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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
PI
Answered by Paul I. • Maths tutor
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Differentiate y=sin(x)*x^2.

Using the chain rule, we let u = sin(x) and v = x^2. Then dy/dx = u dv/dx + v du/dx. dv/dx = 2x and du/dx = cos(x). So dy/dx = sin(x) 2x + x^2 cos(x).
LM
Answered by Lucy M. • Maths tutor
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Expand and simplify (5a-2b)(3a-4b)

=(5a x 3a) + (5a x -4b) + (3a x -2b) + (-2b x -4b) =15a^2 - 20ab - 6ab + 8b^2 =15a^2 - 26ab + 8b^2
PI
Answered by Paul I. • Maths tutor
11040 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) = 1 We can derive, ...
MR
Answered by Matthew R. • Maths tutor
15137 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 will b...
LW
Answered by Luke W. • Maths tutor
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