TS Intermediate Mathematics 1st Year Model Paper 2023

SECTION A

Note: (i) Attempt all questions.

(ii) Each question carries two marks.

1. Find the slope of the straight line passing through the points (2, 7) and (5, 13).Answer: The formula for the slope
   
   $m$

   m of a straight line passing through two points

   $(x_1, y_1)$

   (x1​,y1​) and

   $(x_2, y_2)$

   (x2​,y2​) is:

   $$m = \frac{y_2 – y_1}{x_2 – x_1}$$

   m=x2​−x1​y2​−y1​​

   Given the points

   $(2, 7)$

   (2,7) and

   $(5, 13)$

   (5,13), we substitute into the formula:

   $$m = \frac{13 – 7}{5 – 2} = \frac{6}{3} = 2$$

   m=5−213−7​=36​=2

   So, the slope of the line is

   $m = 2$

   m=2.

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2. Transform the following straight line equation from normal form to slope-intercept form:
   
   $$x + 4y = 8$$

   x+4y=8

   Answer: To convert the given equation to slope-intercept form

   $y = mx + b$

   y=mx+b, solve for

   $y$

   y.

   $$x + 4y = 8$$

   x+4y=8

   Subtract

   $x$

   x from both sides:

   $$4y = -x + 8$$

   4y=−x+8

   Divide both sides by 4:

   $$y = -\frac{1}{4}x + 2$$

   y=−41​x+2

   Therefore, the equation in slope-intercept form is

   $y = -\frac{1}{4}x + 2$

   y=−41​x+2, where the slope

   $m = -\frac{1}{4}$

   m=−41​ and the y-intercept

   $b = 2$

   b=2.

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3. Find the centroid of the tetrahedron whose vertices are
   
   $(0, -4)$

    

    

    

    

    

   $(0,-4)$,

   $(1, 5)$

    

    

   $(1,5)$,

   $(2, 3)$

    

    

   $(2,3)$, and

   $(3, 1)$

    

    

   $(3,1)$.

   Answer: The centroid of a tetrahedron is the average of the coordinates of its vertices. Given the vertices

   $(x_1, y_1) = (0, -4)$

   (x1​,y1​)=(0,−4),

   $(x_2, y_2) = (1, 5)$

   (x2​,y2​)=(1,5),

   $(x_3, y_3) = (2, 3)$

   (x3​,y3​)=(2,3), and

   $(x_4, y_4) = (3, 1)$

   (x4​,y4​)=(3,1), the centroid

   $(x, y)$

   (x,y) is calculated as:

   $$x = \frac{x_1 + x_2 + x_3 + x_4}{4} = \frac{0 + 1 + 2 + 3}{4} = \frac{6}{4} = 1.5$$

   x=4x1​+x2​+x3​+x4​​=40+1+2+3​=46​=1.5

   $$y = \frac{y_1 + y_2 + y_3 + y_4}{4} = \frac{-4 + 5 + 3 + 1}{4} = \frac{5}{4} = 1.25$$

   y=4y1​+y2​+y3​+y4​​=4−4+5+3+1​=45​=1.25

   Therefore, the centroid of the tetrahedron is

   $(1.5, 1.25)$

   (1.5,1.25).

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4. Write the equation of the plane
   
   $4x – 5y + 3z = 5$

    

    

    

    

    

   $4x-5y+3z=5$ in the intercept form.

   Answer: The intercept form of the equation of a plane is given by:

   $$\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$$

   ax​+by​+cz​=1

   For the plane equation

   $4x – 5y + 3z = 5$

   4x−5y+3z=5, we find the intercepts by setting two of the variables to zero and solving for the third:

   • For the x-intercept, set
     
     $y = 0$

      

      

      

      

      

     y=0 and

     $z = 0$

      

      

     z=0:

     $$4x = 5 \quad \Rightarrow \quad x = \frac{5}{4}$$

     4x=5⇒x=45​

   • For the y-intercept, set
     
     $x = 0$

      

      

      

      

      

     x=0 and

     $z = 0$

      

      

     z=0:

     $$-5y = 5 \quad \Rightarrow \quad y = -1$$

     −5y=5⇒y=−1

   • For the z-intercept, set
     
     $x = 0$

      

      

      

      

      

     x=0 and

     $y = 0$

      

