Recent content by njo

Homework Statement \int_0^5 \int_0^2 \int_0^{4-y^2}\ \, dxdydx Change order to dydxdz Homework EquationsThe Attempt at a Solution I'm confused mainly because the limits are mostly numbers, not functions. I graphed the limits in @D and #d and this is what I got: \int_0^{4-y^2} \int_0^5...

The tangent line is z = 10y+5 therefore the parametric equations of this line are: y = t +1 z = 10t + 15

That makes sense. Although I get an answer less than A2*B2*sin(θ2) - A1*B1*sin(θ1) Δθ = .5pi/180 ≅ .0087 ∴ 10sin45°(±0.1) + 15sin45°(±0.1) + 150cos45°(±.pi/360) ≅ ±2.6934

A = area = xysinθ dA = (∂A/∂x)dx + (∂A/∂y)dy + (∂A/∂θ)dθ So the margin for the area is about 5ft^2. How is my differential equation incorrect? I took each partial and multiplied by the rate of change.

Sorry I didn't see your whole post on my phone at first. That's a typo. All should be +- 0.1 ft not .01 ft

What changes? the differential equation is wrong right?

Homework Statement 2 adjacent sides of a parallelogram measure 15ft and 10ft w/ max errors of ±0.1ft angle is 45° w/ max error of ±0.5° What is the maximum error in the calculated value of the area or the parallelogram? Homework Equations A = area = xysinθThe Attempt at a Solution x = 15ft...

So I find the gradient of w = 2x^2 + 5y^2 - z + 2 which would be <4x, 10y, -1>. I substitute P into the gradient? and that's how I get the coefficient in front of the parameter?

Oops, I know better than that. z = 10t + 15 y = t + 1 x = 2

dz/dy = 10y z = 10y + 15 y = t + 1 x = 2 Is this correct?

z = 2x^2 + 5y^2 +2 That's the equation. Typo.

Homework Statement z = 2x^2 + 5y^2 +2 C is cut by the plane x = 2 Find parametric eqns of the line tangent to C @ P(2, 1, 15) Homework Equations z = 5y^2 + 10 dz/dx = 10y dz/dx (1) = 10 The Attempt at a Solution z = 10y + 15 y = t + 1 if the slope is 10/1 then delta z = 10 and delta y = 1...

(5+yzsin(xz)-4y+6z^2x^2)/(-yxsin(xz)-4zx^3) = ∂z/∂x Pretty sure this is right. Just messed up on my algebra. Thank you so much. The internet is great.

-y*sin(xz)*(z+x(∂z/∂x))+4y-4zx^3(∂z/∂x)-6z^2x^2 = 5 This is what I have before rearranging and factoring for ∂z/∂x

Homework Statement ∂z/∂x of ycos(xz)+(4xy)-2z^2x^3=5x[/B] Homework Equations n/a The Attempt at a Solution ∂z/∂x=(5+yz-4y+6z^2x^2)/(-yxsin(xz)-4zx^3)[/B] Is this correct? Just trying to make sure that's the correct answer. I appreciate the help. I can post my work if need be. Thanks