Differentiation (6)

Question 1

Find the gradient of the tangent to the curve y=sin^2x - 2 cos 2x at the point where x=pi/12 radians

Answer 1

u = sin x and y= u^2
dy/du = 2u
du/dx = cosx
dy/dx = 2 sinx cosx

u = 2x and y = 2 cos u
dy/du = -2 sin u
du/dx = 2
dy/dx = -4 sin 2x

Overall, dy/dx = 2 sinx cosx + 4 sin 2x
dy/dx = sin 2x + 4 sin 2x = 5 sin2x

Sub x=pi/12 into dy/dx to get the gradient
5 sin 2(pi/12) = 2.5

Question 2

Find the equation of the normal to curve x = sin 4y that passes through the point (0,(pi/4)). Give your answer in the form y = mx + c, where m and c are constants to be found

Answer 2

dx/dy = 4 cos 4y
dy/dx = 1/(4 cos 4y)

At (0,(pi/4)), y=pi/4 and so dy/dx = 1/(4 cos pi) = -1/4
Normal: -1 / -1/4 = 4

y - (pi/4) = 4(x - 0)
y = 4x + pi/4

Question 3

Find dy/dx for y = sin^3 (2x^2). Simplify your answer where possible

Answer 3

y = u^3 and u = sin (2x^2)
dy/du = 3u^2 = 3 sin^2 (2x^2) 
du/dx = 4x cos (2x^2)

dy/dx = 12x sin^2 (2x^2) cos (2x^2)

Question 4

A curve with equation y = e^x sin x has 2 turning points in the interval -pi <= x <= pi.

Find the value of x at each of these turning points and determine the nature of the turning points

Answer 4

u = e^x and v = sin x
du/dx = e^x and dv/dx = cos x
(e^x * cos x) + (sin x * e^x) = e^x (cos x + sin x)

At the turning points, dy/dx = 0
e^x (cos x + sin x) = 0
e^x = 0 AND (cos x + sin x) = 0

Only (cos x + sin x) can equal 0
sin x = -cos x
sin x / cos x = -1
tan x = -1

There are two solutions -pi <= x <= pi for tan x = -1
So the turning points are x=(-pi/4) and x=(3pi/4)

To determine the nature of these stationary points:
dy/dx = e^x (cos x + sin x)
u = e^x and v = (cos x + sin x)
du/dx = e^x and dv/dx = (cos x - sin x)
(e^x * (cos x - sin x)) + (e^x * (cos x + sin x)) = 2e^x cos x

When x = (-pi/4), d^2y/dx^2 = 0.645 > 0, so it’s a minimum point
When x = (3pi/4), d^2y/dx^2 = -14.9 < 0, so it’s a maximum point

Question 5

Show that if f(x) = cot x, then f’‘(x) = (2 cos x)/(sin^3 x)

Hence show that ((pi/2),0) is a point of inflection of the graph y = cot x

Answer 5

f(x) = cot x
f'(x) = -cosec^2x
y = -u^2 and u=cosecx
dy/du = -2u and du/dx = -cosecxcotx

So f’‘(x) = (-2u)(-cosecxcotx) = 2 cosec^2x cot x
Write in terms of sin and cos:
2(1/sin^2x) * (cosx/sinx) = (2cosx/sin^3x)

At x=(pi/2), f’‘(x)=(2cos(pi/2))/(sin(pi/2)^3) = 0
Either side of x = pi/2 , sin^3x is positive and cosx changes sign from positive to negative. So f’‘(x) changes from positive to negative which means ((pi/2),0) is a point of inflection

Question 6

The curve C is defined by the parametric equations x = 3z - cos3z AND y = 2 sin z. -pi <= z <= pi

Show that the gradient of C at the point (pi+1, ¬3) is 1/3

Find the equation of the normal to C when z = pi/6

Answer 6

dy/dz = 2 cos z
dx/dz = 3 + 3 sin 3z

dy/dx = dy/dz / dx/dz = (2 cos z)/(3 + 3 sin 3z)