      

     y=0:

     $$3z = 5 \quad \Rightarrow \quad z = \frac{5}{3}$$

     3z=5⇒z=35​

   Therefore, the intercept form of the equation is:

   $$\frac{x}{\frac{5}{4}} + \frac{y}{-1} + \frac{z}{\frac{5}{3}} = 1$$

   45​x​+−1y​+35​z​=1

   Simplifying:

   $$\frac{4x}{5} – y + \frac{3z}{5} = 1$$

   54x​−y+53z​=1

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5. Compute the following limits:(a)
   
   $\lim_{x \to 2} \frac{x – 2}{x^2 – 4}$

   limx→2​x2−4x−2​

   Answer: First, observe that

   $x^2 – 4 = (x – 2)(x + 2)$

   x2−4=(x−2)(x+2), so the limit becomes:

   $$\lim_{x \to 2} \frac{x – 2}{(x – 2)(x + 2)}$$

   x→2lim​(x−2)(x+2)x−2​

   Cancel out

   $(x – 2)$

   (x−2) (for

   $x \neq 2$

   x=2):

   $$\lim_{x \to 2} \frac{1}{x + 2} = \frac{1}{2 + 2} = \frac{1}{4}$$

   x→2lim​x+21​=2+21​=41​

   Therefore, the limit is

   $\frac{1}{4}$

   41​.

6.Compute the following limits: x→2lim​x−2x2−4​

1. Factor the numerator:We notice that
   
   $x^2 – 4$

   x2−4 is a difference of squares, which can be factored as:

   $$x^2 – 4 = (x – 2)(x + 2)$$

   x2−4=(x−2)(x+2)

2. Rewrite the expression:Substituting the factorization into the limit expression, we get:
   
   $$\frac{x^2 – 4}{x – 2} = \frac{(x – 2)(x + 2)}{x – 2}$$

   x−2x2−4​=x−2(x−2)(x+2)​

3. Cancel out common factors:For
   
   $x \neq 2$

   x=2, we can cancel out

   $(x – 2)$

   (x−2) from the numerator and the denominator:

   $$\frac{(x – 2)(x + 2)}{x – 2} = x + 2$$

   x−2(x−2)(x+2)​=x+2

4. Substitute
   
   $x = 2$

    

    

    

    

    

   $x=2$ into the simplified expression:

   Now, substitute

   $x = 2$

   x=2 into the remaining expression:

   $$x + 2 = 2 + 2 = 4$$

   x+2=2+2=4

Final Answer:

$$\lim_{x \to 2} \frac{x^2 – 4}{x – 2} = 4$$

 

 

 

 

 

x→2lim​x−2x2−4​=4

7. If
   
   $y = \log(\tan x)$

    

    

    

    

    

   $y=\log (\tan x)$, find

   $\frac{dy}{dx}$

    

    

   dxdy​.

   Answer: To find

   $\frac{dy}{dx}$

   dxdy​, we apply the chain rule.

   $$y = \log(\tan x)$$

   y=log(tanx)

   The derivative of

   $\log u$

   logu is

   $\frac{1}{u} \cdot \frac{du}{dx}$

   u1​⋅dxdu​, and the derivative of

   $\tan x$

   tanx is

   $\sec^2 x$

   sec2x.

   Therefore:

   $$\frac{dy}{dx} = \frac{1}{\tan x} \cdot \frac{d}{dx}(\tan x)$$

   dxdy​=tanx1​⋅dxd​(tanx)

   $$\frac{dy}{dx} = \frac{1}{\tan x} \cdot \sec^2 x$$

   dxdy​=tanx1​⋅sec2x

   Using the identity

   $\frac{\sec^2 x}{\tan x} = \frac{1}{\sin x \cos x}$

   tanxsec2x​=sinxcosx1​, we get:

   $$\frac{dy}{dx} = \frac{\sec^2 x}{\tan x}$$

   dxdy​=tanxsec2x​

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8. Find the derivative of
   
   $p(x)^{-4} (3x – 4x^3)$

    

    

    

    

    

   p(x)−4(3x−4x3).

   Answer: To differentiate this function, we apply the product rule. Let

   $u(x) = p(x)^{-4}$

   u(x)=p(x)−4 and

   $v(x) = 3x – 4x^3$

   v(x)=3x−4x3.

   The product rule states that

   $\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)$

   dxd​[u(x)v(x)]=u′(x)v(x)+u(x)v′(x).

