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The tangent to a point P (p, pi/2) on the curve x=(4y-sin2y)^2 hits the y axis at point A, find the coordinates of this point.

p=4pi 2 differentiating with respect to y we have dx/dy = 2(4y-sin2y)(4-2cos2y) substituting in the value of y =pi/2 we have dx/dy = 24pi, which means dy/dx =1/pi24using (y-y_1)=m(x-x_1) we have y-pi/2=1/24p...
GN
Answered by George N. • Maths tutor
4115 Views

The parametric equations of a curve are: x = cos2θ y = sinθcosθ. Find the cartesian form of the equation.

x = cos2θ y = sinθcosθcos2θ = cos 2 θ - sin 2 θ cos 2 θ + sin 2 θ = 12cos 2 θ = 1 + cos2θ cos 2 θ = 1/2(1 + x)2sin 2 θ = 1 - cos2θ sin 2 θ = 1/2 (1 - x)y 2 = sin 2 θcos 2 θy 2 = ( 1/2(1 + x)) . (1/2 (1 - x))...
AN
Answered by Amelia N. • Maths tutor
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find dy/dx of x^1/2 + 4/(x^1/2) + 4

This can be rewritten as x 1/2 + 4x -1/2 + 4, hence dy/dx = (1/2)x -1/2 - 2x -3/2
OO
Answered by Olufayojimi O. • Maths tutor
3702 Views

(a) By using a suitable trigonometrical identity, solve the equation tan(2x-π/6)^2 =11-sec(2x-π/6)giving all values of x in radians to two decimal places in the interval 0<=x <=π .

say (2x-π/6)=qtan 2 q = sec 2 q-1so sec 2 q-1= 11 - sec q sec 2 q + sec q -12 = 0(sec q -3) (sec q + 4) = 0sec q = 3 or -4because 0&lt;=x&lt;= πso -π/6 &lt;= q &lt;= 11π/6q = 1.23 or 1.82 or 4.46 or 5.05so x...
ZZ
Answered by Zhaohui Z. • Maths tutor
9170 Views

Solve ln(2x-3) = 1

e ln(2x-3) = e 1 2x-3 = e x = (e+3)/2
JC
Answered by Jonathan C. • Maths tutor
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