Find the value of z at (pi+1, ¬3):
y = 2 sin z = ¬3, for -pi <= z <= pi
z = pi/3 or 2pi/3

if z = pi/3, then sub into x = 3z - cos3z to get x = pi - cos (pi). Which equals pi + 1
if z = 2pi/3 then sub into x = 3z - cos3z to get x = 2pi - cos(2pi) = 2pi - 1
So at (pi + 1,¬3). z = pi/3

Find the equation of the normal to C
When z = pi/6, x = pi/2 - cos(pi/2) = pi/2 - 0 = pi/2
And y = 2 sin pi/6 = 2 * 1/2 = 1
So z = pi/6 is stationary at the point (pi/2,1)

z=pi/6, dy/dx = (2cos(pi/6))/(3 + 3sin(pi/6)) = (2(¬3/2))/(3 + 3(1)) = ¬3/6
Gradient of the normal: -1 / (¬3/6) = -2¬3

y - 1 = (-2¬3)(x - pi/2)
y = -2¬3x + 1 + pi¬3

Question 7

Product rule?

Answer 7

u(dv/dx) + v(du/dx)

Question 8

u(dv/dx) + v(du/dx)

Answer 8

Product rule?

Question 9

The equation of the curve C is 6x^2y - 7 = 5x - 4y^2 - x^2

The line T has the equation y = c and passes through a point on C where x = 2. Find c, given that c > 0

T also crosses C at point Q. Find the coordinates of Q

Answer 9

6x^2y - 7 = 5x - 4y^2 - x^2 = 6(2)^2c - 7 = 5(2) - 4c^2 - (2)^2
=> 24c - 7 = 6 - 4c^2 => 4c^2 + 24c - 13 = 0
(2c+13)(2c-1) = 0
c = -6.5 OR c = 0.5
C>0, so c=0.5

Q is another point on C where y=0.5
6x^2y - 7 = 5x - 4y^2 - x^2 = 6(0.5)^2c - 7 = 5(0.5) - 4c^2 - (0.5)^2
=> 3x^2 -7 = 5x - 1 - x^2 => 4x^2 - 5x - 6 = 0
(x-2)(4x+3) = 0
x=2 OR x=-0.75
x!=2 as x=2 at the other point where T crosses C. So coordinates for Q are (-0.75,0.5)

Question 10

The curve C has the equation 3e^x + 6y = 2x^2y. Use implicit differentiation to find an expression for dy/dx

Answer 10

3e^x + 6y = 2(x^2)y
(d/dx)3e^x + (d/dx)6y = (d/dx)2(x^2)y
3e^x + 6(dy/dx) = (2x^2)(d/dy)y(dy/dx) + y(d/dx)(2x^2)
3e^x + 6(dy/dx) = (2x^2)(dy/dx) + 4xy
(2x^2)(dy/dx) - 6(dy/dx) = 3e^x - 4xy
(dy/dx) = (3e^x - 4xy)/(2x^2 - 6)

Question 11

How to find stationary points

Answer 11

dy/dx = 0

Question 12

dy/dx = 0

Answer 12

How to find stationary points

Question 13

Use implicit differentiation to find dy/dx if 2x^2y + y^3 = 6x^2 + 5

Answer 13

(d/dx)2x^2y + (d/dx)y^3 = (d/dx)6x^2 + (d/dx)5
(d/dx)2x^2y + (d/dx)y^3 = 12x

(d/dx)2x^2y + 3y^2(dy/dx) = 12x
2x^2(d/dx))y + y(d/dx)2x^2 + 3y^2(dy/dx) = 12x
2x^2(dy/dx) + 4yx +  3y^2(dy/dx) = 12x
(dy/dx)(2x^2 + 3y^2) = 12x - 4xy
dy/dx = (12x - 4xy)/(2x^2 + 3y^2)