# Calculus!! HELP!!

1) The sides of the triangle shown increase in such a way that (dz/dt=1) and (dx/dt=(3dy/dx))
At the instant when x = 12 and y = 5, what is the value of dx/dt?

2) Let f(x) = x^3 − 4.
Which of these is the equation for the normal line to this curve at the point (2,4).
A) y=1/2x-25/6
B) y=-1/12x+25/6
C) y= 12x-25/6
D) y=-12x-4
E) y= -12x-2

3) Which of the following could be the units for dy/dx if y is the surface area of a tumor and x is the radius of the tumor?

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1. Prob. 1. Not sure...

Prob. 2.
The slope of the curve is df/dx = 3x^2.
At (x,y) = (2,4), df/dx = 12.
The normal line must have slope -1/12, which suggestscanswer b.

Prob. 3.
(Or cm, whatever units your book specifies).

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2. 1) The sides of the triangle shown increase in such a way that (dz/dt=1) and (dx/dt=(3dy/dx))
At the instant when x = 12 and y = 5, what is the value of dx/dt?
=========================
You do not say but I suspect this is a 5,12,13 right triangle. If I am guessing right about what you mean then

and x=12, y = 5, z = 13
x^2+y^2= z^2
2xdx+2ydy = 2zdz
dz= 1 dt
dx/dt = 3 dy/dx

24 dx +10 dy = 26 dz
24 + 10dy/dx = 26 dz/dx
24 + (10/3) dx/dt = 26 dt/dx
=26/dx/dt
(10/3)(dx/dt)^2+24 dx/dt -26 = 0

I get dx/dt = .956 or -8.16

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