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How do I find the equation of the tangent of a curve at a specific point.

The gradient is the rate of change at a specific point on the curve. Since the tangent is a straight line that touches the curve only once at a specific point, the gradient of the curve and the tangent will ...
AS
Answered by Aref S. • Maths tutor
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Use the substitution u = 6 - x^2 to find the value of the integral of (x^3)/(sqrt(6-x^2)) between the limits of x = 1 and x = 2 (AQA core 3 maths

When integrating by substitution the first thing to do is change the limits of the integral by subbing them into the equation for u. This gives u = 5 as the lower limit and u = 2 for the upper limit. The nex...
CC
Answered by Christopher C. • Maths tutor
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Integration

Integration is a simple way of "quickly" adding. In cases where you have a curve under a graph for example a velocity-time graph, the area under the curve will give you the distance. If you know th...
NA
Answered by Neal A. • Maths tutor
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Mechanics 1: How do you calculate the magnitude of impulse exerted on a particle during a collision of two particles, given their masses and velocities.

Conservation of momentum is the key principle always used in problems of these type. Here, it asks for the impulse, which is a force measured in Newtons, and it is defined as the change in momentum. F=m(v-u)...
IH
Answered by Ina H. • Maths tutor
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A curve is described by the equation (x^2)+4xy+(y^2)+27=0. The tangent to the point P, which lies on the curve, is parallel to the x-axis. Given the x-co-ordinate of P is negative, find the co-ordinates of P.

Firstly, implicitly differentiate the function to find dy/dx in terms of y and x: dy=dx = (-x-2y)/(2x+y). Secondly, set this equal to zero to obtain an expression of y in relation to x or vice versa: x=-2y. ...
RP
Answered by Rahul P. • Maths tutor
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