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# 4c

Question | Answer | |||||||
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We can find the vector equation of a three dimensional plane given either three points in | the plane, two vectors and | a point on the plane, a point on the plane and | two points or the equation of the plane in some | formulation. The vector equation of a plane is of the form c=[]=[] | (0 ss)r0(arrow overhead)+μv(arrow overhead) = | λw(overhead). By some means obtain both | v(arrow overhead) and w(arrow overhead). If you have three points in the plane A,B and C then you can take r0 to be any of | (arrows overhead)OA,OB or OC with (arrow overhead)v = |

Partial fractions: a question may seek for the the expression of fractions such as 1/x²-2x-8 in the form | A/Cx+F + B/Ex+F [these are partial fractions); as x²+2x-8 factorises in (x+4)(x-2): 1/x²+2x-8 = 1/(x+4)(x-2) = A/ | x+4 + B/x-2; to get 1=A(x+o)+B(x+p), multiply through by | (x+4)(x-2) to get 1=A(x-2)+B(x+4) after cancellation. Eliminate A by having x equal to | 2 to get b= | 1/6, then eliminate B by having x equal to | -4 to get A= | -0.5. Thus 1/x²+2x-8 = | -1/2(x+4) + 1/6(x-2). |

If we have a denominator which includes an quadratic term which does not factorise, include, in the answer a term | Dx+E/Ax²+Bx+C. In general we have one term for every term in the denominator which does not factorise, and for that term, the highest exponent of x in the numerator is less than the highest exponent of x in the denominator by | 1. When trying to find the value of variables, variables of which the value is known may be replaced by their | value. | |||||

To put 5(x + 2)/(x+1)(x+6) into partial fractions using the cover up method: cover up the x + 6 and substitute, into what's left; | -6, giving 5(-6 + 2)/(-6+1) = -20/-5 = 4. This tells you that one of the partial fractions is | 4/(x + 6). Now cover up (x + 1) and substitute, into what's left: x + 6). Now cover up (x + 1) and substitute, into what's left: | -1, to discover that the other partial fraction is | 1/(x + 1). Note: you can split partial fractions into four groups of problems: ones with: denominators of the form '(ax+b)(cx+d)', denominators with a repeated factor (i.e. | [ax+b][cx+d]²), denominators with one quadratic factor (i.e. [ax+b][cx²+d]); the fourth type is a 'top-heavy' algebraic fraction, which occurs when the highest power of the highest power of the numerator is ... | larger than or equal to the largest power of the denominator. | ||

If the final fraction has a denominator of the form (ax+b)(cx+d) (let the numerator be f[x]), the partial fractions are of the form | A/ax+b + B/cx+d, we just need to find | A and B. Next, [] the two fractions | add--, giving | A(cx+d)+B(ax+b)/(ax+b)(cx+d); as it has the same denominator as f(x)/(ax+b)(cx+d), put f(x) = | A(cx+d)+B(ax+b). To find A, put x = | -b/a, then | rearrange; to find B, put x = | -d/c (then rearrange). Note: substitution of letters for their values is allowed: example: if A=3, Ax² + B(x-1) equals 3x² + B(x-1). |

When the final fraction has a denominator of the form (ax+b)(cx+d)², write the partial fractions as | A/ax+b + B/(cx+d)² + C/cx+d. Next, [] the fractions | add--, and equate the top lines. You can then find values of the letters, by changing the value of x; but if that doesn't work for a variable, pick an x^n (advisably the [] one multiplied by the variable you seek) | largest--, then replace each term with the amount of [] it has | x^n's (thereafter rearrange) example: knowing A=2 and B=1, x²+12x-7 = A(x-1)²+B(x+2)+C(x+2)(x-1) gives 1=A+C: 1=2+C: C=-1; alternatively, using x, 5x²+5x+2=Ax²+C(x-2) gives | 5=C. | |||

