• Title: Geometric Picture for the Gradient

  • Series: Multivariable Calculus

  • YouTube-Title: Multivariable Calculus 9 | Geometric Picture for the Gradient

  • Bright video: https://youtu.be/P0EyEdEcS3c

  • Dark video: https://youtu.be/R4gbrX6sZuU

  • Quiz: Test your knowledge

  • PDF: Download PDF version of the bright video

  • Dark-PDF: Download PDF version of the dark video

  • Print-PDF: Download printable PDF version

  • Thumbnail (bright): Download PNG

  • Thumbnail (dark): Download PNG

  • Subtitle on GitHub: mc09_sub_eng.srt missing

  • Timestamps (n/a)
  • Subtitle in English (n/a)
  • 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 $\binom{1}{1}$.

  • Last update: 2024-10

  • Back to overview page