Question Paper of Calculus - Rates of Change

Question 1:

$T h e v o l u m e o f a s p h e r e i s i n c r e a \sin g a t t h e r a t e o f 5 c m^{3} p e r s e c o n d . A t w h a t r a t e i s t h e s u r f a c e a r e a i n c r e a \sin g w h e n t h e r a d i u s i s 20 c m ?$

Question 2:

$a) A p a r t i c l e i s m o v i n g a l o n g t h e x - a x i s . I t s s p e e d v m / s a t p o s i t i o n x m e t e r s i s g i v e n b y v = \sqrt{5 x - x^{2}} F i n d t h e a c c e l e r a t i o n w h e n x = 2 .$

Question 3:

$F i n d t h e p a r t i c u l a r s o l u t i o n s t o t h e d i f f e r e n t i a l e q u a t i o n \frac{d y}{d x} = \frac{2 x - 1}{y^{2}} g i v e n y = \sqrt[3]{3} w h e n x = 0 .$

Question 4:

$T h e a r e a b o u n d e d b y t h e l i n e y = x - 1 a n d t h e c u r v e y = 4 {(x - 1)}^{2} i s r o t a t e d a b o u t t h e x - a x i s . (i) S k e t c h t h e b o u n d e d a r e a, n o t i n g a n y i n t e r c e p t s . (i i) F i n d t h e v o l u m e g e n e r a t e d .$

Question 5:

$U s e t h e s u b s t u t i o n u = 1 + x t o \int_{1}^{2} \frac{1 - x}{(1 + x^{3})} d x$

Question 6:

$e) S o l v e \frac{5}{x + 2} \leq 1$

Question 7:

$F i n d t h e e x a c t o f \int_{0}^{1} \frac{x}{x^{2} + 1} d x F i n d \frac{d}{d x} (e^{x^{2}} . \cos^{2} x) F i n d t h e c o o r d i n a t e s o f t h e p o i n t P w h i c h A (- 3, 8) a n d B (2, 1) e x t e r n a l l y i n t h e r a t i o o f 7 : 2 . U s e t h e s u b s t i t u t i o n x = 5 \sin θ t o e v a l u a t e \int_{- 5}^{5} \frac{d x}{\sqrt{25 - x^{2}}} S o l v e : \frac{3 x - 2}{x + 3} > 1$

Question 8:

$F i n d \int_{0}^{\frac{π}{2}} \sin^{2} 2 x d x D i f f e r e n t i a t e {(\tan^{- 1} x)}^{2} . H e n c e e v a l u t e \int_{- 1}^{\sqrt{3}} \frac{\tan^{- 1} x}{1 + x^{2}} d x$

Question 9:

$F i n d \int x \sqrt{2 x - 1} d x u \sin g t h e s u b s t i t u t i o n u = 2 x - 1$

Question 10:

$A p o p u l a t i o n o f 1200 f e r a l c a t s i s r e l e a s e d i n t o d a r l i n g P o i n t . T h e r a t e o f i n c r e a s e o f t h e f e r a l c a t p o p u l a t i o n i s : \frac{d P}{d t} = \frac{P}{30} (1 - \begin{matrix} P \\ 10000 \end{matrix}) w h e r e P i s t h e f e r a l c a t p o p u l a t i o n a n d t i s t h e n u m b e r o f m o n t h s . (i) S h o w t h a t \frac{10000}{P (10000 - P)} = \frac{1}{10000 - P} (i i) H e n c e s o l v e t h e d i f f e r e n t i a l e q u a t i o n \frac{d P}{d t} = \frac{P}{30} (1 - \frac{P}{10000}) f o r P i n t e r m s o f t . (i i i) H e n c e f i n d t h e l i m i t i n g f e r a l c a t p o p u l a t i o n . (i v) H o w m a n y m o n t h s d o e s i t t a k e s f o r t h e p o p u l a t i o n t o d o u b l e ?$