If the denominator of the final fraction is in the form (ax+b)(cx²+d) (let the numerator be f), write the partial fractions as | A/(ax+b) + Bx+C/(cx²+d). Next, [] the fractions | add--, and equate the top lines. Finally, find the values of [...] | A, B and C. | |||||

You may get an improper algebraic fraction f(x)/g(x) where the highest indice on f(x) is greater than or equal to the highest on g(x). In such a case, [] to get a [] and a [] | divide to get a quotient (q) and a remainder (r), hence express f(x)/g(x) = | q + r/g(x). Note: you may be able to split r/g(x) into partial fractions. | ||||||

In general if one expands (1+x)^n where |x|<1, then the terms of the expansion get smaller and smaller, and so (often) can be ignored beyond the first [] terms | three--. If n is negative or fractional, simplify what you want to expand to just | (1+x)^n, in which case the expansion, that is, (1+x)^n = | 1 + nx + n(n-1)x²/2! + n(n-1)(n-2)x^3/3! + ... . If the expansion given is (1+f[x]), where f(x) is not x, in the equation, replace x with | f(x); in the expansion (1+f[x])^n (f[x] may equal x), the expansion, using the given formula, the expansion is valid for | -1<f(x)<1. If the expansion is given as (a+x)^n, where a is an integer greater than 1, write the expansion as []*[] | a^n * (1+[x/a])^n. | ||

The movement from A to B may be given the vector | AB (arrow overhead), while the movement from B to A may be given by the vector | BA (arrow overhead). AB (arrow overhead) = BA (arrow overhead) * [] | -1 --. A vector maybe given using three numbers (2 in 2D) in a column, generically | (x,y,z); -(x,y,z) = | (-x,-y,-z). Note: a single letter may be used to label a vector; the letter is often shown in lowercase and in bold type; YOU should write 'a' with an [] in place of 'a' to show that you mean a vector | underbar--. To add one vector (a [underbar]) to another (b [underbar]), position them so that one [] where the other [] | starts--finishes--, a+b can hence be represented by the vector from the start of a to the end of b. | |

The numbers used in a column vector are the c[] of the vector | component--, adding vectors corresponds to adding their | components. | ||||||

The magnitude of a vector v, written as []v[] or []v[] | |v| or ||v|| (so as to not confuse it with modulus), is the [] of the vector | length--. | ||||||

If the vector v extends into the region of the graph with negative values of x, and the angle between v and the negative x-axis is θ, then the angle between v and the direction of the positive x-axis is | 180-θ. [#A40#]. | |||||||

Any lowercase letter may be used to label a vector, however, the use of i and j is reserved for two special vectors. i is a [] vector | unit-- (i.e. it has length of | 1) in the direction of the | positive x axis. j is a [] vector | unit--, in the direction of the | positive y axis. In general, if v = (a,b), then (using i and j) v = | ai + bj. Note: vector expressions in terms of i and j may be simplified in the usual way. In general, where k is a constant and v is a vector, kv is parallel to | v, but k has to be [], because for vectors to be parallel, they have to have the same [] | positive--direction--. |

Let the origin be O: the vector from O to A (a) is the [] vector of A | position--. Let there be another vector, but from O to B (b). Hence AB (arrow overhead) = | b-a. | ||||||

The unit vector in the direction of (vector) v is | v/|v| (i.e. divide v by its magnitude). Example: v=3i+4j, |v|=5, thus the unit vector is v/5 = (3i-4j)/5. Note if you are asked to find a vector of magnitud a in the direction of v, multiply the unit vector (in the direction of v) by | a. | ||||||

The unit vector, the vector component in the third dimension is | 'k'. If a = (x,y,z): (using the unit vectors) a = | xi+yj+zk. | ||||||

A vector that isn't fixed to the origin is called a | free vector; a free vector has the same [] and [] as it would if it were a position vector | magnitude and direction--, but its tail/start (which would have been the origin) can be | anywhere. | |||||

If vector a = (x,y,z), |a| = | √(x²+y²+z²). | |||||||

In general if a and b are the position vectors of two points A and B, then the position vector of the midpoint of AB (M) is m = | (a+b)/2 (=[b-a]/2 + a). | |||||||

