Rates of Change (1 Viewer)

[GaDaMIt]

I'm a preliminary student and ive just started a new exercise under the derivative topic in the cambridge book - Rates of Change

im just a little stuck at the moment on the first question

"Given the y=x^3 + x, differentiate with respect to time, using the chain rule"

Firstly i didnt use the chain rule .. i just differentiated the terms and got

dy/dx=3x^2+1

but the answer is

dy/dt=(3x^2+1)dx/dt

someone care to explain this to me? im not all that confident with derivatives at the moment, and any help will be greatly appreciated

 

[STx]

Using the chain rule:
dy/dt=[dy/dx]*[dx/dt]
=[3x²+1]*[dx/dt]

 

[littleboy]

dy/dx = dy/dt . dt/dx

since dy/dx = 3x^2 +1

so 3x^2 + 1 = dy/dt.dt/dx

move to other side:

dy/dt = (3x^2 + 1). dx/dt

 

[GaDaMIt]

Using the chain rule:
dy/dt=[dy/dx]*[dx/dt]
=[3x²+1]*[dx/dt]

lol jesus im dumb so obvious haha thanks heaps

 

[GaDaMIt]

Another question...

A spherical balloon is to be filled with water so that its surface area increases at a constant rate of 1cm^2/s. Find the volume when the volume is increasing at 10cm^3/s?

 

[SoulSearcher]

Let S stand for the surface area, and V stand for the volume of the balloon.
dS/dt = 1 cm²/s
dV/dt = 10 cm³/s
S = 4*pi*r²
dS/dr = 8*pi*r
V = 4*pi*r³/3
dV/dr = 4*pi*r²
dr/dt = dS/dt * dr/dS = dV/dt * dr/dV
dS/dt * dr/dS = 1 * (1/8*pi*r)
= 1/(8*pi*r)
dV/dt * dr/dV = 10 * (1/4*pi*r²)
= 5/(2*pi*r²)
Since dS/dt * dr/dS = dV/dt * dr/dV
1/(8*pi*r) = 5/(2*pi*r²)
40*pi*r = 2*pi*r²
40r = 2r²
r = 20
Therefore volume is
4/3 * pi * 20³
= 32000pi/3 cm³