part (b)
+
+$
+gradient f = gradient (x^2 y^3 z^4) \
+gradient f = partial / (partial x) (x^2 y^3 z^4) unit(x)
++ partial / (partial y) (x^2 y^3 z^4) unit(y)
++ partial / (partial z) (x^2 y^3 z^4) unit(z) \
+#rect($ gradient f &= 2 x y^3 z^4 unit(x) + 3 x^2 y^2 z^4 unit(y) + 4 x^2 y^3 z^3 unit(z) $)
+$
+
+==== Part (c)
+
+$
+gradient f = gradient (e^x sin(y) ln(z)) \
+gradient f = partial / (partial x) lr((e^x sin(y) ln(z)), size: #200%) unit(x)
++ partial / (partial y) lr((e^x sin(y) ln(z)), size: #200%) unit(y)
++ partial / (partial z) lr((e^x sin(y) ln(z)), size: #200%) unit(z) \
+#rect($ gradient f &= e^x unit(x) + cos(y) unit(y) + 1/z unit(z) $)
+$
+
+=== Problem 1.12
+
+==== Part (a)
+
+The top of the hill is the maximum of $h$. The gradient at this point is zero, so we can find the
+maximum by computing $gradient h$ and finding the values of $x$ and $y$ for which the resulting
+vector is $vec(0)$. In the case of multiple points with zero gradient, we take the point which
+maximizes $h$.
+
+First, we compute the gradient of $h$.
+
+$
+gradient h =& partial / (partial x) (10 (2 x y - 3 x^2 - 4 y^2 - 18 x + 28 y + 12)) unit(x) \
++& partial / (partial y) (10 (2 x y - 3 x^2 - 4 y^2 - 18 x + 28 y + 12)) unit(y) \
+gradient h =& 20 (-3 x + y - 9) unit(x) + 20 (x - 4 y + 14) unit(y)
+$
+
+Now we set $gradient h$ equal to $vec(0)$ and solve the resulting system of equations.
+
+$
+20 (-3 x + y - 9) = 0 \
+20 ( x - 4 y + 14) = 0 \
+implies y = 4 x + 9 \
+x - 3 (4 x + 9) + 14 = 0 implies x = -13/12 \
+y = 3 (-13/12) + 9 = 5 3/4
+$
+
+There is only one solution, and we assume that this point is a maximum instead of a minimum or
+saddle point.
+
+So the top of the hill is located $#rect($1 1/12 "miles west, " 5 3/4 "miles north"$)$ of South
+Hadley.
+
+==== Part (b)
+
+The value of $h$ at the top of the hill is is
+
+$
+h(-13/8, 33/8) &= 10 (2 x y - 3 x^2 - 4 y^2 - 18 x + 28 y + 12)
+&approx 44.3
+$
+
+So the hill is $#rect($"about" 44.3 "feet"$)$ high.
+
+==== Part (c)
+
+To find the slope and steepest direction at the given point, we evaluate the gradient there and find
+its magnitude and direction.
+
+$
+(gradient h)_(x=1,y=1) &= 20 (-3 dot 1 + 1 - 9) unit(x) + 20 (1 - 4 dot 1 + 14) unit(y) = -220 unit(x) + 220 unit(y) \
+abs((gradient h)_(x=1,y=1)) &= sqrt((-220)^2 + 220^2) approx 221 \
+$
+
+The gradient at the given point may be converted directly into a bearing relative to geographic
+north.
+
+$
+tan^(-1) ((-220)/220) + 2 pi approx 5.50 approx 315 degree
+$
+
+So the steepest slope at the given point is $#rect($"about" 221 "feet per mile at a bearing of" 315
+degree "."$)$
+
+For completeness, we also compute the unit vector of the gradient at the given point.
+
+$
+(gradient h)/abs(gradient h)|_(x=1,y=1) = (-220 unit(x) + 220 unit(y))/(sqrt((-220)^2 + 220^2))
+approx -sqrt(2)/2 unit(x) + sqrt(2)/2 unit(y)
+$
+
+=== Problem 1.13
+
+$
+sepVec = (x - x') unit(x) + (y - y') unit(y) + (z - z') unit(z) \
+sepMag = sqrt((x - x')^2 + (y - y')^2 + (z - z')^2)
+$
+
+==== Part (a)
+
+$
+gradient (sepMag^2) &= 2 sepVec \
+partial / (partial x) sepMag^2 unit(x) + partial / (partial y) sepMag^2 unit(y) + partial / (partial z) sepMag^2 unit(z)
+&= 2 sepVec \
+2 (x - x') unit(x) + 2 (y - y') unit(y) + 2 (z - z') unit(z) &= 2 sepVec \
+2 sepVec &= 2 sepVec && qed
+$
+
+==== Part (b)
+
+$
+gradient (1/sepMag) = -sepUnit/sepMag^2 \
+partial / (partial x) 1/sepMag unit(x) + partial / (partial y) 1/sepMag unit(y) + partial / (partial z) 1/sepMag unit(z)
+= -((x - x') unit(x) + (y - y') unit(y) + (z - z') unit(z))/(sqrt((x - x')^2 + (y - y')^2 + (z - z')^2))
+1/((x - x')^2 + (y - y')^2 + (z - z')^2) \
+-(2 (x - x'))/(2 ((x - x')^2 + (y - y')^2 + (z - z')^2)^(3/2)) unit(x) \
+-(2 (y - y'))/(2 ((x - x')^2 + (y - y')^2 + (z - z')^2)^(3/2)) unit(y) \
+-(2 (z - z'))/(2 ((x - x')^2 + (y - y')^2 + (z - z')^2)^(3/2)) unit(z)
+= -sepVec / sepMag^3 \
+-((x - x') unit(x) + (y - y') unit(y) + (z - z') unit(z))/(((x - x')^2 + (y - y')^2 + (z - z')^2)^(3/2)) = -sepVec / sepMag^(3/2) \
+-sepVec / sepMag^3 = -sepVec / sepMag^3 qed
+$
+
+==== Part (c)
+
+$
+gradient sepMag^n =& partial / (partial x) sepMag^n unit(x) + partial / (partial y) sepMag^n unit(y)
++ partial / (partial z) sepMag^n unit(z) \
+=& (n (x - x'))/(2 ((x - x')^2 + (y - y')^2 + (z - z')^2)^(n/2)) unit(x) \
+&+ (n (y - y'))/(2 ((x - x')^2 + (y - y')^2 + (z - z')^2)^(n/2)) unit(y) \
+&+ (n (z - z'))/(2 ((x - x')^2 + (y - y')^2 + (z - z')^2)^(n/2)) unit(z) \
+=& n/2 ((x - x') unit(x) + (y - y') unit(y) + (z - z') unit(z)) / ((x - x')^2 + (y - y')^2 + (z - z')^2)^(n/2) \
+$
+
+$
+#rect($ gradient sepMag^n = n/2 sepVec / sepMag^(n) $)
+$
+
+=== Problem 1.14
+
+=== Problem 1.15
+
+=== Problem 1.16
+
+=== Problem 1.17
+
+=== Problem 1.18
+
+=== Problem 1.19
+
+=== Problem 1.20
+
+=== Problem 1.21
+
+=== Problem 1.22
+
+=== Problem 1.23
+
+=== Problem 1.24
+
+=== Problem 1.25
+
+=== Problem 1.26
+
+=== Problem 1.27
+
+=== Problem 1.28
+
== Integral Calculus
== Curvilinear Coordinates
