# Differentiation (5) Flashcards

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

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

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

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

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

Substitute x = 9

y = 2(9)^(1/2) + 27(9)^-1

y = 9 units of coal

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

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

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

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

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

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

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

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

dy/dx of 4ln3x

y = 4 ln u and u = 3x dy/du = 4/u = 4/(3x) du/dx = 3 dy/dx = 12/(3x) = 4/(x)

y = 4 ln u and u = 3x dy/du = 4/u = 4/3x du/dx = 3 dy/dx = 12/3x = 4/x

dy/dx of 4ln3x

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

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

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

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

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

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

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

dy/dx = 4^x ln 4

As x=0 is 4^0, which is 1

So dy/dx = ln 4 at point (0,1)

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

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.)

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

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)