Chapter 2.9, Problem 38E

To determine

To find:The slopes of tangents to the given relations at given points.

Answer to Problem 38E

.

The slopes of tangents at given points are given in table below:

x	slope
-0.5	- 1
+0.5	- 1
-1	+8
+1	- 8
0	undefined

The graph of the curve is shown in figure.

Explanation of Solution

Given information:

The given implicit relation is $y^2+4xy+4x^4=0$ , the slopes of tangents at points $x=-0.5$ , $x=0.5$ $x=-1$ $x=1$ is required. The slope at of tangent line at $x=0$ is also to be discussed.

Method/Formulaused:

The slope of tangent line ( m ) at point $(a,b)$ to the curve $y=f\left(x\right)$ is given by $m={\left[\frac{dy}{dx}\right]}_{\left(a,b\right)}$ .

Calculations:

The given relation that represents a curve is

  $y^2+4xy+4x^4=0$   ... (1)

Differentiating with respect to ‘ x’

  $2y\frac{dy}{dx}+4\left(x\frac{dy}{dx}+y\right)+16x^3=0$  ... (2)

  $\Rightarrow \left(2y+4x\right)\frac{dy}{dx}=-\left(16x^3+4y\right)$

  $\Rightarrow \frac{dy}{dx}=-\frac{\left(16x^3+4y\right)}{\left(2y+4x\right)}=-\frac{\left(y+8x^3\right)}{\left(y+2x\right)}$  ... (3)

Now, from relation (1) $y^2+4xy+4x^4=0$

  $\Rightarrow y=\frac{-4x\pm \sqrt{16x^2-16x^4}}{2}=\frac{-4x\pm 4x\sqrt{1-x^2}}{2}=-2x\pm 2x\sqrt{1-x^2}$ .

Now, for $x=-0.5$ , we have $y=2x\left(-1\pm \sqrt{1-x^2}\right)=2\left(-0.5\right)\left(-1\pm \sqrt{1-{\left(-0.5\right)}^2}\right)$

  $=-1\left(-1\pm \sqrt{0.75}\right)=\left(1\pm \sqrt{0.75}\right)=0.1340,1.8660$ .

Hence, the slope of tangent at $x=-0.5$ are

  ${\left[\frac{dy}{dx}\right]}_{x\text{=}-0.5}=-\frac{\left(y+8x^3\right)}{\left(y+2x\right)}=-\frac{\left(0.1340+8{\left(-0.5\right)}^3\right)}{\left(0.1340+2\left(-0.5\right)\right)},-\frac{\left(1.8660+8{\left(-0.5\right)}^3\right)}{\left(1.8660+2\left(-0.5\right)\right)}$

  ${\left[\frac{dy}{dx}\right]}_{x\text{=}-0.5}=-1,-1$   ... (4)

Now, for $x=0.5$ , we have $y=2x\left(-1\pm \sqrt{1-x^2}\right)=2\left(0.5\right)\left(-1\pm \sqrt{1-{\left(0.5\right)}^2}\right)=\left(-1\pm \sqrt{0.75}\right)=-0.1340,-1.8660$

Hence, the slope of tangent at $x=0.5$ are

  ${\left[\frac{dy}{dx}\right]}_{x\text{=}0.5}=-\frac{\left(y+8x^3\right)}{\left(y+2x\right)}=-\frac{\left(-0.1340+8{\left(0.5\right)}^3\right)}{\left(-0.1340+2\left(0.5\right)\right)},-\frac{\left(-1.8660+8{\left(0.5\right)}^3\right)}{\left(-1.8660+2\left(0.5\right)\right)}$

  ${\left[\frac{dy}{dx}\right]}_{x\text{=}0.5}=-\frac{\left(-0.1340+1\right)}{\left(-0.1340+1\right)},-\frac{\left(-1.8660+1\right)}{\left(-1.8660+1\right)}=-1,-1$  ... (5)

Now, for $x=-1$ , we have

  $y=2x\left(-1\pm \sqrt{1-x^2}\right)=2\left(-1\right)\left(-1\pm \sqrt{1-{\left(-1\right)}^2}\right)=-2\left(-1\pm \sqrt{0}\right)=2,2$

Hence, the slope of tangent at $x=-1$ are

  ${\left[\frac{dy}{dx}\right]}_{x\text{=}-1}=-\frac{\left(y+8x^3\right)}{\left(y+2x\right)}=-\frac{\left(2+8{\left(-1\right)}^3\right)}{\left(2+2\left(-1\right)\right)},-\frac{\left(2+8{\left(-1\right)}^3\right)}{\left(2+2\left(-1\right)\right)}=\infty ,\infty$   ... (6)

Now, for $x=1$ , we have

  $y=2x\left(-1\pm \sqrt{1-x^2}\right)=2\left(1\right)\left(-1\pm \sqrt{1-{\left(1\right)}^2}\right)=2\left(-1\pm \sqrt{0}\right)=-2,-2$

Hence, the slope of tangent at $x=1$ are

  ${\left[\frac{dy}{dx}\right]}_{x\text{=}1}=-\frac{\left(y+8x^3\right)}{\left(y+2x\right)}=-\frac{\left(-2+8{\left(1\right)}^3\right)}{\left(-2+2\left(1\right)\right)},-\frac{\left(-2+8{\left(1\right)}^3\right)}{\left(-2+2\left(1\right)\right)}=-\infty ,-\infty$   ... (7)

Now, for $x=0$ , we have

  $y=2x\left(-1\pm \sqrt{1-x^2}\right)=0$

Hence, the slope of tangent at $x=0$ are

  ${\left[\frac{dy}{dx}\right]}_{x\text{=}0.5}=-\frac{\left(y+8x^3\right)}{\left(y+2x\right)}=\frac{0}{0}$ ,

that is undefined.

The graph of the curve and tangents at given points are shown in figure here.

  

Conclusion:

The slopes of tangents at given points are given in table below:

x	slope
-0.5	- 1
+0.5	- 1
-1	+8
+1	- 8
0	undefined

The graph of the curve is shown in figure.