Top answers

Find the equation of the tangent to the unit circle when x=sqrt(3)/2 (in the first quadrant)

Unit circle: x 2 + y 2 = 1 when x = sqrt(3)/2: y 2 = 1 - (sqrt(3)/2) 2 y 2 = 1 - 3/4 y 2 = 1/4 y = 1/2 or -1/2 (first quadrant, so y is positive, i.e. y = 1/2) find gradient at (sqrt(3)/2, 1/2): x 2 + y 2 = ...
KJ
Answered by Kiran J. • Maths tutor
4475 Views

Integral of (2(x^3)-7)/((x^4)-14x)

Set f(x)= (x^4)-14x. f’(x)=4(x^3)-14=2(2(x^3)-7). Thus we can write (2(x^3)-7)/((x^4)-14x)=(1/2)f’(x)/f(x). The integral of f’(x)/f(x)=ln|f(x)|+c. Thus the integral of (2(x^3)-7)/((x^4)-14x) is (1/2)(ln|f(x)...
IK
Answered by Issy K. • Maths tutor
3693 Views

A curve has parametric equations: x = 3t +8, y = t^3 - 5t^2 + 7t. Find the co-ordinates of the stationary points.

First differentiate: dx/dt = 3, dy/dt = 3t 2 - 10t + 7 Using the chain rule: dy/dx = dy/dt * dt/dx = (3t 2 - 10t + 7)/3 At stationary points, the gradient is equal to zero: 3t 2 - 10t + 7 = 0 Solve for t usi...
RB
Answered by Robbie B. • Maths tutor
6096 Views

Integrate tan (x) with respect to x.

I = ∫ Tan (x) dx= ∫ (sin(x)) / (cos(x)) dx We see that this is close to the standard integral ∫ F'(x) / F(x) dx = Ln (F(x)) + C So first we must rewrite the Integral as: I = - ∫ (-sin(x)) / (cos(x)) dx (Taki...
MH
Answered by Matthew H. • Maths tutor
13215 Views

Find the tangent to y = x^2 - 4x + 9 at the point (3,15)

First find dy/dx: dy/dx = 4x - 4 And thus at (3,15): dy/dx = 12 - 4 = 8 = m (as m is the gradient of a curve) So using y - y 1 = m(x - x 1 ) where (x 1 ,y 1 ) = (3,15): y - 15 = 8(x - 3) y = 8x- 9
SH
Answered by Scott H. • Maths tutor
3533 Views