2 Related Rates Questions (Calc I)

[hell0]

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.

 

[BvU]

Hello Hell0,

Do I understand the center of the ladder is the point (x,y) and the initial condition is that the center is at (1,1) at t=0 ?
And then what ? It slides down ? Not if y(1) = y(0) = 1 !
Or does it rotate around its feet and fall over backwards in such a way that x'(1) = 1 when y(1) is y(0) = 1 again ?

 

[hell0]

Hello Hell0,

Do I understand the center of the ladder is the point (x,y) and the initial condition is that the center is at (1,1) at t=0 ?
And then what ? It slides down ? Not if y(1) = y(0) = 1 !
Or does it rotate around its feet and fall over backwards in such a way that x'(1) = 1 when y(1) is y(0) = 1 again ?

Good question. Honestly that's all the info I have. Nice professor I have, huh? Anyway I made the sides of the triangle be 1 and 1, and found the hypotenuse to be sqrt(2). Then used the theorem again but this time found x(1) via [x(1)]^2+(1/2)^2=sqrt(2)^2 since y(1) is given in the problem and found that x(1) is sqrt(7/4) or sqrt(7)/2. Then after taking the derivative of the Pythagorean Theorem I get 2x(t)x'(t)+2y(t)y'(t)=0. Plugging in I get 2(sqrt(7)/2)(1)+2(1/2)(y'(1))=0 and find y'(1) to be -sqrt(7)?

 

[haruspex]

y sub 0 = 1. Given that y(1) = 1

Assuming that means y(0)=1 and y(1)=1, as BvU notes, that is not consistent with the ladder sliding down. Please check you have those data correct.
It is unclear whether the x and y represent the coordinates of the ladder's midpoint or the variable ordinates of its endpoints, but since that just makes a factor of 2 difference everywhere it might not matter.