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Vector CalculusQuestion and Answers: Page 1

Question Number 207712    Answers: 1   Comments: 0

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Question Number 204273    Answers: 1   Comments: 0

f(x)=(1/((x−1)^(ln((2/4))) )) Domain f(x) =?

$$\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{1}}{\left(\mathrm{x}−\mathrm{1}\right)^{\mathrm{ln}\left(\frac{\mathrm{2}}{\mathrm{4}}\right)} } \\ $$$$\mathrm{Domain}\:\mathrm{f}\left(\mathrm{x}\right)\:=? \\ $$

Question Number 203900    Answers: 0   Comments: 0

Let A ∈ R^(N×N) be a symmetric positive definite matrix and b ∈ R^N a vector. If x ∈ R^N , evaluate the integral Z(A,b) = ∫e^(−(1/2)x^T Ax + b^T x) dx as a function of A and b.

$${Let}\:{A}\:\in\:{R}^{{N}×{N}} \:{be}\:{a}\:{symmetric}\:{positive} \\ $$$${definite}\:{matrix}\:{and}\:{b}\:\in\:{R}^{{N}} \:{a}\:{vector}. \\ $$$${If}\:{x}\:\in\:{R}^{{N}} ,\:{evaluate}\:{the}\:{integral} \\ $$$${Z}\left({A},{b}\right)\:=\:\int{e}^{−\frac{\mathrm{1}}{\mathrm{2}}{x}^{{T}} {Ax}\:+\:{b}^{{T}} {x}} {dx}\:{as}\:{a}\:{function} \\ $$$${of}\:{A}\:{and}\:{b}. \\ $$

Question Number 201947    Answers: 1   Comments: 0

Question Number 201144    Answers: 2   Comments: 0

(((14)/(15)))^6 ×(((45)/(28)))^6 =

$$\left(\frac{\mathrm{14}}{\mathrm{15}}\right)^{\mathrm{6}} ×\left(\frac{\mathrm{45}}{\mathrm{28}}\right)^{\mathrm{6}} = \\ $$

Question Number 200929    Answers: 0   Comments: 2

A ball lies on the function z=xy at the point (1,2,2). Find the point in the xy−plane where the ball will touch it. Calculus 2 problem.

$$\mathrm{A}\:\mathrm{ball}\:\mathrm{lies}\:\mathrm{on}\:\mathrm{the}\:\mathrm{function}\:{z}={xy}\:\mathrm{at} \\ $$$$\mathrm{the}\:\mathrm{point}\:\left(\mathrm{1},\mathrm{2},\mathrm{2}\right).\:\mathrm{Find}\:\mathrm{the}\:\mathrm{point}\:\mathrm{in} \\ $$$$\mathrm{the}\:{xy}−\mathrm{plane}\:\mathrm{where}\:\mathrm{the}\:\mathrm{ball}\:\mathrm{will} \\ $$$$\mathrm{touch}\:\mathrm{it}. \\ $$$$\mathrm{Calculus}\:\mathrm{2}\:\mathrm{problem}. \\ $$

Question Number 196918    Answers: 1   Comments: 0

Question Number 196412    Answers: 0   Comments: 1

Prove that ∫^( +∞) _( i) (cos2t+isin2t)e^(−t^2 ) dt= ((√π)/(2e))

$$\mathrm{Prove}\:\mathrm{that}\:\underset{\:{i}} {\int}^{\:+\infty} \left(\mathrm{cos2t}+\mathrm{isin2t}\right)\mathrm{e}^{−\mathrm{t}^{\mathrm{2}} } \mathrm{dt}=\:\frac{\sqrt{\pi}}{\mathrm{2e}} \\ $$

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If the area enclosed between the curves y=x² and the line y = 2x is rotated round the x-axis through 4 right angles, find the volume of the solid generated

If the area enclosed between the curves y=x² and the line y = 2x is rotated round the x-axis through 4 right angles, find the volume of the solid generated

Question Number 182183    Answers: 1   Comments: 0

∫_0 ^( 1) e^a a^n da=? n≥1 n∈N

$$\int_{\mathrm{0}} ^{\:\mathrm{1}} \boldsymbol{{e}}^{\boldsymbol{{a}}} \boldsymbol{{a}}^{\boldsymbol{{n}}} \boldsymbol{{da}}=?\:\:\:\:\:\:\:\:\boldsymbol{{n}}\geqslant\mathrm{1}\:\:\:\:\boldsymbol{{n}}\in\boldsymbol{{N}} \\ $$

Question Number 181678    Answers: 0   Comments: 0

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Question Number 174807    Answers: 0   Comments: 0

