Question

Determine a equação da reta tangentes a curva y = $\sqrt{100-x^{2}$ , no ponto P = 6, 8.

Answer 1

$\large\boxed{\begin{array}{l}\rm y=\sqrt{100-x^2}~em~P(6,8).\\\sf~vamos~calcular~a~derivada\\\sf e~em~seguida~substituir~no~ponto.\\\sf Feito~isto,usa-se~a~equac_{\!\!,}\tilde ao\,da\,reta\\\sf na~forma~ponto-coeficiente~angular\\\rm y=y_p+f'(x_p)_\cdot(x-x_p)\\\rm y_p\longrightarrow y\,do\,ponto\,dado\\\rm x_p\longrightarrow x\,do\,ponto\,dado.\\\rm f'(x_p)\longrightarrow derivada\,no\,ponto\,x_p\end{array}}$

$\large\boxed{\begin{array}{l}\rm y=\sqrt{100-x^2}~P(6,8)\\\rm f(x)=\sqrt{100-x^2}\\\rm f'(x)=\dfrac{1}{\backslash\!\!\!\!2\sqrt{100-x^2}}\cdot(-\backslash\!\!\!\!\!2x)=-\dfrac{x}{\sqrt{100-x^2}}\\\\\rm f'(6)=-\dfrac{6}{\sqrt{100-6^2}}\\\\\rm f'(6)=-\dfrac{6}{\sqrt{100-36}}\\\\\rm f'(6)=-\dfrac{6}{\sqrt{64}}\\\\\rm f'(6)=-\dfrac{6\div2}{8\div2}\\\\\rm f'(6)=-\dfrac{3}{4}\end{array}}$

$\large\boxed{\begin{array}{l}\rm y=8+\bigg(-\dfrac{3}{4}\bigg)(x-6)\\\\\rm y=8-\dfrac{3}{4} (x-6)\\\\\rm y=8-\dfrac{3}{4}x+\dfrac{9}{2}\bullet4\\\\\rm 4y=32-3x+18\\\rm 3x+4y-50=0\longrightarrow equac_{\!\!,}\tilde ao\,da\,reta\,tangente\end{array}}$

Answer 2

Resposta:

3x + 4y - 50 = 0

Passo-a-passo

$f(x)=\sqrt{100-x^2} \\\\f'(x)=\frac{-2x}{2\sqrt{100-2x} } \\\\f'(x)=-\frac{x}{\sqrt{100-x^2} } \\\\m=-\frac{6}{\sqrt{100-36} } \\\\m=-\frac{6}{\sqrt{64} } \\\\m=-\frac{6}{8}=-\frac{3}{4} \\\\ y-8=-\frac{3}{4} (x-6)\\\\4y-32=-3x+18\\\\3x+4y-50=0$