Answer to Question #289338 in Calculus for Ace

Question

Determine the equations of the tangent and normal to:

1. 𝑇ℎ𝑒 𝑝𝑎𝑟𝑎𝑏𝑜𝑙𝑎 𝑦² = 6𝑥 − 3 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑙𝑖𝑛𝑒 𝑥 + 3𝑦 = 7

2. 𝑇ℎ𝑒 𝑒𝑙𝑙𝑖𝑝𝑠𝑒 𝑥² + 2𝑦² = 9 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑡𝑜 𝑡ℎ𝑒 𝑙𝑖𝑛𝑒 4𝑥 − 𝑦 = 6

3. 𝑇ℎ𝑒 𝑐𝑢𝑟𝑣𝑒 (𝑥 − 𝑦) ² + 4𝑥 − 1 = 0 𝑎𝑡 (1,0)

4. 𝑇ℎ𝑒 𝑐𝑢𝑟𝑣𝑒 𝑦 = 2𝑥/𝑥+2 𝑎𝑡 𝑥 = 2

Locate the critical points and determine whether the given functions are either

maxima or minima using First Derivate Test and Second Derivative Test:

5. 𝑦 = 𝑥 ²(𝑥 − 2)²

6. 𝑦 = 𝑥(𝑥² − 6𝑥 − 15) + 50

7. 9a³y = x(4a − x)³

Answer 1

$1)\qquad\\ \text{We are given,}\\ \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad y^2=6x-3\\ \text{From the given line } x+3y=7\\ \text{We have}\\ \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad y=-\frac{x}{3}+\frac{7}{3}\\ \text{Therefore, the Slope of this line is }m=-\frac{1}{3}.\\ \qquad \text{The Slope perpendicular to the Slope of the given line (m$\perp$) is }\mathbf{m_{\perp}}=3.\\ \text{Next, differentiating }y^2=6x+3 \text{ implicitly wrt } x:\\ \Rightarrow2yy\prime=6\\ \Rightarrow y\prime=\frac{3}{y}=\mathbf{m_{tan}}, \text{where $\mathbf{m_{tan}}$ is the Slope of a tangent line to the curve } y^2=6x-3.\\ \text{Next, } \mathbf{m_{\perp}}=\mathbf{m_{tan}}\\ \Rightarrow3=\frac{3}{y}\\ \Rightarrow y=1\\ \text{Therefore }y^2=6x-3\\ \Rightarrow 1^2=6x-3\Rightarrow x=\frac{2}{3}\\ \text{Therefore, the eqn. of the tangent passing through ($\frac{2}{3},1$) is}\\ \frac{y-1}{x-\frac{2}{3}}=m_\perp=3\\ \Rightarrow y=3x-1\\ \text{Therefore, the eqn. of the tangent line is}\\ \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad y=3x-1\\ \qquad\\ \text{Next,}\\ \mathbf{m_{normal}}=-\frac{1}{\mathbf{m_{tan}}}=-\frac{y}{3}\\ \mathbf{m_{normal}}=\mathbf{m_{\perp}}\\ \Rightarrow3=-\frac{y}{3}\Rightarrow y=-9\\ \text{So, }y^2=6x-3\Rightarrow(-9)^2=6x-3\Rightarrow x=14\\ \text{Therefore, the eqn. of the normal passing through (14,-9) is}\\ \frac{y+9}{x-14}=m_\perp=3\Rightarrow y=3x-51.\\ \quad\\ \quad\\ \quad\\ 2)\qquad\\\text{We are given,}\\ \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad x^2+2y^2=9\\ \text{From the given line } 4x-y=6\\ \text{We have}\\ \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad y=4x-6\\ \text{Therefore, the Slope of this line is }m=4.\\ \qquad \text{The Slope perpendicular to the Slope of the given line (m$\perp$) is }\mathbf{m_{\perp}}=-\frac{1}{4}.\\ \text{Next, differentiating }x^2+2y^2=9 \text{ implicitly wrt } x:\\ \Rightarrow 2x+4yy\prime=0\\ \Rightarrow y\prime=\frac{-x}{2y}=\mathbf{m_{tan}}, \text{where $\mathbf{m_{tan}}$ is the Slope of a tangent line to the curve } x^2+2y^2=9.