A spherical iron ball covered with ice

[inferi]

hi everyone this is a related rates question no one in the class could solve it. I could not find how to find the equation that relate all the things in the question.the question is:

A spherical iron ball 8 in. in diameter is coated with a layer of ice of uniform thickness. If the ice melts at the rate of 10 in^3/min, answer the following questions.

(a) How fast is the thickness of the ice decreasing when the ice is 2 in. thick?

(b) How fast is the outer surface area of the ice decreasing?



anyone can help?
thank you

 

[atqamar]

a) The volume of a sphere with respect to the radius is $$V=(\frac{4\pi}{3})r^3$$. Solve for $$r$$ then differentiate with respect to time.
b) The surface area of a sphere with respect to the radius is $$\alpha=4\pi r^2$$. Write the surface area in terms of the volume and again, differentiate with respect to time.

EDIT: I just re-read the problem and the fact that the ice is melting, not the iron-ball, changes the way you should consider the 2 in. thickness of the ball. The radius of the iron-ball (4 in.) should be added to 2 in. ice to find the rate.

 

[drecxz]

is the answer for this one 1/14.4 cm/min? ^^

 

[atqamar]

a) You must realize that the rate of melting given is not for the iron ball + ice... It is only for the ice. Write a formula for $$V_i_c_e$$, given the total radius $$r_b_a_l_l_+_i_c_e$$ is $$4 + w$$; when $$w$$ is the width (thickness) of the ice. Then differentiate with respect to time.
b) This is a bit more complicated. Again, start with the formula for surface area, $$\alpha=4\pi (4 + w)^2$$. Solve for $$w$$ with respect to $$V$$, then differentiate with respect to time.

PS. Your answer is $$\pi$$ times greater than the answer.

 

[drecxz]

hmmm. i don't seem to understand ... could you please elaborate a) ? i get the same answer

 

[atqamar]

a)
$$V_i_c_e = V_e_n_t_i_r_e _b_a_l_l - V_i_r_o_n = \frac{4\pi}{3}(4+w)^3 - \frac{4\pi}{3}(4)^3 = \frac{4\pi}{3}((4+w)^3 - (4)^3)$$

Differentiate with respect to time:
$$\frac{dV_i_c_e}{dt} = \frac{4\pi}{3}(3(4+w)^2\frac{dw}{dt})$$

$$-10 = \frac{4\pi}{3}(3(4+2)^2\frac{dw}{dt})$$

$$\frac{dw}{dt} = -\frac{5}{72\pi} = -\frac{1}{45.2389342} = -0.0221048532\frac{m}{s}$$