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Work out the gradient of the curve y=x^3(x-3) at the point (3,17)

First simplify the equation of the curve y= x^4 - 3x^3 .The gradient is the differential.To differentiate, bring down the power and take one from it.x^4 becomes 4x^3-3x^3 becomes (-3x3)= -9x^2dy/dx= 4x^3 - 9...
SM
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What is the distance between two points with x-coordinates 4 and 8 on the straight line with the equation y=(3/4)x-2

Firstly, to be able to find the distance between the two points we must find the y-coordinates of each point by substituting in the x values. For x=4y=3/4x4-2=1For x=8y=3/4x8-2=4Now that we know the coordina...
LD
2501 Views

l1 and l2 are tangents of a circle. l1 intersects the circle at (3-√3,5) with a gradient of √3, and l2 intersects the circle at (3+√2,4+√2) with a gradient of -1. Find the centre of the circle, and hence find the radius of the circle.

To first find the centre of the circle, the key fact to remember is that the radius of a circle is normal (perpendicular) to any tangent. This means if you take the normals to both of these tangents at their...
JB
3247 Views

How do I determine if a stationary point on a curve is the maximum or minimum?

If you are comfortable with differentiation. You can take the second derviatve of the equation of the cruve and plug in the x value of the curve. Based on this answer you can determine if it's a maximum, min...
ES
2986 Views

Show that 2cos^2(x) = 2 - 2sin^2(x) and hence solve 2cos^2(x) + 3sin(x) = 3 for 0<x<180

You can rearrange the equation to get 2cos^2(x)+2sin^2(x) = 2. This can be factorised to get cos^2(x) + sin^2(x) = 1 which is a known identity. We can use the fact that 2cos^2(x) = 2sin^2(x) - 2 and substitu...
MR
5573 Views