Question

Example 11.13
A tree is growing so that, after t - years its height is increasing at a rate of
Assume that when t = 0, the height is 5 cm.
(i) Find the height of the tree after 4 years.
(ii) After how many years will the height be 149 cm? in 11std book​

Answer 1

Answer:

i)t+4

ii)149+t

please mark it as a brainliest answer

Answer 2

Concept:

A differential equation is an equation that connects the derivatives of one or more unknown functions. In applications, functions are used to represent physical quantities, derivatives are used to describe their rates of change, and the differential equation is used to define a connection between them.

Given:

The tree's height is increasing at a rate of  $\frac{18}{\sqrt{t} }$.

The height of the tree is 5cm at t = 0.

Find:

Height of the tree after 4 years.

In how many years will you be 149 cm tall?

Solution:

The expression of the height of the tree is f(t) = $\int\limits^._. {\frac{18}{\sqrt{t} } } \, dx$

                                                                            = $18*\frac{-1}{2}t^{-3/2} + c$

                                                                            = $-9t^{-3/2} + c$

At t=0, f(t) is 5 cm.

f(0) = -9*0 + c

5 = c

So, the expression is f(t) = -9$t^{-3/2}$ + 5.

(i) The height of the tree after 4 years is f(4) = -9*$(4)^{-3/2}$ + 5.

                                                                         = $-9*\frac{1}{8} + 5$

                                                                         = 3.875 cm

Hence, the height of the tree after 4 years is 3.875 cm.

(ii) The height of the tree is 149 cm.

f(t) = -9$t^{-3/2}$ + 5.

149 = -9$t^{-3/2}$ + 5.

144 = -9$t^{-3/2}$

$-t^{3/2} = \frac{144}{9}\\\\ t^{3} = (-16)^2\\\\t = 256^{1/3}$

t = 6.35 years.

Hence, it takes 6.35 years to be at the height of 149 cm.