We know that vertex form of parabola is given as y = a(x−h)^2 +k From the given diagram of bridge that resembles a parabola, we have a vertex points of (0, 30) and other points due to towers that supports the parabolic−shape cable, which is (200, 150). ∴ y = a(x−0)^2 +30 = ax^2 +30 To find the value of ′a′, let′s use the given points other than vertex 150 = a(200)^2 + 30 150−30 = 40000a a = ((120)/(40,000)) = (3/(1000)) ∴ y = ((3x^2 )/(1000)) +30 Also, since we′re asked to find a function that gives a length of metal rod needed relative to its distance from the midpoint of the bridge, with each rods have an equal distance to each other, then we must consider another variable ′d′ that represents the equal distance of metal rods relative to its decided quantity and variable ′n′ given as positive integer that divides the distance of midpoint to tower. d = ((200)/n) ⇒ nd = 200 Example: Engineers decided to use 8 metal rodus, then we have d = ((200)/8) = 25 To calculate the length of each rods, let′s use the formula above First rod: y = ((3(0∙25)^2 )/(1000)) +30 = 30 ft. Second rod: y = ((3(1∙25)^2 )/(1000)) +30 = 31.875 ft. Third rod: y = ((3(2∙25)^2 )/(1000)) +30 = 37.5 ft. Fourth rod: y = ((3(3∙25)^2 )/(1000)) +30 = 46.875 ft.