\\ \text{Next, } \mathbf{m_{\perp}}=\mathbf{m_{tan}}\\ \Rightarrow -\frac{1}{4}=-\frac{x}{2y}\\ \Rightarrow y=2x\\ \text{Therefore }x^2+2y^2=9\\ \Rightarrow x^2+2(2x)^2=9\Rightarrow x=-1,1\\ \text{At x=-1, y=-2}\\ \text{At x=1, y=2}\\ \quad\text{The eqn. of the tangent passing through ($-1,-2$), is}\\ \frac{y+2}{x+1}=m_\perp=-\frac{1}{4}\\ \Rightarrow y=-\frac{x}{4}-\frac{9}{4}\\ \quad\text{Also, the eqn. of the tangent passing through ($1,2$), is}\\ \frac{y-2}{x-1}=m_\perp=-\frac{1}{4}\\ \Rightarrow y=-\frac{x}{4}+\frac{9}{4}\\ \text{Therefore, the eqn. of the tangent lines are}\\ \qquad\qquad\qquad\qquad\qquad\qquad y=-\frac{x}{4}-\frac{9}{4},\ \ y=-\frac{x}{4}+\frac{9}{4}\\ \qquad\\ \text{Next,}\\ \mathbf{m_{normal}}=-\frac{1}{\mathbf{m_{tan}}}=\frac{2y}{x}\\ \mathbf{m_{normal}}=\mathbf{m_{\perp}}\\ \Rightarrow -\frac{1}{4}=\frac{2y}{x}\Rightarrow y=-\frac{x}{8}\\ \text{So, }x^2+2y^2=9\Rightarrow x^2+2(-\frac{x}{8})^2=9\Rightarrow x=\pm\sqrt{\frac{96}{11}}\\ \text{At }x=\sqrt{\frac{96}{11}}, y=-\sqrt{\frac{3}{22}}\\ \text{At }x=-\sqrt{\frac{96}{11}}, y=\sqrt{\frac{3}{22}}\\ \quad\text{The eqn. of the tangent passing through ($\sqrt{\frac{96}{11}},-\sqrt{\frac{3}{22}}$), is}\\ \frac{y+\sqrt{\frac{3}{22}}}{x-\sqrt{\frac{96}{11}}}=m_\perp=-\frac{1}{4}\\ \Rightarrow y=-\frac{x}{4}-\sqrt{\frac{6}{11}}\\ \quad\text{Also, the eqn. of the tangent passing through ($-\sqrt{\frac{96}{11}},\sqrt{\frac{3}{22}}$), is}\\ \frac{y-\sqrt{\frac{3}{22}}}{x+\sqrt{\frac{96}{11}}}=m_\perp=-\frac{1}{4}\\ \Rightarrow y=-\frac{x}{4}+\sqrt{\frac{6}{11}}\\ \qquad\qquad\qquad\qquad\qquad\qquad y=-\frac{x}{4}-\sqrt{\frac{6}{11}},\ \ y=-\frac{x}{4}+\sqrt{\frac{6}{11}}\\ \quad\\ \quad\\ \quad\\ 3)\quad\\ \text{We are given }(x-y)^2+4x-1=0\\ \text{differentiating }(x-y)^2+4x-1=0 \text{ implicitly wrt x;}\\ \Rightarrow2(x-y)(-y\prime)+4=0\\ \Rightarrow y\prime=\frac{2}{x-y}\\ \text{At } (1,0);\\ m_{tan}=y\prime=\frac{2}{1-0}=2\\ \text{Thus, the equation of the tangent of the curve at (1, 0) is;}\\ \frac{y-0}{x-1}=2=m_{tan}\Rightarrow y=2(x-1)\\ \quad\\ \text{Next, }m_{tan}\times m_{normal}=-1\Rightarrow m_{normal}=\frac{-1}{2}\\ \text{Thus, the equation of the normal of the curve at (1, 0) is;}\\ \frac{y-0}{x-1}=\frac{-1}{2}=m_{normal}\Rightarrow y=-\frac{x}{2}+\frac{1}{2}\\ \quad\\ \quad\\ \quad\\ 4)\quad\\ \text{We are given }y=\frac{2x}{x+2}\\ \text{differentiating }y=\frac{2x}{x+2} \text{ implicitly wrt x;}\\ \Rightarrow y\prime=\frac{4}{(x+2)^2}\\ \text{At }x=2;\\ m_{tan}=y\prime=\frac{1}{4}, y=1\\ \text{Thus, the equation of the tangent of the curve passing through (2, 1) is;}\\ \frac{y-1}{x-2}=\frac{1}{4}=m_{tan}\Rightarrow y=\frac{x}{4}+\frac{1}{2}\\ \quad\\ \text{Next, }m_{tan}\times m_{normal}=-1\Rightarrow m_{normal}=-4\\ \text{Thus, the equation of the normal of the curve at (2, 1) is;}\\ \frac{y-1}{x-2}=-4=m_{normal}\Rightarrow y=-4x+9\\ \quad\\ \quad\\ \quad\\ 5)\\ y=x^2(x-2)^2, \text{differentiating yields:}\\y\prime=4x(x-1)(x-2)\\ \quad\text{At critical point, }y\prime=0 \Rightarrow x=0,1,2 \text{ are the critical points.