   First, differentiate

   $u(x) = p(x)^{-4}$

   u(x)=p(x)−4:

   $$u'(x) = -4p(x)^{-5} \cdot p'(x)$$

   u′(x)=−4p(x)−5⋅p′(x)

   Now, differentiate

   $v(x) = 3x – 4x^3$

   v(x)=3x−4x3:

   $$v'(x) = 3 – 12x^2$$

   v′(x)=3−12x2

   Now applying the product rule:

   $$\frac{d}{dx} [p(x)^{-4} (3x – 4x^3)] = -4p(x)^{-5} \cdot p'(x) \cdot (3x – 4x^3) + p(x)^{-4} \cdot (3 – 12x^2)$$

   dxd​[p(x)−4(3x−4x3)]=−4p(x)−5⋅p′(x)⋅(3x−4x3)+p(x)−4⋅(3−12x2)

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9. Find the slope of the tangent to the curve
   
   $y = 5x^2 – 1$

    

    

    

    

    

   y=5x2−1 at

   $x = -1.5$

    

    

   $x=-1.5$.

   Answer: To find the slope of the tangent, we need to compute the derivative of

   $y = 5x^2 – 1$

   y=5x2−1.

   The derivative of

   $y$

   y is:

   $$\frac{dy}{dx} = 10x$$

   dxdy​=10x

   Now, substitute

   $x = -1.5$

   x=−1.5 into the derivative:

   $$\frac{dy}{dx} = 10(-1.5) = -15$$

   dxdy​=10(−1.5)=−15

   Therefore, the slope of the tangent at

   $x = -1.5$

   x=−1.5 is

   $-15$

   −15.

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10. Find
    
    $dy$

     

     

     

     

     

    $dy$ and

    $\Delta y$

     

     

    $\Delta y$ when

    $y = f(x) = x^2 + x$

     

     

    y=f(x)=x2+x and

    $x = 10$

     

     

    $x=10$,

    $\Delta x = 0.1$

     

     

    $\Delta x=0.1$.

Answer:

First, let’s calculate the derivative

$\frac{dy}{dx}$

dxdy​:

$$y = x^2 + x$$

y=x2+x

$$\frac{dy}{dx} = 2x + 1$$

dxdy​=2x+1

Now, calculate

$\frac{dy}{dx}$

dxdy​ at

$x = 10$

x=10:

$$\frac{dy}{dx} = 2(10) + 1 = 21$$

dxdy​=2(10)+1=21

Now, we can find

$dy$

dy, which is approximately

$\frac{dy}{dx} \cdot \Delta x$

dxdy​⋅Δx:

$$dy = 21 \times 0.1 = 2.1$$

dy=21×0.1=2.1

Now, compute

$y$

y at

$x = 10$

x=10:

$$y = 10^2 + 10 = 100 + 10 = 110$$

y=102+10=100+10=110

Now, compute

$\Delta y$

Δy:

$$\Delta y = f(10 + 0.1) – f(10) = (11^2 + 11) – 110 = 121 + 11 – 110 = 22$$

Δy=f(10+0.1)−f(10)=(112+11)−110=121+11−110=22

Therefore:

• 
  
  $dy = 2.1$

   

   

   

   

   

  dy=2.1

• 
  
  $\Delta y = 22$

   

   

   

   

   

  Δy=22

11. The ends of the hypotenuse of a right-angled triangle are fixed. Find the equation of the locus of its third vertex.

Answer: Let the fixed points be

$A(x_1, y_1)$

A(x1​,y1​) and

$B(x_2, y_2)$

B(x2​,y2​), and let the third vertex be

$C(x, y)$

C(x,y). Since

$\triangle ABC$

△ABC is a right-angled triangle, we use the fact that the hypotenuse is the diameter of the circumcircle of the triangle. Therefore, the locus of point

$C$

C lies on a circle with the midpoint of

$AB$

AB as the center and the radius as half the length of

$AB$

AB.