$$\:\mathrm{We}\:\mathrm{know}\:\mathrm{that}\:\mathrm{vertex}\:\mathrm{form}\:\mathrm{of}\:\mathrm{parabola}\:\mathrm{is}\:\mathrm{given}\:\mathrm{as} \\ $$$$\:\mathrm{y}\:=\:\mathrm{a}\left(\mathrm{x}−\mathrm{h}\right)^{\mathrm{2}} +\mathrm{k} \\ $$$$\: \\ $$$$\:\mathrm{From}\:\mathrm{the}\:\mathrm{given}\:\mathrm{diagram}\:\mathrm{of}\:\mathrm{bridge}\:\mathrm{that}\:\mathrm{resembles}\:\mathrm{a}\:\mathrm{parabola}, \\ $$$$\:\mathrm{we}\:\mathrm{have}\:\mathrm{a}\:\mathrm{vertex}\:\mathrm{points}\:\mathrm{of}\:\left(\mathrm{0},\:\mathrm{30}\right)\:\mathrm{and}\:\mathrm{other}\:\mathrm{points}\:\mathrm{due}\:\mathrm{to}\:\mathrm{towers} \\ $$$$\:\mathrm{that}\:\mathrm{supports}\:\mathrm{the}\:\mathrm{parabolic}−\mathrm{shape}\:\mathrm{cable},\:\mathrm{which}\:\mathrm{is}\:\left(\mathrm{200},\:\mathrm{150}\right). \\ $$$$\: \\ $$$$\:\therefore\:\:\mathrm{y}\:=\:\mathrm{a}\left(\mathrm{x}−\mathrm{0}\right)^{\mathrm{2}} +\mathrm{30}\:=\:\mathrm{ax}^{\mathrm{2}} +\mathrm{30} \\ $$$$\: \\ $$$$\:\mathrm{To}\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:'\mathrm{a}',\:\mathrm{let}'\mathrm{s}\:\mathrm{use}\:\mathrm{the}\:\mathrm{given}\:\mathrm{points}\:\mathrm{other}\:\mathrm{than}\:\mathrm{vertex} \\ $$$$\:\mathrm{150}\:=\:\mathrm{a}\left(\mathrm{200}\right)^{\mathrm{2}} \:+\:\mathrm{30} \\ $$$$\:\mathrm{150}−\mathrm{30}\:=\:\mathrm{40000a} \\ $$$$\:\mathrm{a}\:=\:\frac{\mathrm{120}}{\mathrm{40},\mathrm{000}}\:=\:\frac{\mathrm{3}}{\mathrm{1000}} \\ $$$$\: \\ $$$$\:\therefore\:\:\mathrm{y}\:=\:\frac{\mathrm{3x}^{\mathrm{2}} }{\mathrm{1000}}\:+\mathrm{30} \\ $$$$\: \\ $$$$\:\mathrm{Also},\:\mathrm{since}\:\mathrm{we}'\mathrm{re}\:\mathrm{asked}\:\mathrm{to}\:\mathrm{find}\:\mathrm{a}\:\mathrm{function}\:\mathrm{that}\:\mathrm{gives}\:\mathrm{a}\:\mathrm{length}\:\mathrm{of}\:\mathrm{metal}\:\mathrm{rod} \\ $$$$\:\mathrm{needed}\:\mathrm{relative}\:\mathrm{to}\:\mathrm{its}\:\mathrm{distance}\:\mathrm{from}\:\mathrm{the}\:\mathrm{midpoint}\:\mathrm{of}\:\mathrm{the}\:\mathrm{bridge},\:\mathrm{with}\:\mathrm{each}\:\mathrm{rods} \\ $$$$\:\mathrm{have}\:\mathrm{an}\:\mathrm{equal}\:\mathrm{distance}\:\mathrm{to}\:\mathrm{each}\:\mathrm{other},\:\mathrm{then}\:\mathrm{we}\:\mathrm{must}\:\mathrm{consider}\:\mathrm{another}\:\mathrm{variable}\:'\mathrm{d}' \\ $$$$\:\mathrm{that}\:\mathrm{represents}\:\mathrm{the}\:\mathrm{equal}\:\mathrm{distance}\:\mathrm{of}\:\mathrm{metal}\:\mathrm{rods}\:\mathrm{relative}\:\mathrm{to}\:\mathrm{its}\:\mathrm{decided}\:\mathrm{quantity}\:\mathrm{and} \\ $$$$\:\mathrm{variable}\:'\mathrm{n}'\:\mathrm{given}\:\mathrm{as}\:\mathrm{positive}\:\mathrm{integer}\:\mathrm{that}\:\mathrm{divides}\:\mathrm{the}\:\mathrm{distance}\:\mathrm{of}\:\mathrm{midpoint}\:\mathrm{to}\:\mathrm{tower}. \\ $$$$\: \\ $$$$\:\mathrm{d}\:=\:\frac{\mathrm{200}}{\mathrm{n}}\:\:\Rightarrow\:\:\mathrm{nd}\:=\:\mathrm{200} \\ $$$$\: \\ $$$$\:\mathrm{Example}: \\ $$$$\:\mathrm{Engineers}\:\mathrm{decided}\:\mathrm{to}\:\mathrm{use}\:\mathrm{8}\:\mathrm{metal}\:\mathrm{rodus},\:\mathrm{then}\:\mathrm{we}\:\mathrm{have}\:\mathrm{d}\:=\:\frac{\mathrm{200}}{\mathrm{8}}\:=\:\mathrm{25} \\ $$$$\:\mathrm{To}\:\mathrm{calculate}\:\mathrm{the}\:\mathrm{length}\:\mathrm{of}\:\mathrm{each}\:\mathrm{rods},\:\mathrm{let}'\mathrm{s}\:\mathrm{use}\:\mathrm{the}\:\mathrm{formula}\:\mathrm{above} \\ $$$$\:\mathrm{First}\:\mathrm{rod}:\:\:\mathrm{y}\:=\:\frac{\mathrm{3}\left(\mathrm{0}\centerdot\mathrm{25}\right)^{\mathrm{2}} }{\mathrm{1000}}\:+\mathrm{30}\:=\:\mathrm{30}\:\mathrm{ft}. \\ $$$$\:\mathrm{Second}\:\mathrm{rod}:\:\:\mathrm{y}\:=\:\frac{\mathrm{3}\left(\mathrm{1}\centerdot\mathrm{25}\right)^{\mathrm{2}} }{\mathrm{1000}}\:+\mathrm{30}\:=\:\mathrm{31}.\mathrm{875}\:\mathrm{ft}. \\ $$$$\:\mathrm{Third}\:\mathrm{rod}:\:\:\mathrm{y}\:=\:\frac{\mathrm{3}\left(\mathrm{2}\centerdot\mathrm{25}\right)^{\mathrm{2}} }{\mathrm{1000}}\:+\mathrm{30}\:=\:\mathrm{37}.\mathrm{5}\:\mathrm{ft}. \\ $$$$\:\mathrm{Fourth}\:\mathrm{rod}:\:\:\mathrm{y}\:=\:\frac{\mathrm{3}\left(\mathrm{3}\centerdot\mathrm{25}\right)^{\mathrm{2}} }{\mathrm{1000}}\:+\mathrm{30}\:=\:\mathrm{46}.\mathrm{875}\:\mathrm{ft}. \\ $$$$\: \\ $$

Question Number 167221    Answers: 1   Comments: 0

The plane y=1 slices the surface z=arctan(((x+y)/(1−xy))) in a curve C. Find the slope of the tangent line to C at x=2

$${The}\:{plane}\:{y}=\mathrm{1}\:{slices}\:{the}\:{surface}\: \\ $$$${z}={arctan}\left(\frac{{x}+{y}}{\mathrm{1}−{xy}}\right) \\ $$$${in}\:{a}\:{curve}\:{C}. \\ $$$${Find}\:{the}\:{slope}\:{of}\:{the}\:{tangent}\:{line}\:{to} \\ $$$${C}\:{at}\:{x}=\mathrm{2} \\ $$

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