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Maths
A Level

How do you find the normal to a curve at a given co-ordinate?

  1. You first find the gradient of the tangent to the curve at this given co-ordinate by differentiating the given equation of the curve, and then, assuming the equation of the curve is in terms of x, r...
Answered by Srabon I. Maths tutor
3716 Views

Find the equation of the line that is perpendicular to the line 3x+5y=7 and passes through point (-2,-3) in the form px+qy+r=0

Gradient of line 3x+5y=7: 5y=-3x+7, y=-3/5x+7/5 gradient = -3/5 Gradient of perpendicular line: 5/3 Perpendicular line with points: y+3=5/3(x+2), 3y+9=5(x+2), 3y+9=5x+10, 5x-3y+1=0

Answered by Paul I. Maths tutor
8427 Views

Differentiate y=sin(x)*x^2.

Using the chain rule, we let u = sin(x) and v = x^2. Then dy/dx = udv/dx + vdu/dx. dv/dx = 2x and du/dx = cos(x). So dy/dx = sin(x)2x + x^2cos(x).

Answered by Lucy M. Maths tutor
3446 Views

How to find the derivative of arctan(x)

Let y = arctan(x). Then x = tan(y).

Differentiate using the chain rule and rearrange: d(x)/dx = d(tany)/dx So 1 = sec^2(y) * dy/dx dy/dx = 1/sec^2(y)

But from identity sin^2(y) + cos^2(y) = ...

Answered by Matthew R. Maths tutor
12643 Views

How do I do this question: A small stone is projected vertically upwards from the point A with speed 11.2 m/s. Find the maximum height above A reached by the stone.

[Draw Diagram] We are given the inital velocity u=11.2m/s and we know the acceleration due to gravity is 9.8m/s^2. We need to find the distance 's'. When the stone is at its highest point the velocity wil...

Answered by Luke W. Maths tutor
5314 Views

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