Question

The graph of x^2=-2+y+5cosy is shown for y=11

Answer 1

(a)

We have been given an equation $x^{2}=-2+y+5cosy$. Upon taking derivative of this equation with respect to x on both the sides, we get:

$2x=0+\frac{dy}{dx}+5(-siny)\frac{dy}{dx}\\2x=(1-5siny)\frac{dy}{dx}\Rightarrow \frac{dy}{dx}=\frac{2x}{1-5siny}$

(b)

In order to write the equation of tangent line, we need to find the slope of the line. We know that point (3,11) lies on the graph. Therefore, we write the slope of the tangent as:

$\text{Slope}=\frac{dy}{dx}=\frac{2(3)}{1-5sin(11)}=1$

Therefore, equation of tangent line is:

$y-y_{1}=m(x-x_{1})\Rightarrow y-11=1(x-3)\\y-11=x-3\\y=x+8$

(c)

We know that slope of tangent is given as $\frac{dy}{dx}=\frac{2x}{1-5siny}$. The tangent will be vertical when denominator of slope is zero, that is:

$1-5siny=0\Rightarrow siny=\frac{1}{5}\Rightarrow y=arcsin(\frac{1}{5})\\y=1.732,1.525, 3.063 \text{ etc.}$