}\\ \text{\textbf{FIRST DERIVATIVE TEST:}}\\ y\prime(-1)=-24\Rightarrow decreasing\ on\ (-\infty,0) \\ y\prime(\frac{1}{2})=\frac{3}{2}\Rightarrow increasing\ on\ (1,2)\\ y\prime(\frac{3}{2})=-\frac{3}{2}\Rightarrow decreasing\ on\ (0,1) \\ y\prime(3)=24\Rightarrow increasing\ on\ (2,\infty)\\ \text{Since switching from decreasing to increasing and increasing to decreasing at a point}\\ \Rightarrow \text{the point is a local minimum point and local maximum point respectively, thus}\\ \text{x = 1 is a local maximum point and x = 0, 2 are local minimum points.}\\ \text{Thus, the function has local maximum at (1, 1) and local minimum at (2, 0),(0, 0)}.\\ \quad\\ \text{\textbf{SECOND DERIVATIVE TEST:}}\\ y\prime\prime=12x^2-24x+8\\ \text{Plug in the critical numbers:}\\ y\prime\prime(0)=8\Rightarrow0 \text{ is a minimum point}\\ y\prime\prime(1)=-4\Rightarrow1 \text{ is a maximum point}\\ y\prime\prime(2)=8\Rightarrow0 \text{ is a minimum point}\\ \text{Thus, the function has local maximum at (1, 1) and local minimum at (2, 0),(0, 0)}.\\ \quad\\ \quad\\ \quad\\ 6)\\ y=x^3-6^2-15x+50, \text{differentiating yields:}\\y\prime=3(x-5)(x+1)\\ \quad\text{At critical point, }y\prime=0 \Rightarrow x=5,-1 \text{ are the critical points.}\\ \text{\textbf{FIRST DERIVATIVE TEST:}}\\ y\prime(-2)=21\Rightarrow increasing\ on\ (-\infty,-1) \\ y\prime(0)=-15\Rightarrow decreasing\ on\ (-1,5)\\ y\prime(9)=120\Rightarrow increasing\ on\ (5,\infty) \\ \text{Since switching from decreasing to increasing and increasing to decreasing at a point}\\ \Rightarrow \text{the point is a local minimum point and local maximum point respectively, thus}\\ \text{x = -1 is a local maximum point and x = 5 is a local minimum point.}\\ \text{Thus, the function has local maximum at (-1, 58) and local minimum at (5, -50)}.\\ \quad\\ \text{\textbf{SECOND DERIVATIVE TEST:}}\\ y\prime\prime=6x-12\\ \text{Plug in the critical numbers:}\\ y\prime\prime(-1)=-18\Rightarrow0 \text{ is a maximum point}\\ y\prime\prime(5)=18\Rightarrow1 \text{ is a minimum point}\\ \text{Thus, the function has local maximum at (-1, 58) and local minimum at (5, -50)}.\\ \quad\\ \quad\\ \quad\\ 7)\\ y=\frac{x(4a-x)^3}{9a^3}, \text{differentiating yields:}\\y\prime=\frac{(4a-x)^2(4a-4x)}{9a^3}\\ \quad\text{At critical point, }y\prime=0 \Rightarrow x=a, 4a \text{ are the critical points.}\\ \quad\\ \text{Determine the 2nd derivative:}\\ 9a^3 y'' = -12(4a - x)(2a - x)\\ \quad\\ x = 4a\Rightarrow\\ y=0\\ \text{Check if Maxima or Minima.}\\ 9a^3 y''(4a) = -12(4a - 4a)(2a - 4a)\Rightarrow y'' = 0 \ (\text{inflection point})\\ \text{Hence, (4a, 0) is neither maximum nor minimum}\\ \quad\\ x=a \Rightarrow y=3a\\ \text{Check if Maxima or Minima}\\ y''(a) = (-)\ (\text{the curve is concave downward})\\ \text{Hence, (a, 0) is maximum.}$