Let a = (xa,ya,za) and b=(xb,yb,zb) (a,b ss), then the angle θ, at the origin - between a and b, is given by | cos θ = (xaxb + yayb + zazb)/|a|*|b|. | |||||||

The scalar product of two vectors, a and b, is written as | a.b; where θ is the angle between them, the scalar product is defined as the value of either | |a||b|cosθ or xaxb+yayb+zazb (a,b ss); note: the former found through rearrangement of cos θ = (xaxb + yayb + zazb)/|a|*|b|, a = (xa,ya,za) and b=(xb,yb,zb) (a,b ss). | ||||||

The scalar product of vectors a and b can be given by a.b = |a||b| cosθ. If θ=90°, then the vectors are perpendicular, at which point cosθ = | 0. Hence a and b are perpendicular vectors if either [...] or [...] equal 0. | |a||b|cosθ or a.b(=xaxb+yayb+zazb [a,b ss]). | ||||||

In general, you can write any straight line in vector form if you know a [] on the line | point--(p1,p2)/(p1,p2,p3), and a [] which is [] to the line | vector--parallel--(d1,d2)/(d1,d2,d3). Then the equation is given in 2d by r = [...] and in 3d by [...] | (p1,p2) + λ(d1,d2)--(p1,p2,p3) + λ(d1,d2,d3); changing [] will take you to any position, [], on the line | λ--r--. Note: r is the [] [] of any point on the line | position vector--. | |||

(1,2,3 ss) If you have a line with equation of the form (p1,p2,p3) + λ(d1,d2,d3), and a point (a1,a2,a3), you can the value of λ at that point by equating a | line (e.g. you can equate the first line to get p1 + λd1 = a1 and knowing p1, d1 and a1, deduce λ). | |||||||

If a line passes through the x axis, then there must be a coordinate on the line of the form | (a,0,0); with the y axis, there would be a coordinate on the line of the form | (0,a,0); and with the z axis, there would be a coordinate of the form | (0,0,a). To check that a line passes (for example) the x-axis, let []=[] | (p1,p2,p3) + λ(d1,d2,d3) = (a,0,0), first equate the [] []s | y coordinates, then rearrange for | λ, then equate the [] [] and rearrange for [] | z coordinates--λ--; if both values are the same, then there is a single value for which 'a' exists, thus it does pass the x-axis. Note: the technique is easily adaptable. | |

(1,2,3 ss) If the general vector equation of a line, r=(p1,p2,p3) + λ(d1,d2,d3), is given in terms of i, j and k, then the equation becomes | r = p1i + p2j + p3k + λ(d1i+d2j+d3k). | |||||||

With the vector equation of a line, the vector in the equation that's parallel to the line can go in either | direction, the scalar (usually written as λ) can be changed to get to any point on the line. Thus (a,b,c) + λ(d1,d2,d3) and (d,e,f) + μ(nd1,nd2,nd3), where n is a constant, are | parallel. | ||||||

2D lines either [], are [] or are the [] line | intersect--parallel--same--. However, in 3D, two lines can not parallel but will never intersect; they are called | skew lines. | ||||||

If you want to see if two 3D lines intersect: equate the [] and [] coordinates | x and y coordinates, to get simultaeneous equations, then solve to find out the value of the | scalars. Then equate the | z coordinates, then [] in the [...] | substitue in the value of the scalars. If the two sides are equal, then the lines | intersect, otherwise, they don't. Note: if they intersect you can input the value of either of the scalars into the equation of the line it belongs to to get the coordinates of the intersection. | |||

To find the angle between two lines: find the angle between the two | directions, thus find the [] [] | direction vectors. | ||||||

Let there be a line on a graph - given a point, p, the shortest distance to the line is the length of the line [] which [] P to the line | line segment--joins-- , that is, where the line segment is at [] [] to the line | right angles--. | ||||||

