Recent content by Unknown9

Wait wait wait, okay, when I plug in x... how can I plug in a negative number into that...? -3^3/2 into x^- 1/3?

So I got z = sqr( (150-x)^2 + y^2) Then I differentiated that and got. dz/dt = 1/2( (150-x)^2 + y^2)^-1/2 * (-2(dx/dt)(150-x) + 2y(dy/dt)) when I plugged in the numbers, which I used x as 140 y as 100, and dx/dt as 35 and dy/dt as 25, I got dz/dt = 1/(2*sqrt(10,100)) * 4300 am I...

Haha, nevermind, I just divided both of the numbers and got .6, so they're both the same answer.

So I tried doing the problem from scratch again, and this time I received a different answer :P. y/2 = 12/(12-x) Differentiated it right away 12(dy/dt) - x(dy/dt) + y(dx/dt) = 0 12(dy/dt) - x(dy/dt) = -y(dx/dt) dy/dt = -y(dx/dt)/12-x Then I received 4.8/8 as the rate of change.

Calculus1:Related rates I'll give this a better title so people know what I'm talking about :). Homework Statement At noon, ship A is 150km west of ship B. Ship A is sailing east at 35km/h and ship B is sailing north at 25km/h. HOw fast is the distance between the ships changing at 4:00pm...

That wouldn't change my answer would it?

Okay, I was looking at it again and here's what I've come up with.So y/2 = 12/(12-x) I solved for y y = 24/(12-x) Then I differentiated it dy/dt = -24(12-x)-2 * -1(dx/dt) simplified dy/dt = (24*(dx/dt))/(12-x)2 then plug in dx/dt which is 1.6 and plug in x which is 4 dy/dt = 38.4/64 m/s...

Okay, so I drew the picture and I got the height of the shadow where the man is 4 m away to be 3m. with 2/8 = y/12

quick question about the question, I copied the exact question, the question isn't saying that the shadow at the beginning is 12 m high right? It's saying that the distance between the light and the building is 12 m apart?

Homework Statement A spotlight on the ground shines on a wall 12 m away. If a man 2m tall walks from the spotlight toward the building at a speed of 1.6m/s, how fast is the length of his shadow on the building decreasing when he is 4m from the building? Homework Equations x2 + y2 = z2 The...

So then I have (dz/dt) = (1/z) * (35x + 25y), so would the xyz be the distance? Like 150 as x and...?

What? You mean to tell me you won't do my work for me? :O Here's what I got so far. dx/dt = 35km/h,,, dy/dt = 25km/h,,, dz/dt = ? Then we'll use the z2 = x2 + y2 Then we implicit differentiate it in regards to time. 2z(dz/dt) = dx(dx/dt) + 2y(dy/dt) Solve for dz/dt (dz/dt) =...

I think I got it, does the slope end up being -3*31/2? Otherwise I did it incorrectly :P

Homework Statement At noon, ship A is 150km west of ship B. Ship A is sailing east at 35km/h and ship B is sailing north at 25km/h. HOw fast is the distance between the ships changing at 4:00pm. Homework Equations ?? The Attempt at a Solution I was able to draw a picture of the...

Homework Statement Implicit Differentiation:Use implicit differentiation to find an equation of the tangent line to the curve at the given point. x2/3 + y2/3 = 4 at (-3*31/2,1) Homework Equations ? None? The Attempt at a Solution 2/3 * x-1/3 + 2/3y-1/3*y' = 0 then after a few...