The function $z(x,y)$ is given by $z(x,y)=4x^2y^{\frac{1}{3}}$. Determine the tangent line to the level curve through the point $(5,1)$.
Antwoord 1 correct
Correct
Antwoord 2 optie
$t(x)=-\frac{6}{5}x+6\frac{1}{5}$
Antwoord 2 correct
Fout
Antwoord 3 optie
$t(x)=\frac{11}{25}x-\frac{6}{5}$
Antwoord 3 correct
Fout
Antwoord 4 optie
$t(x)=6\frac{1}{5}x-\frac{6}{5}$
Antwoord 4 correct
Fout
Antwoord 1 optie
$t(x)=-\frac{6}{5}x+7$
Antwoord 1 feedback
Correct: $$\begin{align*}
\dfrac{z'_x(x,y)}{z'_y(x,y)}&= \dfrac{8xy^{-\frac{1}{3}}}{\frac{4}{3}x^2y^{-\frac{2}{3}}}\\
& = \dfrac{6y}{x}
\end{align*}$$
$$\begin{align*}
\textrm{slope} &=- \dfrac{6\cdot 1}{5}\\
&=-\dfrac{6}{5}.
\end{align*}$$
Hence, $t(x)=-\frac{6}{5}x+b$. Moreover, $t(5)=-\frac{6}{5} \cdot 5+b=1$ gives $b=7$.
Consequently, $t(x)=-\frac{6}{5}x+7$.
Go on.
\dfrac{z'_x(x,y)}{z'_y(x,y)}&= \dfrac{8xy^{-\frac{1}{3}}}{\frac{4}{3}x^2y^{-\frac{2}{3}}}\\
& = \dfrac{6y}{x}
\end{align*}$$
$$\begin{align*}
\textrm{slope} &=- \dfrac{6\cdot 1}{5}\\
&=-\dfrac{6}{5}.
\end{align*}$$
Hence, $t(x)=-\frac{6}{5}x+b$. Moreover, $t(5)=-\frac{6}{5} \cdot 5+b=1$ gives $b=7$.
Consequently, $t(x)=-\frac{6}{5}x+7$.
Go on.
Antwoord 2 feedback
Wrong: $x=5$ and $y=1$, not the other way around.
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Antwoord 3 feedback
Wrong: The slope of the tangent line is the number preceeding $x$.
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Try again.
Antwoord 4 feedback
Wrong: The slope of the tangent line is the number preceeding $x$.
Try again.
Try again.