Question

question 5
what is the slope of the tangent line to the curve $x^{2}+y^{2}=1$ at the point $(\frac{3}{5},\frac{4}{5})$?
$m = -\frac{3}{4}$
$m = \frac{3}{4}$
$m = \frac{4}{3}$
$m = -\frac{4}{3}$

Explanation:

Step1: Differentiate implicitly

Differentiate $x^{2}+y^{2}=1$ with respect to $x$. Using the chain - rule, we have $2x + 2y\frac{dy}{dx}=0$.

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

First, isolate $\frac{dy}{dx}$ in the equation $2x + 2y\frac{dy}{dx}=0$. Subtract $2x$ from both sides: $2y\frac{dy}{dx}=-2x$. Then divide both sides by $2y$ (assuming $y
eq0$) to get $\frac{dy}{dx}=-\frac{x}{y}$.

Step3: Substitute the point

We want to find the slope at the point $(\frac{3}{5},\frac{4}{5})$. Substitute $x = \frac{3}{5}$ and $y=\frac{4}{5}$ into $\frac{dy}{dx}=-\frac{x}{y}$. So $\frac{dy}{dx}=-\frac{\frac{3}{5}}{\frac{4}{5}}=-\frac{3}{4}$.

Answer:

A. $m =-\frac{3}{4}$