P4 Pure Mathematics 4

Example:Express 5 − x + 2x²(1 − x)(1 + x²) in partial fractions.

Solution:The degree of the numerator, 2, is less than the degree of the denominator, 3, so we do not need long division and can write

Example:Express x² − 7x + 22(2x − 1)(x − 3)² in partial fractions.

Solution:The degree of the numerator, 2, is less than the degree of the denominator, 3, so we do not need long division and can write

Example:Express x³ + x² − 9x − 3x² − 9 in partial fractions.

Solution:Firstly the degree of the numerator is not less than the degree of the denominator so we must divide top by bottom.

Factorise to give x² – 9 = (x – 3)(x + 3) and write

Example: x = 2 + t,  y = t² – 3 is the parametric equation of a curve. Find

(i)the points where the curve meets the x-axis,

(ii)the points of intersection of the curve with the line y = 2x + 1.

Solution:

Example: Find whether the curves x = 2t + 3,  y = t² – 2 and x = s – 1,  y = s – 3 intersect.

If they do give the point of intersection, otherwise give reasons why they do not intersect.

Solution:

If they intersect there must be values of t and s (not necessarily the same), which make their x-coordinates equal, so for these values of t and s

The y-coordinates must also be equal for the same values of t and s

Or  curves intersect at  t = –1 giving  (1, –1).  Check  s = 2 giving  (1, –1).

Example: Find the Cartesian equation of the curve given by y = t² – 3,  x = t + 2.

Solution:

which is the Cartesian equation of a parabola with vertex at (2, –3)

Example: Find the Cartesian equation of the curve given by

y = 3sin t + 2,  x = 3cos t – 1.

Solution: Re-arranging we have

which is the Cartesian equation of a circle with centre  (–1, 2)  and radius 3.

Example: Find the Cartesian equation of the curve given by y = 3tan t,  x = 4sec t.

Hence sketch the curve.

Solution: Re-arranging we have

Example: Find the area between the curve  y = t² – 1,  x = t³ + t, the  x-axis and the lines

x = 0  and  x = 2.

Solution: The area is  A = 2∫0 y dx .

Firstly we need to find  y  and  dxdt in terms of  t.

Secondly we are integrating with respect to  t  and so the limits of integration must be for values of  t.

Example: Expand (1 + 3x)^–2, giving the first four terms, and state the values of x for

which the series is convergent.

Solution:

This series is convergent when |3x| < 1   ⇔   |x| < 1/3.

Example: Use the previous example to find an approximation for  10⋅9997² .

Solution: Notice that  10⋅9997²  =  0⋅9997 ^–2  =  (1 + 3x)^–2 when x = – 0⋅0001.

So writing  x = – 0⋅0001  in the expansion    1 – 6x  +  27x²  –  108x³

0⋅9997 ^–2  ≈  1 + 0⋅0006 + 0⋅00000027 + 0⋅000000000108  =  1⋅000600270108

The correct answer to 13 decimal places is  1⋅0006002701080not bad eh?

Example: Expand (4 − x)¹², giving all terms up to and including the term in x³, and state

for what values of x the series is convergent.

Solution: As the formula holds for  (1 + x)ⁿ we first re-write

This expansion converges for  |x4| < 1   ⇔   |x| < 4.

Example: Find the expansion of   3x − 1x² − x − 6   in ascending powers of x up to x².

Solution: First write in partial fractions

Examples:   The following examples use the chain rule (or implicit differentiation)

Example:   Find the gradient of, and the equation of, the tangent to the curve

x² + y² – 3xy  =  –1    at the point   (1, 2).

Solution:   Differentiating   x² + y² – 3xy  =  –1   with respect to   x   gives

Equation of the tangent is   y – 2   =   4(x – 1)

Example:   A curve has parametric equations   x  =  t² + t,    y = t³ – 3t.

(i)    Find the equation of the normal at the point where   t = 2.

(ii)   Find the points with zero gradient.

Solution:

(i)    When   t  =  2,   x = 6   and   y = 2.

Thus the gradient of the normal at the point   (6, 2)   is   −59

and its equation is   y – 2   =   −59 (x – 6)     ⇒     5x + 9y = 48.

(ii)   gradient = 0 when   dydx  =  3t² − 32t + 1  =  0

Example:     Find the derivative of   y = 3^x2.

Solution:    dydx = 3^x2 ln 3 × d(x²)dx   = 3^x2 ln 3 × 2x

Example:     Find the derivative of   y = 5^{sin x}.

Solution:    dydx = 5^{sin x} ln 5 × d(sin x)dx   = 5^{sin x} ln 5 × cos x

Example:   A spherical snowball is melting at a rate of 96  cm³ s^–1 when its radius is 12  cm.