The midpoint of

$AB$

AB is:

$$M \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$$

M(2x1​+x2​​,2y1​+y2​​)

The radius is:

$$r = \frac{1}{2} \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2}$$

r=21​(x2​−x1​)2+(y2​−y1​)2​

Thus, the equation of the locus of

$C$

C is:

$$\left( x – \frac{x_1 + x_2}{2} \right)^2 + \left( y – \frac{y_1 + y_2}{2} \right)^2 = \frac{(x_2 – x_1)^2 + (y_2 – y_1)^2}{4}$$

(x−2x1​+x2​​)2+(y−2y1​+y2​​)2=4(x2​−x1​)2+(y2​−y1​)2​

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12. When the axes are rotated through an angle
    
    $\frac{\pi}{6}$

     

     

     

     

    6π​, find the transformed equation of

    $x^2 + 2\sqrt{3}xy – y^2 = 2a^2$

     

     

    x2+23​xy−y2=2a2.

Answer: The transformation of coordinates due to a rotation by an angle

$\theta = \frac{\pi}{6}$

θ=6π​ is given by:

$$x’ = x \cos \theta + y \sin \theta$$

x′=xcosθ+ysinθ

$$y’ = -x \sin \theta + y \cos \theta$$

y′=−xsinθ+ycosθ

For

$\theta = \frac{\pi}{6}$

θ=6π​, we have:

$$\cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}, \quad \sin \frac{\pi}{6} = \frac{1}{2}$$

cos6π​=23​​,sin6π​=21​

Substituting into the transformation equations, we get:

$$x’ = \frac{\sqrt{3}}{2} x + \frac{1}{2} y$$

x′=23​​x+21​y

$$y’ = -\frac{1}{2} x + \frac{\sqrt{3}}{2} y$$

y′=−21​x+23​​y

Now substitute the new coordinates into the equation

$x^2 + 2\sqrt{3}xy – y^2 = 2a^2$

x2+23​xy−y2=2a2, and simplify the result to get the transformed equation in terms of

$x’$

x′ and

$y’$

y′.

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13. If
    
    $Q(h, k)$

     

     

     

     

    $Q(h,k)$ is the foot of the perpendicular from

    $P(x_1, y_1)$

     

     

    P(x1​,y1​) on the straight line

    $ax + by + c = 0$

     

     

    $ax+by+c=0$, then prove that:

$$\frac{h – x_1}{a} = \frac{k – y_1}{b} = \frac{-(ax_1 + by_1 + c)}{a^2 + b^2}$$

ah−x1​​=bk−y1​​=a2+b2−(ax1​+by1​+c)​

Answer: The point

$Q(h, k)$

Q(h,k) lies on the line and is the foot of the perpendicular from

$P(x_1, y_1)$

P(x1​,y1​). The formula for the coordinates of the foot of the perpendicular from a point to a line is derived using the fact that the vector

$\overrightarrow{PQ}$

PQ​ is perpendicular to the line. Using the projection formula, we find:

$$\frac{h – x_1}{a} = \frac{k – y_1}{b} = \frac{-(ax_1 + by_1 + c)}{a^2 + b^2}$$

ah−x1​​=bk−y1​​=a2+b2−(ax1​+by1​+c)​

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14. Compute:

$$\lim_{x \to 0} \frac{1 – \cos(2mx)}{\sin^2(nx)} \quad \text{where} \quad m, n \in \mathbb{Z}$$

x→0lim​sin2(nx)1−cos(2mx)​wherem,n∈Z

Answer: Using the standard small angle approximations

$\cos x \approx 1 – \frac{x^2}{2}$

cosx≈1−2x2​ and

$\sin x \approx x$

sinx≈x as

$x \to 0$

x→0, we approximate:

$$\cos(2mx) \approx 1 – 2m^2x^2$$

cos(2mx)≈1−2m2x2

$$\sin^2(nx) \approx n^2x^2$$

sin2(nx)≈n2x2

Substituting these approximations into the expression:

$$\frac{1 – \cos(2mx)}{\sin^2(nx)} \approx \frac{2m^2x^2}{n^2x^2} = \frac{2m^2}{n^2}$$

sin2(nx)1−cos(2mx)​≈n2x22m2x2​=n22m2​

Therefore, the limit is:

$$\lim_{x \to 0} \frac{1 – \cos(2mx)}{\sin^2(nx)} = \frac{2m^2}{n^2}$$

x→0lim​sin2(nx)1−cos(2mx)​=n22m2​

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15. Find the derivative of the function
    
    $\tan(2x)$

     

     

     

     

    $\tan (2x)$ from the first principle.

