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Express the equation cosecθ(3 cos 2θ+7)+11=0 in the form asin^2(θ) + bsin(θ) + c = 0, where a, b and c are constants.

We must first use the identity cosecθ = 1/sinθ. Now the equation becomes (1/sinθ)(3 cos 2θ+7)+11=0. Since we know that the question is asking for the answer in the form of asin²θ + bsinθ + c = ...

Answered by George L. • Maths tutor

6444 Views

Explain why for any constant a, if y = a^x then dy/dx = a^x(ln(a))

So let's start with taking the natural log on both sides of y=a^x, giving us ln(y) = ln(a^x). Using the laws of logarithms we can write this as ln(y) = xln(a).Next, we differentiate bo...

Answered by James M. • Maths tutor

12078 Views

Find the coordinates and determine the nature of the stationary points of curve y=(2/3)x^3+2x^2-6x+3

Stationary points occur when dy/dx=0, therefore determine dy/dx first:
dy/dx= 2x²+ 4x - 6
2) solve dy/dx=0 for two values of x (using quadratic formula, if necessary):
(x+...

Answered by Barbora V. • Maths tutor

3352 Views

Solve for x and y: 2x +5y + 5= 0 , 2y + 31= 5x

For a question like this you should aim to eliminate either x OR y from one equation in order to deduce the value of the other. 1) 2x +5y + 5= 0 , 2) 2y + 31= 5x
Rearrange equation 2) so that 2y +31...

Answered by Tutor114325 D. • Maths tutor

5422 Views

The normal to the curve x*(e^-y) + e^y = 1 + x, at the point (c,lnc), has a y-intercept c^2 + 1. Determine the value of c.

We firstly need to find the slope of the line normal to the curve at the point (c, lnc). To do this, we will differentiate implicitly the equation of the curve to find the slope of the tangent. By implici...

Answered by Katerina A. • Maths tutor

6909 Views