unit 6: quiz: related rates and application of derivatives
score: 20/100 2/15 answered
question 3
find $\frac{dz}{dt}$ at $(x,y) = (3,3)$ and $z^2 = x^2 + y^2$ if $\frac{dx}{dt} = 1$ and $\frac{dy}{dt} = 1$.
$\frac{dz}{dt} =$

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Accepted Answer

{ "quality": "clear", "question_count": 1, "skills_matched": [ "step_cot", "blank_filling_formatter", "explain_then_answer", "quick" ], "needs_drawing": false, "concepts_used": [ "Related Rates", "Implicit Differentiation" ], "new_concepts": [], "current_concepts": [ "Related Rates", "Implicit Differentiation" ] } ## Differentiate the relation with respect to time $$ \begin{aligned} z^2 &= x^2 + y^2 \ 2z \frac{dz}{dt} &= 2x \frac{dx}{dt} + 2y \frac{dy}{dt} \ z \frac{dz}{dt} &= x \frac{dx}{dt} + y \frac{dy}{dt} \end{aligned} $$ ## Determine the value of z at the given point $$ \begin{aligned} z^2 &= 3^2 + 3^2 = 18 \ z &= \sqrt{18} = 3\sqrt{2} \quad (z > 0) \end{aligned} $$ ## Substitute known values and solve for dz/dt $$ \begin{aligned} 3\sqrt{2} \frac{dz}{dt} &= 3(1) + 3(1) \ 3\sqrt{2} \frac{dz}{dt} &= 6 \ \frac{dz}{dt} &= \frac{6}{3\sqrt{2}} = \sqrt{2} \end{aligned} $$ Find $\frac{dz}{dt}$ at $(x,y) = (3,3)$ and $z^2 = x^2 + y^2$ if $\frac{dx}{dt} = 1$ and $\frac{dy}{dt} = 1$. $\frac{dz}{dt} =$ $\sqrt{2}$ { "subject": "Mathematics", "question_type": "Fill-in-the-blank", "knowledge_point": [ "Mathematics", "Calculus", "Related Rates" ] }

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Explanation:

Differentiate the relation with respect to time

Determine the value of z at the given point

Substitute known values and solve for dz/dt

Answer: