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Course: Multivariable calculus>Unit 2

Lesson 7: Partial derivatives of vector-valued functions

Partial derivative of a parametric surface, part 1

When a vector-valued function represents a parametric surface, how do you interpret its partial derivative? Created by Grant Sanderson.

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• At , why is the partial dt vector stretched to something that points along the curve? Why is it necessarily tangent to the curve in the output space?
• 1. stretching only change the magnitude of vector, and has no effect on the direction
2. it tangent to the curve because derivative is the change of v while t goes to infinitesimal
• The input space in not actually the t-s space, but the x-y space; Garant says "I'm actually gonna cheat a little bit". Is this kind of "cheating" possible because the most basic parametrisation of a surface, or of a curve, can be obtained by using t instead of x and s instead of y?
• Yes, I think it is reasonable to think about it that way. Personally, I don't like how Grant describes this as "cheating", because in my class we would generally just define the axes as t, s, and z and call that good. I think Grant is putting a little too much emphasis on notation.
Hope this helps!
• At when we confirm the output vector of the partial derivative [2, 1, -1], is that output vector a position vector with the origin of (0,0) pointing to (2,1,-1), or is the output vector pointing from (0, 1, 0) (the red dot)?
• The (2; 1; -1) vector is pointing from the red dot, because this shows how the red dot will "change" along the red line.
• at isn't the point on the 2d t axis valued as 1? Why when the dv/ds is -2 for t, it only ends up on the s axis? Shouldn't it go beyond the y axis and be on -1?
• Are you referring to partial v/partial s in the x direction? If so, partial v/ partial s is the slope of the curve as s changes. In order to see it you keep t constant. The -2 is visualized as the slope in the output space with a constant input variable.

This might help you think about it. In single variable calculus, imagine the graph y=x^2. At the input value x=1, the output value is y=1, and dy/dx=2. The dy/dx cannot be seen as "my input value, 1, moves to dy/dx, 2," or "when I move 1 in the x direction, I move 2 in the y direction." If this were true, the point (1,2) or (2,3) would be on the graph, but it's not. Instead we visualize this dy/dx=2 as the slope of the tangent line.

Also, the next video explains the partial v/partial s and shows the slope.
• if the gradient is the direction of steepest ascent, then shouldn't partial with respect to t be pointing uphill?