A missile fired from ground level rises x metres vertically upwards in t seconds and x = 100 t - 25 2 t 2 . Find : the initial velocity of the missile.

Given, A missile fired from ground level rises x metres vertically upwards in t seconds and x = 100 t - 25 2 t 2 We know that velocity is the rate of change of displacement. Velocity, v = d x d t = 100 - 25 2 2 t ⇒ v initial = 100 - 0 [ ∵ t = 0 at initial velocity] ∴ Initial velocity v = 100 m / sec

A missile fired from ground level rises x metres vertically upwards in t seconds and x = 100 t - 25 2 t 2 . Find : the initial velocity of the missile.

Given, A missile fired from ground level rises x metres vertically upwards in t seconds and x = 100 t - 25 2 t 2 We know that velocity is the rate of change of displacement. Velocity, v = d x d t = 100 - 25 2 2 t ⇒ v initial = 100 - 0 [ ∵ t = 0 at initial velocity] ∴ Initial velocity v = 100 m / sec

A missile fired from ground level rises x metres vertically upwards in t seconds and x = 100 t - 25 2 t 2 . Find : the time when the height of the missile is a maximum.

Given, A missile fired from ground level rises x metres vertically upwards in t seconds and x = 100 t - 25 2 t 2 Given, x = 100 t - 25 2 t 2 Velocity, v = d x d t At maximum height, v = 0 . ⇒ d x d t = 0 ⇒ 100 - 25 t = 0 ⇒ t = 4 sec ∴ Time taken by missile fired to reach maximum height is 4 sec .

Show that the volume of the largest right circular cone that can be inscribed in a sphere of radius ' a ' is 8 27 (volume of the sphere).

Given: sphere of radius is a Let x be the radius of the inscribed right circular cone and h be its height, then OM 2 + MC 2 = OC 2 [ From Pythagoras theorem ] ⇒ ( h - a ) 2 + x 2 = a 2 ⇒ x 2 = 2 a h - h 2 . . . . . . . . . ( 1 ) If V is the volume of the inscribed cone, then V = 1 3 π x 2 h = 1 3 π ( 2 a h - h 2 ) h = π 3 ( 2 a h 2 - h 3 ) [ ∵ x 2 = 2 a h - h 2 ] Differentiating w.r.t. h , we get d V d h = π 3 ( 4 a h - 3 h 2 ) and d 2 V d h 2 = π 3 ( 4 a - 6 h ) Now for maximum volume, d V d h = 0 ⇒ π 3 ( 4 a h - 3 h 2 ) = 0 ⇒ h ( 4 a - 3 h ) = 0 ⇒ h = 0 , 4 a 3 but h > 0 ⇒ h = 4 a 3 Also, for h = 4 a 3 , d 2 V d h 2 = π 3 ( 4 a - 6 · 4 a 3 ) = - 4 3 π a < 0 ⇒ V is maximum when h = 4 a 3 Therefore, maximum volume = π 3 h 2 ( 2 a - h ) = π 3 ( 4 a 3 ) 2 ( 2 a - 4 a 3 ) = π 3 · 16 a 2 9 · 2 a 3 = 8 27 · 4 3 π a 3 = 8 27 of the volume of sphere. [ ∵ Volume of sphere of radius a = 4 3 π a 3 ] .

Find the length of the curve x a 2 3 + y b 2 3 = 1 .

Given that, the equation of the curve is x a 2 3 + y b 2 3 = 1 Now, the length of the curve is L = ∫ x = a x = b 1 + d y d x 2 d x Therefore, L = 4 × length of curve from x = 0 to x = a , so L = 4 × ∫ 0 a 1 + d y d x 2 d x Now, y b 2 3 = 1 - x a 2 3 ⇒ y b = 1 - x a 2 3 3 2 ⇒ y = b 1 - x a 2 3 3 2 Now, differentiate y with respect to x , then d y d x = b × 3 2 1 - x a 2 3 1 2 d d x 1 - x a 2 3 ⇒ d y d x = b × 3 2 1 - x a 2 3 1 2 2 3 x a - 1 3 - 1 a ⇒ d y d x = - b a x a - 1 3 1 - x a 2 3 1 2 ⇒ d y d x 2 = b 2 a 2 x a - 2 3 1 - x a 2 3 ⇒ d y d x 2 = b 2 a 2 x a - 2 3 - 1 Therefore, the length of the curve is L = 4 × ∫ 0 a 1 + b 2 a 2 x a - 2 3 - 1 d x Lets assume a = b , then L = 4 × ∫ 0 a 1 + x a - 2 3 - 1 d x ⇒ L = 4 × ∫ 0 a x a - 1 3 d x ⇒ L = 4 × 3 a 2 x a 2 3 0 a ⇒ L = 4 × 3 a 2 a a 2 3 - 0 ⇒ L = 4 × 3 a 2 = 6 a Hence, the required length of the curve is 6 a units.

