Calculus

Find the gradient of the curve with equation

2x² − 4xy + 3y² = 3, at the point (2, 1).

• 2

• 3

• 4

• 5

The equation of a curve is x²y + y² = 6x

the equation of the tangent to the curve at the point with coordinates (1, 2)

• 3x – 5y + 8 = 0

• 2x – 5y + 8 = 0

• 2x – 5y - 8 = 0

• 3x – 5y - 8 = 0

A solid rectangular block has a square base of side x cm. The height of the block is h cm and the total surface area of the block is 96 cm^2. Find the stationary value of V

• V = 84

• V = 74

• V = 64

• V = 54

The diagram shows the curve y = (x − 2)² and the line y + 2x = 7, which intersect at points A and B. Find the area of the shaded region.

• $10\ \frac{5}{7}$

• $10\ \frac{3}{4\ }$

• $10\ \frac{1}{3\ }$

• $10\ \frac{2}{3}$

 $\int_{ }^{ }\left(x^{3\ }+\frac{1}{x^3}\right)\ dx$  

•  $\frac{x^{4\ }}{4}-\frac{x^{-2}}{2\ }+c$  

•  $\frac{x^{4\ }}{4}+\frac{x^{-2}}{2}+c$  

•  $\frac{x^4}{4}-\frac{x^2}{2\ }+c$  

•  $\frac{-x^4}{4}+\frac{x^{2\ }}{2\ }+c$  

the equation of the curve $y=\frac{4}{3x-4}$ 
and P(2, 2) is a point on the curve. The equation of the tangent to the curve at P and the angle that this tangent makes with the x-axis respectively 

• y − 2 = −3(x − 2) and  $108.4^o$  

• y − 2 = 3(x − 2) and  $128.8^o$  

• y − 2 = −3(x + 2) and  $78.3^o$  

• y + 2 = −3(x − 2) and  $45^o$  

The diagram shows part of the curve $y=4\sqrt{x}-x$  
The curve has a maximum point at M and meets the x-axis at O and A. Find the volume obtained when the shaded region is rotated through 360o about the x-axis, givingyour answer in terms of π

• 138.5π

• 137.5π

• 136.5π

• 135.5π

The equation of a curve is $y=4\sqrt{x}+\frac{2}{\sqrt{x}}$  
A point is moving along the curve in such a way that              the x-coordinate is increasing at a constant rate of 0.12 units per second. Find the rate of change of the y-coordinate when x = 4

• 0.105

• 0.125

• 0.345

• 0.546

The diagram shows part of the curve y = −x² + 8x − 10 which passes through the points A and B. The curve has a maximum point at A and the gradient of the line BA is 2. evaluate the area of the shaded region.

• $9\ \frac{1}{3}$

• $10\ \frac{1}{3}$

• $11\ \frac{1}{3}$

• $12\ \frac{1}{3}$

A curve is such that  $\frac{dy}{dx}=\frac{6}{x^2}$  and (􏰎2, 9)􏰏 is a point on the curve. Find the equation of the curve

•  $y=-6x^{-1}+12$  

•  $y=6x^{-1}-12$  

•  $y=-6x^{-2}+12$  

•  $y=6x^{-2}+12$