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A curve has parametric equations -> x = 2cos(2t), y = 6sin(t). Find the gradient of the curve at t = π/3.

First we need to find the derivatives of x and y in terms of t. dx/dt can be found using the chain rule. Differentiating the inside of the bracket gives us 2. Multiplying the outside gives -2sin(2t) (Derivat...
MM
Answered by Matvei M. • Maths tutor
5180 Views

How do I remember the common values of cosx, sinx and tanx?

To remeber the equations us SOH CAH TOA. This is Sinx = Opposite/Hypotenuse, Cos x = Adjacent/Hypotenuse and Tanx = Opposite/Adjacent. For values x=pi/3 and x=pi/6 always start by drawing an equilateral of s...
BL
Answered by Becky L. • Maths tutor
3879 Views

Let y be a function of x such that y=x^3 + (3/2)x^2-6x and y = f(x) . Find the coordinates of the stationary points .

y = x 3 + 1.5x 2 -6x Hence, dy/dx = 3x 2 + 3x - 6 Solve to find x when dy/dx = 0 as gradient is zero at stationary points Substitute the vaules for x back into y to find y coordinates of the stationary point...
MC
Answered by Michael C. • Maths tutor
4376 Views

Two forces P and Q act on a particle. The force P has magnitude 7 N and acts due north. The resultant of P and Q is a force of magnitude 10 N acting in a direction with bearing 120°. Find the magnitude of Q and the bearing of Q.

There are 2 methods to solving this- the visual method and the kinesthetic method. Here I will use the visual one. We start by creating a vector triangle. We are going to use R = P + Q, where R is the result...
YP
Answered by Yaasir P. • Maths tutor
12158 Views

A particle is moving in the with acceleration (2t - 3) ms^-2 and initial velocity 2ms^-1. Find the distance travelled when the velocity has reached 12ms^-1.

(1.) Integrate the expression for acceleration to find an expression for velocity: Velocity v = t^2 - 3t + c When t = 0, velocity = 2. Substituting in to find constant c, 2 = 0 + 0 + c therefore c = 2. v = t...
RF
Answered by Richard F. • Maths tutor
6948 Views