Find the rate at which its surface area is decreasing at that moment.

Solution:   We know that   V = 43 πr³   and that   A = 4π r².

Using the chain rule we have

Using the chain rule again

Example:   The mass of a radio-active substance at time   t   is decaying at a rate which is

proportional to the mass present at time   t. Find a differential equation connecting the

mass   m   and the time   t.

Solution:   Remember that   dmdt   means the rate at which the mass is increasing so in this

case we must consider the rate of decay as a negative increase

Example:    Find   ∫x³ + 3xx²  dx

Solution:    ∫x³ + 3xx²  dx

Example:    Find   ∫ cot² x  dx.

Solution:    cot² x  =  cosec² x  –  1

Example:    Find   ∫ sin² x  dx.

Solution:    sin² x  =  ½ (1 – cos 2x)

Example:    Find   ∫ sin² 3x  dx.

Solution:    sin² 3x  =  ½ (1 – cos 2 × 3x)  =  ½ (1 – cos 6x)

Example:    Find   ∫ sin 3x cos 5x  dx.

Solution:    Using the formula    2 sin A cos B  =  sin(A + B)  +  sin(A – B)

This formula is NOT in the formula booklet – you can use the formulae for sin(A ± B) and add them

Example:    Find   ∫ sin⁴ 3x cos 3x  dx.

Solution:    ∫ sin⁴ 3x cos 3x  dx

If we think of  u = sin 3x , then the integrand looks like  u⁴ dudx  if we ignore the constants, which would integrate to give  ¹⁄₅ u⁵

so we differentiate  u⁵  =  sin⁵ 3x

to give  ddxsin⁵ 3x  =  5sin⁴ 3x× 3 cos 3x  =  15 sin⁴ 3x cos 3x

which is 15 times what we want and so

Example:    Find   ∫x2x² − 3  dx

Solution:    ∫x2x² − 3  dx

If we think of  u = 2x² – 3 , then the integrand looks like  1u dudx  if we ignore the constants, which would integrate to  ln |u|

so we differentiate  ln |u|  =  ln |2x² – 3|

to give  ddxln |2x² − 3|  =  12x² − 3 × 4x  =  4x(2x² − 3)

which is 4 times what we want and so

In general    ∫f′(x)f(x) dx = ln|f(x)| + c

Example:    Find   ∫xe^2x2  dx

Solution:    First consider  ddxe^2x2 = 4x e^2x2 , which is 4 × the integrand

Example:    Find   ∫5^3x dx

Solution:    We know that  ddx5^3x = 5^3x ln 5 × 3,     using the chain rule

Example:    Find   ∫6xx² + x − 2  dx

Solution:    First express   6xx² + x − 2   in partial fractions.

Example:    Find   ∫ x√3x² − 5  dx   using the substitution   u = 3x² – 5.

Example:    Find   ∫ 11 + x²  dx   using the substitution   x = tan u.

Example:    Find   ∫ 3x√x² − 4  dx   using the substitution   u² = x² – 4.

Example:    Find   6∫2 x√3x − 2  dx   using the substitution   u = 3x – 2.

Example:    Find   I = ∫ √16 − x²  dx, and express your answer in as simple a form as possible.

Solution:    This is of the form   ∫ xⁿ√16 − x²  dx   where   n = 0, an even number

Example:    Find   ∫ x sin x  dx

Example:    Find   ∫ ln x  dx

Example:    If   x = tan t   and   y = sin t, find the area under the curve from   x = 0   to   x = 1.

Solution:    The area   =  ∫ y dx   for some limits on x  =  ∫ y dxdt  dt   for limits on t.

Example:    The curve shown has parametric equations

x = t² − 1 ,   y = t³.

The region, R, between x = 0 and x = 8 above the x-axis is rotated about the x-axis through 2π radians. Find the volume generated.

Solution:    V  =  8∫0 π y²  dx.

Example:    Solve the differential equation   dydx  =  3y + xy.

Solution:    dydx  =  3y + xy  =  y(3 + x)

Example:    A radio-active substance decays at a rate which is proportional to the mass of the substance present. Initially 25 grams are present and after 8 hours the mass has decreased to 20 grams. Find the mass after 1 day.

Solution:    Let   m grams be the mass of the substance at time   t.

When   t = 0,   m = 25   ⇒   A = 25

When   t = 8,   m = 20

So when   t = 24,     m = 25e^{–24 × 0⋅027892943} = 12⋅8.

Answer    12⋅8 grams after 1 day .

Example:   Which two of the following vectors are parallel?

Solution:   6−3 = −32 × −42 and so 6−3 is parallel to −42

but 21 is not a multiple of −42 and so cannot be parallel to the other two vectors.