Answer: The derivative of a function

$f(x)$

f(x) from the first principle is given by:

$$f'(x) = \lim_{h \to 0} \frac{f(x + h) – f(x)}{h}$$

f′(x)=h→0lim​hf(x+h)−f(x)​

For

$f(x) = \tan(2x)$

f(x)=tan(2x), we apply the first principle:

$$f'(x) = \lim_{h \to 0} \frac{\tan(2(x + h)) – \tan(2x)}{h}$$

f′(x)=h→0lim​htan(2(x+h))−tan(2x)​

Using the identity

$\tan(A) – \tan(B) = \frac{\sin(A – B)}{\cos(A)\cos(B)}$

tan(A)−tan(B)=cos(A)cos(B)sin(A−B)​, we simplify the expression and find that:

$$f'(x) = 2 \sec^2(2x)$$

f′(x)=2sec2(2x)

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16. If the increase in the side of a square is 2%, then find the approximate percentage of increase in its area.

Answer: Let the side of the square be

$s$

s. The area of the square is

$A = s^2$

A=s2. If the side increases by 2%, then the new side length is

$s + 0.02s = 1.02s$

s+0.02s=1.02s. The new area is:

$$A_{\text{new}} = (1.02s)^2 = 1.0404s^2$$

Anew​=(1.02s)2=1.0404s2

The percentage increase in area is:

$$\frac{1.0404s^2 – s^2}{s^2} \times 100 = 4.04\%$$

s21.0404s2−s2​×100=4.04%

Therefore, the approximate percentage increase in area is

$4.04\%$

4.04%.

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17. Find the lengths of the subtangent and subnormal at a point ‘t’ on the curve.

Answer: For a curve given by

$y = f(x)$

y=f(x), the lengths of the subtangent and subnormal at a point are given by the following formulas:

• Subtangent:
  
  $\text{Subtangent} = \frac{y}{f'(x)}$

   

   

   

   

  Subtangent=f′(x)y​

• Subnormal:
  
  $\text{Subnormal} = \frac{y}{f'(x)} \times \frac{f”(x)}{f'(x)}$

   

   

   

   

  Subnormal=f′(x)y​×f′(x)f′′(x)​

To find these lengths, calculate

$f'(x)$

f′(x) and

$f”(x)$

f′′(x) at the point

$t$

t, and substitute into the formulas.

SECTION C 

Note: (i) Answer any five questions.

(ii) Each question carries seven marks.

18. Find the orthocenter of the triangle with the following vertices (-2, -1), (6, -1), and (2, 5).

Answer: The orthocenter of a triangle is the point where the altitudes (perpendiculars from the vertices to the opposite sides) meet. To find the orthocenter, we need to:

1. Calculate the equations of two altitudes.
2. Find the intersection of these altitudes.

First, find the slope of the side

$(-2, -1)$

(−2,−1) and

$(6, -1)$

(6,−1):

$$\text{slope} = \frac{-1 – (-1)}{6 – (-2)} = 0$$

slope=6−(−2)−1−(−1)​=0

The perpendicular from the vertex

$(2, 5)$

(2,5) to this line will have an undefined slope (vertical line), so the equation of the altitude is

$x = 2$

x=2.

Next, for the side

$(6, -1)$

(6,−1) and

$(2, 5)$

(2,5), calculate the slope:

$$\text{slope} = \frac{5 – (-1)}{2 – 6} = -\frac{3}{2}$$

slope=2−65−(−1)​=−23​

The perpendicular slope is

$\frac{2}{3}$

32​, and the equation of the altitude from

$(-2, -1)$

(−2,−1) is:

$$y + 1 = \frac{2}{3}(x + 2)$$

y+1=32​(x+2)

Simplifying:

$$y = \frac{2}{3}x + \frac{1}{3}$$

y=32​x+31​

Solve these two equations to find the intersection point, which is the orthocenter.

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19. If the equation
    
    $ax^2 + 2hxy + by^2 = 0$

     

    ax2+2hxy+by2=0 represents a pair of straight lines and
    $\theta$

     

    $\theta$ is the angle between the lines, then prove that:

$$\cos(\theta) = \frac{|a – b|}{\sqrt{(a + b)^2 + 4h^2}}$$

cos(θ)=(a+b)2+4h2​∣a−b∣​

Answer: The equation of the pair of lines is

$ax^2 + 2hxy + by^2 = 0$

ax2+2hxy+by2=0. The angle

$\theta$

θ between the lines is given by the formula:

$$\cos(\theta) = \frac{|a – b|}{\sqrt{(a + b)^2 + 4h^2}}$$

cos(θ)=(a+b)2+4h2​∣a−b∣​

This can be derived by considering the general form of the pair of straight lines and finding the relation between the coefficients using the condition for the angle between two lines.

