Question

if $\frac{x^{2}}{36}+\frac{y^{2}}{16}=1$ is the equation of an ellipse, and the point $(3,3.46)$ is on its graph, use implicit differentiation to find the slope of the tangent line at that point. $y(3)=$

Explanation:

Step1: Differentiate both sides of the equation

Differentiate \(\frac{x^{2}}{36}+\frac{y^{2}}{16} = 1\) with respect to \(x\).
Using the sum rule \((u + v)^\prime=u^\prime + v^\prime\), where \(u=\frac{x^{2}}{36}\) and \(v = \frac{y^{2}}{16}\).
For \(u=\frac{x^{2}}{36}\), by the power rule \((x^{n})^\prime=nx^{n - 1}\), \(u^\prime=\frac{2x}{36}=\frac{x}{18}\).
For \(v=\frac{y^{2}}{16}\), using the chain - rule \((f(g(x)))^\prime=f^\prime(g(x))\cdot g^\prime(x)\) (here \(f(u)=\frac{u^{2}}{16}\), \(u = y(x)\)), \(v^\prime=\frac{2y}{16}\cdot y^\prime=\frac{y}{8}y^\prime\).
The derivative of the right - hand side (since the derivative of a constant \(C\) is \(0\)) is \(0\). So, \(\frac{x}{18}+\frac{y}{8}y^\prime=0\).

Step2: Solve for \(y^\prime\)

Subtract \(\frac{x}{18}\) from both sides of the equation \(\frac{x}{18}+\frac{y}{8}y^\prime=0\) to get \(\frac{y}{8}y^\prime=-\frac{x}{18}\).
Then, multiply both sides by \(\frac{8}{y}\) (assuming \(y
eq0\)) to isolate \(y^\prime\). So, \(y^\prime=-\frac{8x}{18y}=-\frac{4x}{9y}\).

Step3: Substitute \(x = 3\) and \(y = 3.46\)

Substitute \(x = 3\) and \(y = 3.46\) into the formula for \(y^\prime\).
\(y^\prime=-\frac{4\times3}{9\times3.46}\).
First, calculate \(4\times3 = 12\) and \(9\times3.46=31.14\).
Then, \(y^\prime=-\frac{12}{31.14}\approx - 0.385\).

Answer:

\(-0.385\)