(1,2,3 ss) To find the shortest distance from the origin to a line r = (p1,p2,p3) + λ(d1,d2,d3). The first thing to do is find the point A, where the perpendicular line hits | the line r. A point A is on the line r, its coordinates will satisfy the line's [] for some value of λ | equation--λ--. Therefore the position vector of the point A must be of the form a = | (p1+λd1,p2+λd2,p3+λd3). As a is perpendicular to the line r, the scalar product of their directions must be | 0, hence [].[] = 0 | (p1+λd1,p2+λd2,p3+λd3).(p1,p2,p3) = 0. Find the value of | λ. To find the coordinates of A, substitute the value of [] into [] | λ into r. The distance is the [] of the vector a | magnitude--. |

When trying to find the shortest distance from any point A(a1,a2,a3) to a line r=(p1,p2,p3) + λ(d1,d2,d3) (i.e. to a point B[b1,b2,b3]).You want to find the vector [] | AB (arrow overhead). As B lies on line r, b = OB (arrow overhead) = ([]) | (p1+λd1,p2+λd2,p3+λd3), for some value of | λ. Thus AB = | (p1-a1+λd1,p2-a2+λd2,p3-a3+λd3). Use the fact that AB .[] = 0 | (d1,d2,d3)--, to find the value of | λ, which is then to be inputed into | AB. To get the distance, find AB's | magnitude. |

To convert the equation of a 2D line from vector to cartesian form, you can: add the [] and [] values seperately | x and y--, and set them equal, respectively, to | x and y to get two [] [] | simultaeneous equation, then create an equation including [] and [] as the only variables (this can be achieved by getting rid of []) | x--y--λ--. | ||||

To convert the equation of a line from vector to cartesian form: add the [], [] and [] values seperately | x, y and z--, and set them equal, respectively, to | x, y and z. Then, in each equation, rearrange to make | λ the subject. Hence [] the three equations | equate--, ignoring | λ (such is the 3D cartesian equation of the line). Looking at the format, you can quickly write down the direction of the line: its components are the | denominators (note: the denominator of a function of x is the x component etc.), and you can get a point on the line by, for each part, subtracting the variable and multiplying the rest by | -1 (note: the part with the x variable maps onto the x-axis). If there are 0λ's for a certain line, write the others as usual then write it later, following an '[]' | 'and' (e.g. x=2+3λ, y=7, z=4+5λ gives x-2/3=z-4/5 and y=7). |

(1,2,3 ss) In general, a line in vector form written as r = (p1,p2,p3) + λ(d1,d2,d3) has cartesian form | x-p1/d1 =y-p2/d2 = z-p3/d3. | |||||||

For the vector equation of a plane you need the [] [] of a point on the plane | position vector--(i.e.[a1,a2,a3]), and 2 non-[] [] [] which [] in the plane | parallel direction vectors--lie-- (i.e. [p1,p2,p3] and [q1,q2,q3]). Where (a1,a2,a3) is the position vector of a point on the plane and (p1,p2,p3) and (q1,q2,q3) are non-parallel direction vectors lying in the plane: the vector equation of a plane is r = | (a1,a2,a3) + λ(p1,p2,p3) + μ(q1,q2,q3). | |||||

To find where the line l meets the plane p (with the equations in vector form: equate the [], [] and [] values | x, y and z--. Solve the consequent simultaeneous equations to find the value of the | scalars. Having found the value of the scalar used in l, | substitute it into l, to get the [] of the intersection points | coordinates--. | ||||

To convert the equation of a plane to cartesian form: individually [] the values of x, y and z | add--, and set them equal to, respectively, [...] | x, y and z. To get three [] equations | simultaeneous--. Then, eliminate the | scalars (μ and λ). Then you can rearrange to get an equation of the form | ax+by+cz+d=0. | |||

(1,2,3 ss) Where a is the position vector of a point on the plane and n is a vector perpendicular to the plane, where n = (n1,n2,n3): the equation of a plane in cartesian form is given as | n1x+n2y+n3z+d=0, where d = | -a.n. Though sometimes, by using r as a general point (x,y,z) and knowing n, it is shortened to | r.n=d. | |||||