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20. Find the values of
    
    $k$

     

    $k$, if the lines joining the origin to the points of intersection of the curve
    $2x^2 + 2xy + 3y^2 = 5x + 2y – 1 = 0$

     

    2x2+2xy+3y2=5x+2y−1=0 and the line
    $x + 2y = k$

     

    $x+2y=k$ are mutually perpendicular.

Answer: First, solve the system of equations for the points of intersection. Then, use the condition that the lines joining the origin to these points are perpendicular, i.e., their slopes should satisfy:

$$m_1 \times m_2 = -1$$

m1​×m2​=−1

By calculating the slopes of the lines joining the origin to the intersection points, find the values of

$k$

k.

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21. Find the angle between two diagonals of a cube.

Answer: The two diagonals of a cube intersect at the center of the cube. To find the angle between them, consider the direction vectors of the diagonals. For a cube with side length

$a$

a, the direction vectors of the diagonals are

$\vec{v_1} = (1, 1, 1)$

v1​​=(1,1,1) and

$\vec{v_2} = (-1, 1, 1)$

v2​​=(−1,1,1). The angle between them is given by:

$$\cos \theta = \frac{\vec{v_1} \cdot \vec{v_2}}{|\vec{v_1}| |\vec{v_2}|}$$

cosθ=∣v1​​∣∣v2​​∣v1​​⋅v2​​​

The dot product is

$\vec{v_1} \cdot \vec{v_2} = -1 + 1 + 1 = 1$

v1​​⋅v2​​=−1+1+1=1, and the magnitudes of

$\vec{v_1}$

v1​​ and

$\vec{v_2}$

v2​​ are

$\sqrt{3}$

3​. Thus:

$$\cos \theta = \frac{1}{3}$$

cosθ=31​

Therefore, the angle is:

$$\theta = \cos^{-1} \left( \frac{1}{3} \right)$$

θ=cos−1(31​)

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22. If
    
    $\sqrt{1 – x^2} + \sqrt{1 – y^2} = a(x – y)$

     

    1−x2​+1−y2​=a(x−y), then show that:

$$\frac{dy}{dx} = \frac{\sqrt{1 – y^2}}{\sqrt{1 – x^2}}$$

dxdy​=1−x2​1−y2​​

Answer: Differentiating the given equation implicitly with respect to

$x$

x:

$$\frac{d}{dx}\left( \sqrt{1 – x^2} + \sqrt{1 – y^2} \right) = \frac{d}{dx}\left( a(x – y) \right)$$

dxd​(1−x2​+1−y2​)=dxd​(a(x−y))

Use the chain rule on each term to differentiate. After simplifying, you should arrive at the required result:

$$\frac{dy}{dx} = \frac{\sqrt{1 – y^2}}{\sqrt{1 – x^2}}$$

dxdy​=1−x2​1−y2​​

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23. Show that the curves
    
    $y^2 = 4(x + 1)$

     

    y2=4(x+1) and
    $y^2 = 36\sqrt{9 – x}$

     

    y2=369−x​ intersect orthogonally.

Answer: First, find the points of intersection of the two curves by equating the equations for

$y^2$

y2. Once the points of intersection are found, check the slopes of the tangents to the curves at these points. If the product of the slopes is

$-1$

−1, the curves intersect orthogonally.

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24. The volume of a cube is increasing at a rate of 9 cubic centimeters per second. How fast is the surface area increasing when the length of the edge is 10 centimeters?

Answer: Let the edge length of the cube be

$s$

s. The volume of the cube is

$V = s^3$

V=s3, and the surface area is

$A = 6s^2$

A=6s2. Differentiating both with respect to time

$t$

t, we get:

$$\frac{dV}{dt} = 3s^2 \frac{ds}{dt}$$

dtdV​=3s2dtds​

$$\frac{dA}{dt} = 12s \frac{ds}{dt}$$

dtdA​=12sdtds​

Given that

$\frac{dV}{dt} = 9$

dtdV​=9 and

$s = 10$

s=10, solve for

$\frac{ds}{dt}$

dtds​, and then find

$\frac{dA}{dt}$

dtdA​.