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.

(a).   y'(1)=0  and    y'(2) = 3

(b).  

(c).  

Step-by-step explanation:

(a). Let the curve is,

So the stationary point or the critical point of the differential function of a single real variable , f(x) is the value  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,    

Differentiating the above equation with respect to x, we get

Applying chain rule,

(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

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

2kbcos(b) = 0

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

And

y'(2) = 3

step-by-step explanation:

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