consider the following curve. y = √(19 + 2x) find y(x). y(x) = at what point on the given curve is the…

Explanation:

Step1: Rewrite the function

Rewrite $y = \sqrt{19 + 2x}=(19 + 2x)^{\frac{1}{2}}$.

Step2: Apply the chain - rule

The chain - rule states that if $y = f(g(x))$, then $y'=f'(g(x))\cdot g'(x)$. Let $u = 19+2x$, so $y = u^{\frac{1}{2}}$. First, find $\frac{dy}{du}=\frac{1}{2}u^{-\frac{1}{2}}$ and $\frac{du}{dx}=2$. Then $y'=\frac{1}{2}(19 + 2x)^{-\frac{1}{2}}\cdot2=\frac{1}{\sqrt{19 + 2x}}$.

Step3: Find the slope of the given line

Rewrite the line $20x + 4y=1$ in slope - intercept form $y=mx + b$ where $m$ is the slope. Solving for $y$ gives $4y=-20x + 1$, so $y=-5x+\frac{1}{4}$. The slope of this line is $m_1=-5$.

Step4: Find the slope of the tangent line

If two lines are perpendicular, the product of their slopes is $- 1$. Let the slope of the tangent line be $m_2$. Since $m_1m_2=-1$ and $m_1=-5$, then $m_2=\frac{1}{5}$.

Step5: Set the derivative equal to the slope of the tangent line

Set $y'=\frac{1}{\sqrt{19 + 2x}}=\frac{1}{5}$. Cross - multiply to get $\sqrt{19 + 2x}=5$.

Step6: Solve for $x$

Square both sides of the equation $\sqrt{19 + 2x}=5$: $19 + 2x = 25$. Then $2x=25 - 19=6$, so $x = 3$.

Step7: Solve for $y$

Substitute $x = 3$ into the original curve equation $y=\sqrt{19+2x}$. So $y=\sqrt{19+2\times3}=\sqrt{19 + 6}=\sqrt{25}=5$.

Answer:

$y'(x)=\frac{1}{\sqrt{19 + 2x}}$; $(x,y)=(3,5)$