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})\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})\n(type an equation. type an exact answer, using \\pi as needed.)

Answer

Explanation:

Step1: Differentiate 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 $\pi\sin y$. We get $6y+6x\frac{dy}{dx}+\pi\cos y\frac{dy}{dx}=0$.

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

Substitute $x = 4$ and $y=\frac{5\pi}{2}$ into $\frac{dy}{dx}$. $\cos(\frac{5\pi}{2}) = 0$, so $\frac{dy}{dx}\big|_{x = 4,y=\frac{5\pi}{2}}=\frac{-6\times\frac{5\pi}{2}}{6\times4+\pi\times0}=\frac{-15\pi}{24}=-\frac{5\pi}{8}$.

Step4: Use the point - slope form to find the tangent line equation

The point - slope form is $y - y_1=m(x - x_1)$, where $(x_1,y_1)=(4,\frac{5\pi}{2})$ and $m =-\frac{5\pi}{8}$. $y-\frac{5\pi}{2}=-\frac{5\pi}{8}(x - 4)$. Expand: $y-\frac{5\pi}{2}=-\frac{5\pi}{8}x+\frac{5\pi}{2}$. The equation of the tangent line is $y=-\frac{5\pi}{8}x + 5\pi$.

Answer:

$y=-\frac{5\pi}{8}x + 5\pi$