# Differentiation (5) Flashcards

1
Q

A steam train travels between A and B at a speed of x miles per hour and burns y units of coal. y is modelled by 2x^(1/2) + 27x^-1, for x>2

Find the speed that gives the minimum coal consumption

A

Find the stationary point of the curve y. Can be solved by differentiating and solving dy/dx=0

```dy/dx = x^(-1/2) - 27x^-2
x^(-1/2) - 27x^-2 = 0
x^(-1/2) = 27x^-2
x^(3/2) = 27
x = 9
So minimum coal consumption is 9mph```
2
Q

A steam train travels between A and B at a speed of x miles per hour and burns y units of coal. y is modelled by 2x^(1/2) + 27x^-1, for x>2

Find d^2y/dx^2 and hence show that the speed gives the minimum coal consumption

A
```dy/dx = x^(-1/2) - 27x^-2
d^2y/dx^2 = 54x^-3 - ((-1/2)x^(-3/2))
Substitute x = 9
54(9)^-3 - ((-1/2)(9)^(-3/2)) = 0.055... > 0
So the stationary point is a minimum```
3
Q

A steam train travels between A and B at a speed of x miles per hour and burns y units of coal. y is modelled by 2x^(1/2) + 27x^-1, for x>2

Calculate the minimum coal consumption

A

Substitute x = 9
y = 2(9)^(1/2) + 27(9)^-1
y = 9 units of coal

4
Q

Percy is building a closed-back bookcase. He uses a total of 72m^2 of wood to make a bookcase that is x metres high, x/2 metres wide and d metres deep

Show that the full capacity of the bookcase is given by: V = 12x - (x^3)/12

A
```Surface area = 2(d*x) + 2(d*(x/2)) + (x*(x/2))
2dx + dx + (x^2)/2 = 3dx + (x^2)/2
As surface area = 72
3dx + (x^2)/2 = 72
6dx + x^2 = 144
d = (144 - x^2)/6x```

Volume = width * height * depth
V = (x^2)/2 * (144-x^2)/6x = (144x^2 - x^4)/12x
(144x^2 - x^4)/12x simplifies to 12x - (x^3)/12

5
Q

Percy is building a closed-back bookcase. He uses a total of 72m^2 of wood to make a bookcase that is x metres high, x/2 metres wide and d metres deep

Find the value of x for which V is stationary. leave answer in surd form

A
```Differentiate V and then solve dV/dx = 0
dV/dx = 12 - (x^2)/4
12 - (x^2)/4 = 0
(x^2)/4 = 12
x^2 = 48
x = ¬48
x = 4¬3```
6
Q

Percy is building a closed-back bookcase. He uses a total of 72m^2 of wood to make a bookcase that is x metres high, x/2 metres wide and d metres deep

Show that this is a maximum point and hence calculate the maximum V

A

Using the solution from dV/dx

d^2V/dx^2 = -(x/2)
So when x = 4¬3, d^2V/dx^2 = -2¬3
d^2V/dx^2 is negative so it’s a maximum point

Sub x=4¬3 into the 12x - (x^3)/12
12(4¬3) - ((4¬3)^3)/12 = 55.4m^3

7
Q

Find the equation of the tangent to the curve y = 3^(-2x) at the point ((1/2),(1/3))

A

u = -2x and y = 3^u
du/dx = -2 and dy/du = (3^u) ln 3
dy/du * du/dx = dy/dx
dy/dx = -2(3^(-2x) ln 3)

At ((1/2),(1/3)), dy/dx = -(2/3) ln 3

y-(1/3) = ((-2/3) ln 3)(x - 1/2)
3y - 1 = ln 3 - (2 ln 3)x
(2 ln 3)x + 3y - (1 + ln 3) = 0

8
Q

dy/dx of 4ln3x

A
```y = 4 ln u and u = 3x
dy/du = 4/u = 4/(3x)
du/dx = 3
dy/dx = 12/(3x) = 4/(x)```
9
Q
```y = 4 ln u and u = 3x
dy/du = 4/u = 4/3x
du/dx = 3
dy/dx = 12/3x = 4/x```
A

dy/dx of 4ln3x

10
Q

The triangular prism shown in the diagram is expanding. The dimensions of the prism after t seconds are given in terms of x. prism is 4x m long, and its cross section is an isoceles triangle with base (3/2)x m and height x m

Show that, if the surface area of the prism after t seconds is Am^2, then A=(35/2)x^2

A

Start by finding the missing side length of the triangular face. Call this s
s = ¬(x^2 + ((3/4)x)^2) = (5/4)x

Find A by adding up the area of each of these faces
A = 2((1/2) * (3/2)x * x) + ((3/2)x * 4x) + 2((5/4)x * 4x)
A = (35/2)x^2

11
Q

The triangular prism shown in the diagram is expanding. The dimensions of the prism after t seconds are given in terms of x. prism is 4x m long, and its cross section is an isoceles triangle with base (3/2)x m and height x m

The surface area of the prism is increasing at a constant rate of 0.07m^2s^-1

Find dx/dt when x=0.5

A
```dA/dt = 0.07
A = (35/2)x^2 => dA/dx = 35x```
```dx/dt = dx/dA * dA/dt
dx/dt = 1/35x * 0.07 = 0.07/(35*0.5) = 0.004ms^-1```
12
Q

The triangular prism shown in the diagram is expanding. The dimensions of the prism after t seconds are given in terms of x. prism is 4x m long, and its cross section is an isoceles triangle with base (3/2)x m and height x m

The surface area of the prism is increasing at a constant rate of 0.07m^2s^-1

If the volume of the prism is Vm^3, find the rate of change of V when x=1.2

A

Find the rate of change of V means dV/dt

```V = (1/2 * (3/2)x * x) * 4x = 3x^3
dV/dx = 9x^2 ```
```dV/dt = dV/dx * dx/dt = 9x^2 * (0.07/35x)
dV/dt  = 9(1.2)^2 * (0.07/35(1.2))
dV/dt = 0.0216ms^-1```
13
Q

Curve A has the equation y=4^x. What are the coordinates of the point on A where dy/dx = ln 4

A

dy/dx = 4^x ln 4
As x=0 is 4^0, which is 1
So dy/dx = ln 4 at point (0,1)

14
Q

Curve B has the equation y=4^(x-4)^3 and find the gradient of B at the point (3,(1/4))

A

Let u=(x-4)^3
dy/dx = dy/du * du/dx
= d/du (4^u) * d/dx((x-4)^3)
= (4^u ln 4) * (3(x-4)^2 * 1)
= (4^(x-4)^3) * 3(x-4)^2 * ln 4
When x = 3: 4^-1 * 3 * ln 4 = 1.04 (3 s.f.)

15
Q

When dy/dx for y=ln(3x+1)sin(3x+1)

A

u=ln(3x+1), v=sin(3x+1)

```du/dx = 3/(3x+1)
dv/dx = 3cos(3x+1)```

Then use the formula u(dv/dx) + v(du/dx)

16
Q

Multiply (- ¬(x^2 + 3) * -3sin3x) by ¬(x^2 + 3)

A

3(x^2 + 3) sin3x

17
Q

3(x^2 + 3) sin3x

A

Multiply (- ¬(x^2 + 3) * -3sin3x) by ¬(x^2 + 3)

18
Q

What’s (sin3x) / ((cos^2)3x)

A

tan3x / cos3x

19
Q

tan3x / cos3x

A

What’s (sin3x) / ((cos^2)3x)