- #1

hell0

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Moved from a technical forum, so homework template missing

1. Suppose we have ladder laying against a wall, with x sub 0 = 1 and y sub 0 = 1. Given that y(1) = 1 and x′(1) = 1, find y'(1).

Okay so using Pythagorean's Theorem x^2 + y^2 = L, I found that the remaining side is sqrt(2). After taking the derivative I got 2x(t)x'(t) + 2y(t)y'(t) = 0. The only thing I'm confused about here is which values to plug in. I keep thinking that there's missing information in the problem or that there's some kind of typo. I've done questions like this one before, but the working of this specific one is so confusing to me.

2. Suppose we are draining an ice cream cone of the completely melted ice cream from a very small hole. Given that h(t) = 4 − 3t, r0 = 3 and V(t) = π/3 * (3/4)^2 * [h(t)]^3 , find V′(1).

This one I'm just stuck on completely :(.

edit: After taking the derivative and plugging in appropriately, i got -27π/16 for this one.

Okay so using Pythagorean's Theorem x^2 + y^2 = L, I found that the remaining side is sqrt(2). After taking the derivative I got 2x(t)x'(t) + 2y(t)y'(t) = 0. The only thing I'm confused about here is which values to plug in. I keep thinking that there's missing information in the problem or that there's some kind of typo. I've done questions like this one before, but the working of this specific one is so confusing to me.

2. Suppose we are draining an ice cream cone of the completely melted ice cream from a very small hole. Given that h(t) = 4 − 3t, r0 = 3 and V(t) = π/3 * (3/4)^2 * [h(t)]^3 , find V′(1).

This one I'm just stuck on completely :(.

edit: After taking the derivative and plugging in appropriately, i got -27π/16 for this one.

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