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y = 2t^2, and x = 3t^3 - 2. Find dy/dx in terms of t.

dy/dt = 4t dx/dt = 9t^2dy/dx = dy/dt * dt/dx= 4/(9t)

Answered by Theo M. • Maths tutor

5747 Views

If I throw a ball, of mass 2kg, straight up in the air, with velocity 10ms-1, how long until it lands? Assume gravity = 10ms-2

Using SUVAT-s = ut + 1/2 at^2Subbing in... 0 = 2t - 5t^2t = 5/2 = 2.5

Answered by Theo M. • Maths tutor

2948 Views

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

Differentiate wrt x to get 3x² + 3y²dy/dx = 3y + 3x dy/dx Rearrange to get dy/dx = (3x² - 3y)/(3x-3y²). Set dy/dx =0 and infer y=x². Substitute in fo...

Answered by Joe B. • Maths tutor

9423 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 nu...

Answered by Harrison M. • Maths tutor

3330 Views

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*...

Answered by Sam F. • Maths tutor

3614 Views