Which one of the following statements is true about the curve y = x 1 3 ?

The curve has a point of inflection in which y ' ' does not exist

The curve has more than one point of inflection

The curve has no point of inflection

The curve has a point of inflection in which y ' ' = 0

Trace the curve y = x 3 .

We have to trace the curve y = x 3 We will first check its domain and range. Since x and y are real numbers, we have x ∈ R and y ∈ R , the domain and range both stretch over all real numbers. Now we will find the intercepts with the coordinate axes. When x = 0 , we get y = 0 When y = 0 , we get x = 0 This means that the curve passes through the origin. Now, let us check its symmetry. Substitute x by - x , then we get: y 1 = - x 3 = - y So, y = x 3 , satisfies the property that f ( - x ) = - f ( x ) So it is an odd function and is symmetric about the origin. Now let us check if the function is asymptotic or not. When x → ∞ , y → ∞ and vice versa. Now, let us check its derivative: d y d x = 3 x 2 Since, 3 x 2 > 0 , ∀ x Hence, d y d x > 0 , ∀ x Hence, the graph is monotonically increasing and will intersect the axes at the origin only. Now, let us check its second derivative: d 2 y d x 2 = 6 x 6 x = 0 for x = 0 At x = 0 , then y = 0 So, ( 0 , 0 ) , is the point of inflection. Using all this information, we draw the following graph:

Let P be a point on the curve y = x 3 and suppose that the tangent line at P intersects the curve again at Q . Prove that the slope at Q is four times the slope at P .

Pick a point P and draw the tangent line at P intersecting again at Q . Let a be the x -co-ordinate of P so that P is the point a , a 3 . We have y = x 3 Differentiate with respect to x , then d y d x = 3 x 2 The slope at P is m = d y d x a , a 3 = 3 a 2 Now Q is the point of intersection of the tangent at P and the curve y = x 3 ...(i) Equation of this tangent line at P is y - a 3 = 3 a 2 x - a ⇒ y - a 3 = 3 a 2 x - 3 a 3 ⇒ y = 3 a 2 x - 2 a 3 ...(ii) Solving (i) and (ii), we get: x 3 = 3 a 2 x - 2 a 3 ⇒ x 3 - 3 a 2 x + 2 a 3 = 0 ⇒ ( x - a ) 2 ( x + 2 a ) = 0 ⇒ x = a or x = - 2 a Now a is the x -co-ordinate of P and - 2 a must be the co-ordinate of Q . Hence, slope at Q is d y d x x = - 2 a = 3 - 2 a 2 = 3 × 4 a 2 = 4 3 a 2 Therefore, slope at Q = 4 times the slope at P .

Show that the function f x = tan - 1 sin x + cos x , x > 0 is strictly increasing in the interval 0 , π 4 .

Given function is f x = tan - 1 sin x + cos x , x > 0 Differentiating f x with respect to x , we get f ' x = d d x tan - 1 sin x + cos x Using chain rule of differentiation d d x f g x = f ' g x × g ' x and d d x tan - 1 x = 1 1 + x 2 . ⇒ f ' x = 1 1 + sin x + cos x 2 d d x sin x + cos x ⇒ f ' x = 1 1 + sin 2 x + cos 2 x + 2 sin x cos x cos x - sin x ⇒ f ' x = cos x - sin x 2 + 2 sin x cos x ⇒ f ' x = cos x - sin x 2 + sin 2 x For strictly increasing function, f ' x > 0 We know that, - 1 ≤ sin 2 x ≤ 1 ⇒ 1 ≤ 2 + sin 2 x ≤ 3 So, 2 + sin 2 x is always positive. For strictly increasing function, cos x - sin x > 0 ⇒ cos x > sin x This is possible only if x ∈ 0 , π 4 . Hence, f x is strictly increasing function in the interval 0 , π 4 .

If w = x + 2 y + z 2 and x = cos t ; y = sin t ; z = t find d w d t by using chain rule. Also find d w d t by substitution of x , y and z in w and hence verify the result.

We have, x = cos t ⇒ d x d t = − sin t y = sin t ⇒ d y d t = cos t z = t ⇒ d z d t = 1 w = x + 2 y + z 2 Differentiate w.r.t ′ t ′ d w d t = d x d t + 2 d y d t + 2 z d z d t = − sin t + 2 cos t + 2 t . . . . . ( 1 ) Substituting, w = cos t + 2 sin t + t 2 Differentiate w.r.t ′ t ′ d w d t = − sin t + 2 cos t + 2 t . . . . . ( 2 ) Hence, proved ( 1 ) = ( 2 )