Answer :

Given the height of a cone increases by k%.


Let x be the height of the cone and Δx be the change in the value of x.


Hence, we have


Δx = 0.01kx


Let us assume the radius, the slant height and the semi-vertical angle of the cone to be r, l and α respectively as shown in the figure below.



From the above figure, using trigonometry, we have




r = x tan(α)


We also have





l = x sec(α)


(i) The total surface area of the cone is given by


S = πr2 + πrl


From above, we have r = x tan(α) and l = x sec(α).


S = π(x tan(α))2 + π(x tan(α))(x sec(α))


S = πx2tan2α + πx2tan(α)sec(α)


S = πx2tan(α)[tan(α) + sec(α)]


On differentiating S with respect to x, we get




We know




Recall that if y = f(x) and Δx is a small increment in x, then the corresponding increment in y, Δy = f(x + Δx) – f(x), is approximately given as



Here, and Δx = 0.01kx


ΔS = (2πxtan(α)[tan(α) + sec(α)])(0.01kx)


ΔS = 0.02kπx2tan(α)[tan(α) + sec(α)]


The percentage increase in S is,




Increase = 0.02k × 100%


Increase = 2k%


Thus, the approximate increase in the total surface area of the cone is 2k%.


(ii) The volume of the cone is given by



From above, we have r = x tan(α).





On differentiating V with respect to x, we get




We know




Recall that if y = f(x) and Δx is a small increment in x, then the corresponding increment in y, Δy = f(x + Δx) – f(x), is approximately given as



Here, and Δx = 0.01kx


ΔV = (πx2tan2α)(0.01kx)


ΔV = 0.01kπx3tan2α


The percentage increase in V is,





Increase = 0.03k × 100%


Increase = 3k%


Thus, the approximate increase in the volume of the cone is 3k%.


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