Recent content by b521

Just to make sure, the slope is 1/3?

x^(1/3) + y^(1/3) = 4 This is the original equation. Sorry about the typo before.

Here's all of my work: (1/3)x^(-2/3) + 1/3y^(-2/3)(dy/dx) = 0 (dy/dx)1/3y^(2/3) = -1/3x^(2/3) dy/dx = -3y^(2/3)/3x^(2/3) dy/dx = -y^(2/3)/x^(2/3) I hope that helps find my problem. This is where I tried plugging in the x value and everything got complicated. 1/(-3√3)^(2/3)

Homework Statement Find the equation of a tangent line at the curve at point (-3√3, 1) x^(1/3) + y^(1/3) = 4Homework Equations Point-slope: y-1=m(x-1) The Attempt at a Solution I took the derivative of that equation and resulted in -y^(2/3)/x^(2/3) When I tried plugging in x and y to...

Ahhh thank you very much!

Homework Statement If z² = x² + y² and dx/dt = 2 and dy/dt = 3 Find dz/dt when x=5 and y=12 Homework Equations I used The Attempt at a Solution But I resulted in: 2z dz/dt = 2x(dx/dt) + 2y(dy/dt) Plugged in x,y,dx/dt, and dy/dt to get: 2z (dz/dt) = 2(5)(2) + 2(12)(3) dz/dt = 80/2z...

Thank you so much!

Homework Statement If y = x³ + 2x and dx/dt = 5, find dy/dt when x = 2. The Attempt at a Solution I'm not really sure how to solve this problem at all. It's the first I've seen like this so I don't know what approach I should take.

Yes, thank you!

Homework Statement Find an equation of the tangent line to the curve at the given point. y = ln(xe^(x²)) P=(1, 1) The Attempt at a Solution I know how to plug in the numbers into the Point-Slope formula once I find the derivative of the function, or slope, I'm just not sure...