Question

Consider the curve of the form y(t) = ksin(bt2) . (a) Given that the first critical point of y(t) for positive t occurs at t = 1 tells us that y '(0) = 1 y(0) = 1 y '(1) = 0 y(1) = 0 Given that the derivative value of y(t) is 3 when t = 2 tells us that y '(3) = 2 y '(0) = 2 y '(2) = 0 y '(2) = 3 (b) Find dy dt = kcos(bt2)·b2t (c) Find the exact values for k and b that satisfy the conditions in part (a). Note: Choose the smallest positive value of b that works.

Answer 1

Answer:

(a).   y'(1)=0  and    y'(2) = 3

(b).  $y'(t)=kb2t\cos(bt^2)$

(c).  $b = \frac{\pi}{2} \text{ and}\ k = \frac{3}{2\pi}$

Step-by-step explanation:

(a). Let the curve is,

$y(t)=k \sin (bt^2)$

So the stationary point or the critical point of the differential function of a single real variable , f(x) is the value $x_{0}$  which lies in the domain of f where the derivative is 0.

Therefore,  y'(1)=0

Also given that the derivative of the function y(t) is 3 at t = 2.

Therefore, y'(2) = 3.

(b).

Given function,    $y(t)=k \sin (bt^2)$

Differentiating the above equation with respect to x, we get

$y'(t)=\frac{d}{dt}[k \sin (bt^2)]\\ y'(t)=k\frac{d}{dt}[\sin (bt^2)]$

Applying chain rule,

$y'(t)=k \cos (bt^2)(\frac{d}{dt}[bt^2])\\ y'(t)=k\cos(bt^2)(b2t)\\ y'(t)= kb2t\cos(bt^2)$  

(c).

Finding the exact values of k and b.

As per the above parts in (a) and (b), the initial conditions are

y'(1) = 0 and y'(2) = 3

And the equations were

$y(t)=k \sin (bt^2)$

$y'(t)=kb2t\cos (bt^2)$

Now putting the initial conditions in the equation y'(1)=0

$kb2(1)\cos(b(1)^2)=0$

2kbcos(b) = 0

cos b = 0   (Since, k and b cannot be zero)

$b=\frac{\pi}{2}$

And

y'(2) = 3

$\therefore kb2(2)\cos [b(2)^2]=3$

$4kb\cos (4b)=3$

$4k(\frac{\pi}{2})\cos(\frac{4 \pi}{2})=3$

$2k\pi\cos 2 \pi=3$

$2k\pi(1) = 3$  

$k=\frac{3}{2\pi}$

$\therefore b = \frac{\pi}{2} \text{ and}\ k = \frac{3}{2\pi}$

Answer 2

The y'(1) =0, y'(2) = 3, and the  $\rm y'(t) = kb2t \ cos(bt^2)$  and value of b and k are $\pi/2$  and  $3/2\pi$ respectively.

It is given that the curve  $\rm y(t) = ksin(bt^2)$

It is required to find the critical point, first derivative, and smallest value of b.

What is a function?

It is defined as a special type of relationship and they have a predefined domain and range according to the function.

We have a curve:

$\rm y(t) = ksin(bt^2)$

Given that the first critical point of y(t) for positive t occurs at t = 1

First, we have to find the first derivative of the function or curve:

$\rm y'(t) = \frac{d}{dt} (ksin(bt^2))$

$\rm y'(t) = k\times2bt\times cos(bt^2)$   [ using chain rule]

$\rm y'(t) = kb2t \ cos(bt^2)$

y(0) = 0

y'(0) = 0

The critical point is the point where the derivative of the function becomes 0 at that point in the domain of a function.

From the critical point y'(1) = 0 ⇒  $\rm kb2 \ cos(b) =0$

k and b can not be zero

$\rm cos(b) = 0$

b = $\rm \frac{\pi}{2}$

and y'(2) =3

$\rm y'(2) = kb2\times 2 \times cos(b\times2^2) =3\\\\\rm 4kb \ cos(4b) =3$(b =$\rm \frac{\pi}{2}$)

$\rm 4k\frac{\pi}{2} \ cos(4\frac{\pi}{2} ) =3\\\\\rm2 \pi kcos(2\pi) = 3$

$\rm2 \pi k\times1) = 3\\\rm k = \frac{3}{2\pi}$

Thus, y'(1) =0, y'(2) = 3, and the  $\rm y'(t) = kb2t \ cos(bt^2)$  and value of b and k are $\pi/2$  and  $3/2\pi$ respectively.

Learn more about the function here:

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