Let y = 4t/(t^2 + 5). Find dy/dt, writing your answer in it's simplest form, and find all values of t for which dy/dt = 0
We use the quotient rule to find dy/dt. Let u = 4t and v = (t^2 + 5). Then, u' = 4 and v' = 2t. Hence,
dy/dt = u'v - v'u / v2 = 4(t^2 + 5) - 4t x 2t / (t^2 + 5)2 = 20 - 4t2 / (t^2 + 5)2. Now, we need to find all t such that dy/dt < 0 i.e.
20 - 4t2 / (t^2 + 5)2 < 0 which rearranges to give t2 > 5, so, t > 51/2 and t < -51/2.
JH
Related Maths A Level answers
All answers ▸The equation of a curve is x(y^2)=x^2 +1 . Using the differential, find the coordinates of the stationary point of the curve.
Write 36% as a fraction in its simplest terms.
The complex numbers Z and W are given by Z=3+3i and W=6-i. Giving your answers in the form of x+yi and showing how you clearly obtain them, find: i) 3Z-4W ii) Z*/W
(C3 question). Find an expression for all stationary points on the curve y=sin(x)cos(x). How many such points are there and why?