The volume of sphere is increaseing at the Constant rate of 10cm²lsee-
Calculate the rate of increase of
the surface area
at the Instant
When the radus is 5cm. What
is the radius of the sphere when
the surface are is increasing at 2cm​

Volume of a sphere:

V = 4/3 π r ³

Differentiate V with respect to time t :

dV/dt = 4π r ² dr/dt

Surface area of a sphere:

A = 4π r ²

Differentiate A with respect to t :

dA/dt = 8π r dr/dt

You're given that the volume is increasing at a constant rate, dV/dt = 10 cm³/s, so

10 cm³/s = 4π r ² dr/dt

Solve for dr/dt (the rate of change of the sphere's radius):

dr/dt = 5/(2π r ²) cm³/s

Substitute this into the equation for dA/dt :

dA/dt = 8π r (5/(2π r ²) cm³/s)

dA/dt = 20/r cm³/s

At the moment the radius is r = 5 cm, the surface area is increasing at a rate of

dA/dt = 20/(5 cm) cm³/s = 4 cm²/s

When the surface area is increasing at a rate of dA/dt = 2 cm²/s, the radius r is such that

2 cm²/s = 20/r cm³/s

which happens when the radius is

r = 20/2 (cm³/s) / (cm²/s) = 10 cm

the value of v.w= -12.

step-by-step explanation:

we are given three vector r,v and w as:

r = < 5, -5, -2> ; v = < 2, -8, -8> ; w = < -2, 6, -5>

we know that r,v and w could also be represented as:

r=5i-5j-2k

v=2i-8j-8k

and w= -2i+6j-5k

where i,j and k are the unit vectors of x,y and z-axis respectively.

also we know that it has the following property:

i.i=1 ; j.j=1; k.k=1

and i.=j.i=i.k=k.i=j.k=k.j=0

i.e. when we multiply two vectors or we find it's dot product we need to only multiply the components of x coordinates with each other; y with each other and z with each other.

hence, v.w= (2i-8j- -2i+6j-5k)

                  = 2×(-2)+(-8)×(6)+(-8)×(-5)

                  = -4-48+40

                  = -12

hence, v.w= -12