Differentiation (4) Flashcards
Find the gradient of the curve y = x^-(1/2) + x^-1 at the point (4,(3/4))
-(1/2)x^-(3/2) - x^-2
Substitute 4 to get the gradient
The curve C is given by the equation y = mx^3 - x^2 + 8x + 2 for a constant M
Find dy/dx
dy/dx = 3mx^2 - 2x + 8
The curve C is given by the equation y = mx^3 - x^2 + 8x + 2 for a constant M
The point P lies on C, and has an X coordinate of 5. The normal to C at P is parallel to the line given by the equation y+4x-3=0
Find the gradient of curve C at P
Rearrange to get y=-4x+3
Gradient is -4, to get the gradient of the normal, do -1 / -4 = 1/4
The curve C is given by the equation y = mx^3 - x^2 + 8x + 2 for a constant M
The point P lies on C, and has an X coordinate of 5. The normal to C at P is parallel to the line given by the equation y+4x-3=0
Find the value of m
When x = 5, the gradient is 3mx^2 - 2x + 8 = 1/4
m(3*5^2) - 2(5) + 8 = 1/4
75m - 10 + 8 = 1/4
75m = 9/4
m = 0.03
The curve C is given by the equation y = mx^3 - x^2 + 8x + 2 for a constant M
The point P lies on C, and has an X coordinate of 5. The normal to C at P is parallel to the line given by the equation y+4x-3=0
The y-coordinate of P
When x=5, y=(0.03*5^3) - (5)^2 + 8(5) + 2
y=20.75
Show that the lines y = (x^3)/3 - 2x^2 - 4x + 86/3 and y=x^(1/2) both go through the point (4,2) and are perpendicular at that point
Differentiating y = (x^3)/3 - 2x^2 - 4x + 86/3
Gets dy/dx = x^2 - 4x - 4
When x = 4, y = (64/3) - 32 - 16 + 86/3 = 2
dy/dx = 16 - 16 - 4 = -4
Differentiating y=x^(1/2)
Gets dy/dx = (1/2)x^-(1/2)
When x = 4, y = 4^(1/2) = 2
dy/dx = (1/2)4^-(1/2) = 1/4
Both curves past through the point (4,2). And are perpendicular as 1/4 * -4 = -1
Find dy/dx for the curve y = 6 + (2/3)x^3 - (5/2)x^2 + 2x
dy/dx = 2x^2 -5x + 2
Find the coordinates of the stationary points of the curve. y = 6 + (2/3)x^3 - (5/2)x^2 + 2x
Stationary points occur at 2x^2 -5x + 2 = 0
(2x - 1)(x-2)=0
x=1/2 and x=2
Substitute x=2 into y = 6 + (2/3)x^3 - (5/2)x^2 + 2x.
y = 6 + (2/3)2^3 - (5/2)2^2 + 2x. y=5(1/3)
Substitute x=1/2 into y = 6 + (2/3)x^3 - (5/2)x^2 + 2x.
y = 6 + (2/3)(1/2)^3 - (5/2)(1/2)^2 + 2(1/2). y=6(11/24)
Determine the nature of each stationary point. y = 6 + (2/3)x^3 - (5/2)x^2 + 2x
Differentiate twice to get d^2y/dx^2 = 4x - 5
When x=2, d^2y/dx^2 = 4(2) - 5 = 3
As 3 is positive, (2,5(1/3)) is a minimum
When x=1/2, d^2y/dx^2 = 4(1/2) - 5 = -3
As -3 is negative ((1/2),6(11/24)) is a maximum
How to approach a question such as differentiate y=(x-1)(3x^2 - 5x - 2)
Expand the brackets and then differentiate
The function f(x)=(1/2)x^4 - 3x has a single stationary point
Find the coordinates of the stationary point
f’(x) = 2x^3 - 3 = 0
2x^3 = 8
x^3 = 4
x=1.14 to 3 s.f.
f(1.1447…) = (1/2)(1.1447…)^4 - 3(1.1447…) = -2.58 to 3 s.f.
(1.14,-2.58)
The function f(x)=(1/2)x^4 - 3x has a single stationary point
Determine the nature of the stationary point
f’‘(x) = 6x^2
f’‘(x) = 6(1.1447…)^2 = 7.862… > 0
7.862 is positive, so it’s a minimum
The function f(x)=(1/2)x^4 - 3x has a single stationary point
State the range of values for which f(x) is increasing
As the stationary point is a minimum, f(x) is increasing on the right of the stationary point x>1.14
The function f(x)=(1/2)x^4 - 3x has a single stationary point
State the range of values for which f(x) is decreasing
As the stationary point is a minimum, f(x) is decreasing on the left of the stationary point x<1.14
Show that f(x) = 2x^6 - 6x^5 + 5x^4 has no points of inflection
f'(x) = 12x^5 - 30x^4 + 20x^3 f''(x) = 60x^4 - 120x^3 + 60x^2
60x^2
(60x^2)((x-1)^2)
Since 60x^2 and (x-1)^2 are both >=0, f’‘(x)>=0 for all x. So curve is never concave and can’t be a point of inflection
For f(x)=x^4 + (5/3)x^3 + x^2 - 2x, find the ranges of x for which f(x) is concave and convex
f’(x)=4x^3 + 5x^2 + 2x - 2
f’‘(x)=12x^2 + 10x + 2
So f’‘(x)>0 => 6x^2 + 5x + 1 > 0
6x^2 + 5x + 1 > 0 => (3x+1)(2x+1)>0
x > -1/3 or x < -1/2
Find dy/dx for the curve given by the equation x=¬(y^2 + 3y) at the point (2,1)
x = u^(1/2) u = y^2 + 3y
dx/du = (1/2)u^(-1/2) = 1/(2 ¬(u)) = 1/(2 ¬(y^2 + 3y))
du/dy = 2y + 3
So dx/dy = (2y + 3)/(2 ¬(y^2 + 3y))
dy/dx = (2 ¬(y^2 + 3y))/(2y + 3)
At the point (2,1), y=1. So (2 ¬((1)^2 + 3(1)))/(2(1) + 3)
= 4/5 = 0.8
Find dy/dx for the curve given by the equation x=¬(y^2 + 3y) at the point (2,1)
Hence find the equation of the tangent to the curve at (2,1) in the form y = mx + c
Gradient = 0.8
y - 1 = 0.8(x - 2)
y = 0.8x - 0.6
Using the chain rule to get dx/dt from dA/dt and dA/dx
dx/dt = dx/dA * dA/dt = 1/(da/dx) * da/dt
dx/dt = dx/dA * dA/dt = 1/(dA/dx) * dA/dt
Using the chain rule to get dx/dt from dA/dt and dA/dx
