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Find the stationary points of the graph x^3 + y^3 = 3xy +35

Differentiate wrt x to get 3x 2 + 3y 2 dy/dx = 3y + 3x dy/dx Rearrange to get dy/dx = (3x 2 - 3y)/(3x-3y 2 ). Set dy/dx =0 and infer y=x 2 . Substitute in for y into original equation and rearrange to get x ...
JB
Answered by Joe B. • Maths tutor
10561 Views

Differentiate y=ln(x)+5x^2, and give the equation of the tangent at the point x=1

First differentiate the equation, giving you, y'=(1/x)+10x. To get the gradient at this point of the curve, plug in x=1, to get a y' value of 11, and a y value of 5. From there you can plug these three numbe...
HM
Answered by Harrison M. • Maths tutor
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Find the stationary points of y = (x-7)(x-3)^2.

This problem requires the use of the product rule.Let u= x-7 ; by differentiation du/dx = 1. Let v = (x-3)^2 ; by differentiation using the chain rule, dv/dx = 2(x-3) Product Rule: dy/dx = u*(dv/dx) + v*(du/...
SF
Answered by Sam F. • Maths tutor
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Question shown in the answer section as a hyperlinked link.

Question tests students knowledge of trigonometric identities and reverse chain rule integration.
SG
Answered by Samuel G. • Maths tutor
3627 Views

A particle of mass m is placed on an slope with an incline 30 degrees. Once released it accelerates down the line of greatest slope at 2 m s^-2. What is the coefficient of friction between the particle and the slope?

Resolve forces perpendicular to the slope. No acceleration in this direction hence resultant force in this direction must equal zero:R = mg * cos(30) Apply Newtons second law in the direction parallel to the...
DH
Answered by Daniel H. • Maths tutor
4068 Views