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Title: Geometric Picture for the Gradient
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Series: Multivariable Calculus
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YouTube-Title: Multivariable Calculus 9 | Geometric Picture for the Gradient
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Bright video: https://youtu.be/P0EyEdEcS3c
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Dark video: https://youtu.be/R4gbrX6sZuU
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Quiz: Test your knowledge
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Dark-PDF: Download PDF version of the dark video
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Print-PDF: Download printable PDF version
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Thumbnail (bright): Download PNG
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Thumbnail (dark): Download PNG
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Subtitle on GitHub: mc09_sub_eng.srt missing
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Timestamps (n/a)
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Subtitle in English (n/a)
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Quiz Content
Q1: Let $\gamma: \mathbb{R} \rightarrow \mathbb{R}^2$ and $f: \mathbb{R}^2 \rightarrow \mathbb{R}$ be two totally differentiable functions where $f\circ \gamma$ is constant. What is the derivative $\frac{d}{dt}( f(\gamma(t)))$?
A1: 0
A2: 1
A3: One needs more information.
Q2: Let $\gamma: \mathbb{R} \rightarrow \mathbb{R}^2$ be given by $\gamma(t) = \binom{t}{t}$. How does the curve look in $\mathbb{R}^2$?
A1: It’s a straight line.
A2: It’s a circle.
A3: It’s a parabula.
Q3: Let $\gamma: \mathbb{R} \rightarrow \mathbb{R}^2$ be given by $\gamma(t) = \binom{t}{t}$. Assume that the image of $\gamma$ lies on a contour line of a function $f: \mathbb{R}^2 \rightarrow \mathbb{R}$. What can one say about the gradient $\mathrm{grad}f(\gamma(t))$?
A1: It’s perpendicular to the vector $\binom{1}{1}$.
A2: It’s perpendicular to the vector $\binom{1}{0}$.
A3: It’s parallel to the vector $\binom{1}{0}$.
A4: It’s parallel to the vector $\binom{1}{1}$.
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Last update: 2024-10