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A curve has parametric equations: x=(t-1)^3 and y= 3t - 8/(t^2). Find dy/dx in terms of t. Then find the equation of the normal at the point on the curve where t=2.

dx/dt = 3(t-1) 2 dy/dt = 3 + 16t -3 dy/dx=(dy/dt)(dt/dx) dy/dx = 3 + 16t -3 / 3(t-1) 2 At t=2 dy/dx= (3 + 16/8) / 3 = 5/3 Gradient of the normal = -3/5with t=2 y-4=0x-1=0 y=mx + c y - 4 = -3/5(x-1) 3x +5y -2...
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Answered by Jasmin H. • Maths tutor
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How can functions be transformed?

A function, y = f(x), with y on the vertical axis and x on the horizontal axis, can be transformed by 3 different ways: It can be stretched (or shrunk)If y = f(ax), the function is stretched by a scale facto...
JM
Answered by Jack M. • Maths tutor
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If y = 15 + 5(x + 2), and x = 6, what is y?

Because we're told that x = 6, we can just rewrite the original equation but replace the x with a 6. So: y = 15 + 5(6 + 2) 6 + 2 = 8, so we can rewrite this again as: y = 15 + 5(8) 5(8) is just another way o...
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Answered by Alfie H. • Maths tutor
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How do I solve simultaneous equations like 2x + 5y = 50 and 3x + y = 23?

With simultaneous equations like these, you first want to get to a point where you have one equation with only one variable. To do this, you must eliminate one of the variables. In this case, if you multiply...
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Answered by Alfie H. • Maths tutor
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How do I solve a quadratic equation like x^2 - 2x - 35 = 0 without using a calculator?

If possible, you should try to factorise a quadratic. To do this, look a the factors of the constant, in this case -35. 35 is clearly divisible by 5, and 35 / 5 = 7, so 5 and 7 are one pair of factors. These...
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Answered by Alfie H. • Maths tutor
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