verify that the given point is on the curve. find the lines that are (a) tangent and (b) normal to the curve…

verify that the given point is on the curve. find the lines that are (a) tangent and (b) normal to the curve at the given point.\n6xy + \\pi\\sin y = 61\\pi, (4, \\frac{5\\pi}{2})\nthe point is on the curve because when 4 is substituted for x and \\frac{5\\pi}{2} is substituted for y, the resulting statement is 61\\pi = 61\\pi, which is a true statement. (type an exact answer, using \\pi as needed.)\na. write the equation of the tangent line to the curve at (4, \\frac{5\\pi}{2})\ny = -\\frac{5\\pi}{8}x + 5\\pi (type an equation. type an exact answer, using \\pi as needed.)\nb. write the equation of the normal line to the curve at (4, \\frac{5\\pi}{2})\n(type an equation. type an exact answer, using \\pi as needed.)

verify that the given point is on the curve. find the lines that are (a) tangent and (b) normal to the curve at the given point.\n6xy + \\pi\\sin y = 61\\pi, (4, \\frac{5\\pi}{2})\nthe point is on the curve because when 4 is substituted for x and \\frac{5\\pi}{2} is substituted for y, the resulting statement is 61\\pi = 61\\pi, which is a true statement. (type an exact answer, using \\pi as needed.)\na. write the equation of the tangent line to the curve at (4, \\frac{5\\pi}{2})\ny = -\\frac{5\\pi}{8}x + 5\\pi (type an equation. type an exact answer, using \\pi as needed.)\nb. write the equation of the normal line to the curve at (4, \\frac{5\\pi}{2})\n(type an equation. type an exact answer, using \\pi as needed.)

Answer

Explanation:

Step1: Differentiate the given equation implicitly

Differentiate $6xy+\pi\sin y = 61\pi$ with respect to $x$. Using the product - rule $(uv)^\prime=u^\prime v + uv^\prime$ for $6xy$ (where $u = 6x$ and $v = y$) and the chain - rule for $\pi\sin y$. The derivative of $6xy$ is $6y+6x\frac{dy}{dx}$, and the derivative of $\pi\sin y$ is $\pi\cos y\frac{dy}{dx}$, and the derivative of $61\pi$ (a constant) is $0$. So, $6y + 6x\frac{dy}{dx}+\pi\cos y\frac{dy}{dx}=0$.

Step2: Solve for $\frac{dy}{dx}$

Factor out $\frac{dy}{dx}$: $\frac{dy}{dx}(6x+\pi\cos y)=- 6y$. Then $\frac{dy}{dx}=\frac{-6y}{6x+\pi\cos y}$.

Step3: Find the slope of the tangent line at the point $(4,\frac{5\pi}{2})$

Substitute $x = 4$ and $y=\frac{5\pi}{2}$ into $\frac{dy}{dx}$. $\cos(\frac{5\pi}{2}) = 0$, so $\frac{dy}{dx}\big|_{x = 4,y=\frac{5\pi}{2}}=\frac{-6\times\frac{5\pi}{2}}{6\times4+\pi\times0}=-\frac{5\pi}{8}$. The equation of a line in point - slope form is $y - y_1=m(x - x_1)$, where $(x_1,y_1)=(4,\frac{5\pi}{2})$ and $m =-\frac{5\pi}{8}$. $y-\frac{5\pi}{2}=-\frac{5\pi}{8}(x - 4)$. $y-\frac{5\pi}{2}=-\frac{5\pi}{8}x+\frac{5\pi}{2}$. $y=-\frac{5\pi}{8}x + 5\pi$.

Step4: Find the slope of the normal line

The slope of the normal line $m_n$ is the negative reciprocal of the slope of the tangent line. Since the slope of the tangent line $m_t=-\frac{5\pi}{8}$, then $m_n=\frac{8}{5\pi}$. Using the point - slope form $y - y_1=m_n(x - x_1)$ with $(x_1,y_1)=(4,\frac{5\pi}{2})$ and $m_n=\frac{8}{5\pi}$. $y-\frac{5\pi}{2}=\frac{8}{5\pi}(x - 4)$. $y=\frac{8}{5\pi}x-\frac{32}{5\pi}+\frac{5\pi}{2}$. $y=\frac{8}{5\pi}x+\frac{- 64 + 125\pi^{2}}{10\pi}$

Answer:

$y=\frac{8}{5\pi}x+\frac{- 64 + 125\pi^{2}}{10\pi}$