Question

l₁ and $l_{2}$are the side lengths of two variable squares S₁ and S₂ respectively. If l₁ = $l_{2}$ + ${l_{2}}^{3}$ +6 then rate of change of the area of S₂ with respect to rate of change of the area of S₁ when $l_{2}$=1 is equal to

• A. ¾
• B. 4/3
• C. 3/2
• D. 1/32

Answer 1

Answer: (D)

1/32

Answer 2

Let ∆₁, and ∆₂ be area of squares S₁ and S₂ then

∆₁ = $\ell _ { 1 } ^ { 2 }$, A₂ = $\ell _ { 2 } ^ { 2 }$

⇒ $\frac { d \Delta _ { 1 } } { d \ell _ { 1 } }$ = 2l₁, $\frac { d \Delta _ { 2 } } { d \ell _ { 2 } }$ = 2l₂

= $\frac { d \Delta _ { 2 } } { d \Delta _ { 1 } } = \frac { \ell _ { 2 } } { \ell _ { 1 } } \cdot \frac { d \ell _ { 2 } } { d \ell _ { 1 } } = \frac { \ell _ { 2 } } { \ell _ { 1 } \left( 1 + 3 \ell _ { 2 } \right) }$

When l₂ = 1, l₁ = 8

⇒