When trying to find the intersection of a line with a plane (with its equation given in cartesian form): in the line, [] the values of x, y and z | add--, and set them equal to, respectively, | x, y and z. Then substitute those values, of x, y and z into | the plane equation to find the value of | λ. Finally put the found value of λ into | the equation of the line, to get the | point of intersection. | ||

A plane has equation ax+by+cz+d=0. To find the shortest distance from the point A(x1,y1,z1) to this plane, we want to find the distance from A to the point where the [] line from A intersects the plane | perpendicular-- (point P, which is also perpendicular to the plane, thus the distance AP). To do this we need to find | P. Thus find the [] of the [] through A and P | equation--line--, then see where it intersects the [] to get [] | plane--P--. Now AP is perpendicular to the plane ax+by+cz+d=0; another vector perpenducular is [] = [] | n = (a,b,c) (i.e. n1,n2,n3). Thus the line through A and P passes through []([]) and has direction [], hence the equation of the line is r = [] | A(x1,y1,z1)--(a,b,c)--(x1,y1,z1) + λ(n1,n2,n3)--. Thereafter find where the line | intersects with the plane. Now find the vector [] and hence [] | AP (arrow overhead)--|AP*| *(arrow overhead)--. |

To check if a point lies on a line (with equation in vector form): [] the line's equation and the position vector of the point | equate--. Then find [] using the [] component | λ--x--. Substitute the value of λ into the [] of [] and [] | components of y and z. If the value works, then | the point lies on the line. | ||||

To check if a point A(x1,y1,z1) lies on a line (with equation in cartesian form): substitute [] into the [] [] | A--line equation--, taking x=[], y=[], z=[] | x1--y1--z1--. If the equation is correct: | the point lies on the line. | |||||

To check if a point A(x1,y1,z1) lies on a plane (with equation in cartesian form): substitute [] into the [] [] | A--plane equation--, taking x=[], y=[], z=[] | x1--y1--z1--. If the equation is correct: | the point lies on the plane. (If you are given a plane with an equation in vector form, you can convert it to cartesian form). | |||||

If you want to see if a point lies on a plane: equate the [] and [] coordinates | x and y coordinates, to get simultaeneous equations, then solve to find out the value of the | scalars. Then equate the | z coordinates, then [] in the [...] | substitute in the value of the scalars. If the two sides are equal, then the | point lies on the plane. | |||

To see if a line lies completely within a plane, first show that they are | parallel: if they are the direction perpendicular to the plane (n1,n2,n3) and the direction of the line (d1,d2,d3) will be | perpendicular, thus have a [] [] of 0 | scalar product--. Thereafter, show also that the line and the plane have a common | point (i.e. show a point in the line lies on | the plane/a point on the plane lies on | the line). | ||

If a line does not lie completely within a plane: then they are either just | parallel or they [] at one [] | intersect--point--. | ||||||

Two planes are parallel if their perpendicular vectors are | parallel. Two planes are perpendicular if their perpendicular vectors are | perpendicuar. | ||||||

The angle between two planes is the same as the angle between their | perpendicular vectors. | |||||||

Perpendicular vectors are also known as n[] | normals. | |||||||

|Parametric function| Parametric curves: in these cases, it is sometimes easier to write the equations for x and y in terms of another | variable, t. To sketch the graph of x=f(t), y=g(t), for each value of t (usually within a given interval) work out the value of [] and [] | x and y; these values pair up to be | coordinates, which are then | plotted. If f(t)=1/t and x or y = gf(t), then you may get a/multiple | asymptote(s). Generally, due to the nature of the functions of t, the graph may only be valid for (a) certain | interval(s). | ||

To translate a parametric function by a vector (a,b) x=f(t) becomes [] and y=g(t) becomes [] | x=f(t)+a--y=g(t)+b--. | |||||||

To stretch a parametric function by a factor of a in the x direction, x=f(t) becomes | x=af(t); to stretch a parametric function by a factor of a in the y direction y=g(t) becomes | y=ag(t). | ||||||

To reflect a parametric function in the y axis, x=f(t),y=g(t) becomes | x=-f(t),y=g(t) ; to reflect a parametric function in the x axis, x=f(t),y=g(t) becomes | x=f(t),y=-g(t). | ||||||

To find where a parametric function crosses the x axis, make y= | 0 and solve for | t. The value(s) of t can them be put into the original | equations. | |||||

To find where a parametric function crosses the y axis, make x= | 0 and solve for | t. The value(s) of t can then be put into the original | equations. | |||||

To convert a parametric function to cartesian form, from x=f(t), get t= | f^-1(x), then put that value of t into | g(t), to get y=h(x). Alternatively, if y=g(t) is less complex, from there, get t= | g^-1(y), then put that value of t into | f(t), to get x=i(y), which can be rearranged to make y the subject. If given two parametric functions, x=f(t) and y=g(t), and asked to convert them to cartesian form: you are allowed to [], [], [] and [] them | add, subtract, multiply and divide-- (i.e. treat them like [] equations | simultaeneous--). Example: x = t + 1/t, y = t - 1/t: adding gives (1)x + y = 2t, subracting: x - y = 2/t; (1)*(2) = (x+y)(x-y)=4;x²-y²=4 | ||

The pair of parametric equations (using θ as the variable) that define a circle of radius 1, with centre at the origin is | x=cosθ, y=sinθ. A circle with radius r, and centre (a,b) is given in parametric form as | x=rcosθ+a, y=rsinθ+b, while in cartesian form it is given as | (x-a)²+(y-b)²=r². An ellipse centre (0,0), with a x-radius of a and a y-radius of b, is given in parametric form as | x=acosθ, y=bsinθ, while in cartesian form it's given as | x²/a² + y²/b² = 1. Note: when trying to find the value of θ where the graph cuts an axis, be aware that there may be multiple solutions (e.g. sin θ = 0.5: θ=30°, θ=180°-30°). | |||

With a parametric equation x=f(t), y=g(t), dy/dx = [], though it's usually written as | dy/dt * dt/dx--dy/dt ÷ dx/dt--where dx/dx≠0--, this is through use of | the chain rule. | ||||||

To find dy/dx when x is a fuction of y (x=f[y]) or there are several occurrence of both x and y (or if one can't rearrange the equation to the form y=f(x) etc.), in these cases it may be best to differentiate | implicitly. To do this, differentiate with respect with to | x; afterwards you can rearrange for | dy/dx; note: due to the chain rule - d(f[y])/dx = | d(f[y])/dy * dy/dx. | ||||

When wanting to find the tangent or normal to a curve with an equation where y is not explicitly a function of x: in these cases typically: differentiate | implicitly and find dy/dx as a function of | x and y, and then get the values of x and y from a [] ([]) | point (x,y), to find, at that point, the | gradient. You may find the equation of the LINE with the equation: | y-y1 = m(x-x1) (1 ss). | |||

The trapezium rule: for a region with n strips of width h, we have: Area ≈ | h/2(y0 + yn + 2(y1 + y2 + ... + y[n-1])). | |||||||

|Integration by Inspection| If you want to integrate a*u^n, where u is a function of x, differentiate | u^(n+1) (let the answer be b*u^n); you can hence deduce that the integral of a*u^n equals | au^(n+1)/b. This method only works if the part multiplied by u^n (a) is a multiple of the [] of u^n | differential--(i.e. the differential multiplied by a constant). | |||||

∫a/x dx = | aln|x| +c; ∫e^(ax) dx = | e^(ax)/a +c; ∫1/ax+b dx = | ln|ax+b|/a +c; ∫f'(x)/f(x) = | ln|f(x)| +c, hence, where h(x)=nf'(x), where n is greater or less than 1, ∫h(x)/f(x)= | nln|f(x)| +c. | |||

∫tan(x) dx = | ln|sec x| +c | |||||||

To make the integration of harder fractions easier: convert them to [] fractions | partial--. | |||||||

∫cos(ax) dx = | sin(ax)/a +c. ∫sin(ax) dx = | -cos(ax)/a +c. ∫sec²x dx = | tan(x) +c; ∫cosec²x dx = | -cot(x) +c. | ||||

d(cosec x)/dx = | -cosec(x)cot(x); d(sec x) = | sec(x)tan(x). | ||||||

Integration of trigonometric functions (especially ones involving exponents greater than one) may be simplified using i[] | identities. | |||||||

With identities, if f(x)=g(x), f(ax)= | g(ax). | |||||||

If asked to integrate trigonometric functions of the form ∫sin(ax)cos(bx) dx: through use of addition formulae*, we can conclude that 2sinAcosB= | sin(A+B)+sin(A-B). *sin(A+B)=sinAcosB+cosAsinB; sin(A-B)=sinAcosB-cosAsinB; sin(A+B)+sin(A-B) = sinAcosB+sinAcosB=2sinAcosB | |||||||

Where 'a' is a constant, an integral of the form ∫1/(a²+x²) dx can be solved using the substitution | x = atan(u); in general ∫1/a²+x² dx = | tan^-1(x/a)/a +c. | ||||||

Where 'a' is a constant, an integral of the form ∫1/√(a²-x²) dx can be solved using the substitution | x = asin(u); in general ∫1/√(a²-x²) dx = | sin^-1(x/a) +c. [#A41#] | ||||||

∫f(x) dx = ∫(f[x]* | dx). Example: ∫cos²x dx/cosx = ∫cosx dx. | |||||||

When integrating some products, inegration by parts may need to be done more than | once. If you have integrated a product to get the integral, which includes one term that is the integral of a product, on that term, you can use | integration by parts. With integration by parts: ∫uv' dx = | uv - ∫vu' dx. | |||||

If I is the integral you want, and I = f(x) + aI, simply rearrange to find the value of | I. | |||||||

An equation which involves a derivative (e.g. dy/dx) is called a | differential equation. | |||||||

If asked to solve an equation of the form dy/dx = f(y), to start, put all the []-terms on [] side | y--one--. Then [] both sides | integrate-- with respect to | x (considering that dy/dx * dx = | dy). Afterwards, make the subject | y. Note: where c is the arbitrary constant: f(c) can be replaced with another representation of a constant (e.g. A). This method is known as the method of | seperation of variables. Example: dy/dx = e^-2y;... ∫e^2y dy = ∫ dx: e^2y/2 = x +c; e^2y = 2x + A; y = ln(2x+A)/2. Note: c can be replace by f(c), where f(c) can yield any constant in the appropriate range (e.g. in some cases c may be replaced by ln c). | ||

To solve a differential equation, get rid of the | derivative. | |||||||

For any growth or decay problem, the solution to dy/dx = ay, where 'a' and 'k' are constants, is | y = ke^(ax); if a<0, then it's a [] problem | decay--, and if a>0, then it's a [] problem | growth--. | |||||

(1,2 ss) If asked to find the particular solution for a differential equation for which y=y1 and x = x1; after getting rid of the derivative, to get an equation with y and x as terms: input | y=y1 and x=x1 to find (usually) | c (the arbitrary constant) (,but it could be any other algebraic term in the equation). | ||||||

When integrating, with integrals, it is permitable to have the arbitrary constant on one | side (e.g. ∫f(x) dx = ∫g(x) dx +c). | |||||||

If dN/dt is proportional to N, then dN/dt = | kn, where k is a | constant. | ||||||

With differential equation (esp. real life examples): let the two variable be x and y, to start, it is usually advisable to solve the | differential equation; rearranging usually puts it in the form y=Ae^kx, where A and k are constants. A (though it can be done for k) is often found by substituting in, at a certain point, the values | of x and y (usually where x=0). The other constant, which has not been found, can be found through another | substituton (of the values of x and y). |