Question

What is the slope of the line tangent to the curve y+2 = (x^2/2) - 2siny at the point (2,0)?

Answer 1

, y+2 = (x^2/2) - 2sin(y) so we are taking the derivative y in respect to x so we have dy/dx use chain rule on y so y' = 2x/2 - 2cos(y)*y'

Now rearrange it to solve for y' y' = 2x/2 - 2cos(y)*y' 0 = x - 2cos(y)y' - y' - x = 2cos(y)y' - y' -x = y'(2cos(y) - 1) -x/(2cos(y) - 1) = y'
we know when f(2) = 0 so thus y = 0 so when f'(2) = -2/(2cos(0)-1)
2/2 = 1
f'(2) = -2/(2cos(0)-1) cos(0) = 1 thus f'(2) = -2/(2(1)-1) = -2/-1 = 2 f'(2) = 2

Answer 2

Answer:

The slope of the line tangent to the given curve at the point (2,0) is 2/3.

Step-by-step explanation:

The given equation is

$y+2=(\frac{x^2}{2})-2\sin y$

we need to find the slope of the line tangent to the given curve at the point (2,0).

$slope=\frac{dy}{dx}_{(2,0)}$

Differentiate given equation with respect to x.

$\frac{dy}{dx}+0=\frac{2x}{2}-2(\cos y)\frac{dy}{dx}$

$\frac{dy}{dx}=x-2\cos y\frac{dy}{dx}$

$\frac{dy}{dx}+2\cos y\frac{dy}{dx}=x$

$(1+2\cos y)\frac{dy}{dx}=x$

Divide both sides by(1+2cos y).

$\frac{dy}{dx}=\frac{x}{(1+2\cos y)}$

Substitute x=2 and y=0.

$\frac{dy}{dx}_{(2,0)}=\frac{2}{1+2\cos (0)}$

$\frac{dy}{dx}_{(2,0)}=\frac{2}{1+2(1)}$

$\frac{dy}{dx}_{(2,0)}=\frac{2}{3}$

Therefore the slope of the line tangent to the given curve at the point (2,0) is 2/3.