Tutor profile: Thiaza T.

If sin θ = 0.625 and cos θ = 0.500 determine the values of cosec θ, sec θ, tan θ and cot θ

cosec θ = 1/sin θ = 1/0.625 = 1.60 sec θ = 1/cos θ = 1/0.500 = 2.00 tan θ = sin θ/cos θ = 0.625/0.500 = 1.25 cot θ = cos θ/sin θ = 0.500/0.625 = 0.80

Find the gradient of the curve y = 3x^4 − 2x^2 + 5x − 2 at the points (0, −2) and (1, 4)

The gradient of a curve at a given point is given by the corresponding value of the derivative. Thus, since y = 3x^4 − 2x^2 + 5x − 2 then the gradient; dy/dx = 12x^3 − 4x + 5 At the point (0, −2), x = 0. Thus the gradient = 12(0)3 − 4(0) + 5 = 5 At the point (1, 4), x = 1. Thus the gradient = 12(1)3 − 4(1) + 5 = 13

Show that y(x)=x^(-3/2) is a solution to 4x^2 (d^2 y)/(dx^2 )+12x dy/dx+3y=0 for x>0.

Show that y(x)=x^(-3/2) is a solution to 4x^2 (d^2 y)/(dx^2 )+12x dy/dx+3y=0 for x>0. We’ll need the first and second derivative to do this dy/dx= -3/2 x^(-5/2) (d^2 y)/(dx^2 )=15/4 x^(-7/2) Plug these as well as the function into the differential equation 4x^2 (-3/2 x^(-5/2) )+12x(15/4 x^(-7/2) )+3y=0 15x^(-3/2)-18x^(-3/2)+3x^(-3/2)=0 0=0 So, y(x)=x^(-3/2) does satisfy the differential equation and hence is a solution.