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The gradient of a curve is given by dy/dx = 3 - x^2. The curve passes through the point (6,1). Find the equation of the curve.

Since we differentiate a function to find the gradient of a curve at any point, we need to reverse that to find the equation of the curve. We do this by integrating with respect to x:If you have a constan...

Answered by Darya N. • Maths tutor

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Points A and B have coordinates (–2, 1) and (3, 4) respectively. Find the equation of the perpendicular bisector of AB and show that it may be written as 5x +3 y = 10.

Formula for a straight line y-y₁=m(x-x₁), where m is the gradient substituting in the values given to find the gradient we get 4-1=m(3+2), therefore m= 3/5the midpoint of the two poi...

Answered by Paige M. • Maths tutor

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Find the values of x for which f(x) is an increasing function given that f(x)=8x-2x^2

When a function is increasing, it’s derivative is positive. So first let us differentiate f(x). To differentiate xⁿwe multiply by n and then reduce the power by 1. So f’(x)=8-2*2x=8-4x. We wan...

Answered by Ruby L. • Maths tutor

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Express 4sin(x)+6cos(x) in terms of Rsin(x+a) where R and a are constants to be determined (a should be given in rad).

R=sqrt(42)=6.48...a=arcos(4/sqrt(42))=0.905...Rsin(x+a)=6.48sin(x+0.905)

Answered by Benjamin U. • Maths tutor

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Integrate ln(x)

Take ln(x) = 1 * ln(x) and integrate by parts. Let u = ln(x) and dv/dx = 1 such that du/dx = 1/x and v = x. Using the integrating by parts formula, you get x ln(x) - integral(1) = x ln(x) - x + c.

Answered by Jason Y. • Maths tutor

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