Example:   Find a vector of length 15 in the direction of 4−3.

Solution:   a = 4−3 has length a = |a| =  √4² + 3² =  5

and so the required vector of length 15 =  3 × 5   is 3a = 3 × 4−3 = 12−9.

Example:   If a and b are not parallel and if

Solution:   Since a and b are not parallel, the coefficients of a and b must ‘balance out’

⇒ 2 = α – 5 ⇒  α = 7 and 1 + β = 3 ⇒  β = 2.

Example:   Find a unit vector in the direction of   −125.

Solution:   a = −125   has length   |a| = a = √12² + 5² = 13,

and so the required unit vector is   ¹/₁₃ × a = ¹/₁₃ × −125 = −1213513.

Example:   A, B are the points (2, 1) and (4, 7). M lies on AB in the ratio 1 : 3. Find the coordinates of M.

Solution:   AB = 26

AM = 14AB = 1426 = 0∙51∙5

OM = OA + AM = 21 + 0∙51∙5

⇒     OM = 2∙52∙5

⇒     M is (2⋅5, 2⋅5)

Example:   In a triangle OBC let M and N be the midpoints of OB and OC. Prove that BC = 2 MN and that BC is parallel to MN.

Solution:   Write the vectors OB as b, and OC as c.

Then OM = ½ OB = ½ b

and ON = ½ OC = ½ c.

To find MN, go from M to O using  –½ b   and then from O to N using  ½ c

Also, to find BC, go from B to O using  –b   and then from O to C using  c

But MN = –½ b + ½ c = ½(c – b) = ½ BC

and BC is twice as long as MN.

Example:   P lies on OA in the ratio 2 : 1, and Q lies on OB in the ratio 2 : 1. Prove that PQ is parallel to AB and that PQ = ²/₃ AB.

Solution:   Let  a = OA,   and  b = OB

and  PQ = −OP + OQ

Example:   Find the values of λ so that a = 3i – 2λj + 2k  and  b = 2i + λj + 6k  are perpendicular.

Solution:   Since a and b are perpendicular  a · b = 0

Example:   Find a vector which is perpendicular to a, 1−12, and b, 311.

Solution:   Let the vector c, pqr, be perpendicular to both a and b.

Adding these equations gives  4p + 3r = 0.

Notice that there will never be a unique solution to these problems, so having eliminated one variable, q, we find p in terms of r, and then find q in terms of r.

and we choose a sensible value of r = 4 to give  −354.

Example:   Find the angle between the vectors

Solution:   First re-write as column vectors (if you want)

a = 4−52    and  b = −12−3

a = |a| = √4² + 5² + 2² = √45 = 3√5 ,   b = |b| = √1² + 2² + 3² = √14

and  a · b = 4−52·−12−3 = –4 – 10 – 6 = –20

a · b = ab cos θ ⇒ –20 = 3√5 × √14 cos θ

Example:   A, (–1, 2, 4), B, (2, 3, 0), and C, (0, 2, -3) form a triangle. Find the angle ∠BAC.

Solution:   ∠BAC = α, which is the angle between the vectors AB and AC.

Note that the angle between BA and AC is the angle β,

which is not the angle requested.

Then proceed as in the example above.

Example:   Find the vector equation of the line through the points M, (2, −1, 4), and N, (–5, 3, 7).

Solution:   We are looking for the line through M  (or N)  which is parallel to the vector MN.

MN = n – m = −537 – 2−14 = −743

Example:   Show that the point P, (–1, 7, 10), lies on the line

Solution:   The x co-ord of P is –1 and of the line is 1 – λ

In the equation of the line this gives   y = −1 + 2 × 4 = 7   and   z = 4 + 2 × 3 = 10

Example:   Find A′ the reflection of the point A (2, 4, 0) in the line ℓ,

Solution:   From the diagram we can see that AA′ is perpendicular to ℓ.

So, if we can find the point B, where AB is perpendicular to ℓ, we will be able to find A′, since AB = BA′.

B is a point on ℓ

b is perpendicular to ℓ, and ℓ is parallel to −121

Example:   Find the intersection of the lines

Solution:   We are looking for values of λ and µ which give the same x and y co-ordinates on each line.

Example:   Investigate whether the lines

and if they do find their point of intersection.

Solution:   If the lines intersect we can find values of λ and µ to give the same x, y and z coordinates in each equation.

We must now check to see if we get the same point for the values of λ and µ

In ℓ₁,  λ = 3  gives the point  (–1, 7, 6);

in ℓ₂,  µ = 2  gives the point  (–1, 7, 7).

The x and y co-ords are equal (as expected!), but the z co-ordinates are different and so the lines do not intersect.