Does anyone know the equation to this graph
2 answers:
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Answer:
(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
It is given that:
cos2x is negative square root 2 divided by 2 in the second and third quadrants, so it follows:
Here theta is given by:
So the solution is given by: