Mike wants to paint his new office with two different colors. If he needs 4 4/5 gallons of red paint and 3 1/10 gallons of brown
1 answer:
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a. The gradient is
b. The gradient at point P(1, 2) is
c. The derivative of at P in the direction of
is
It looks like
so that
Then
First, let's use the given information to determine the function's amplitude, midline, and period.
Then, we should determine whether to use a sine or a cosine function, based on the point where x=0.
Finally, we should determine the parameters of the function's formula by considering all the above.
Determining the amplitude, midline, and period
The midline intersection is at y=5 so this is the midline.
The maximum point is 1 unit above the midline, so the amplitude is 1.
The maximum point is π units to the right of the midline intersection, so the period is 4 * π.
Determining the type of function to use
Since the graph intersects its midline at x=0, we should use thesine function and not the cosine function.
This means there's no horizontal shift, so the function is of the form -
a sin(bx)+d
Since the midline intersection at x=0 is followed by a maximumpoint, we know that a > 0.
The amplitude is 1, so |a| = 1. Since a >0 we can conclude that a=1.
The midline is y=5, so d=5.
The period is 4π so b = 2π / 4π = 1/2 simplified.
<span>= Solution </span>
Then, we should determine whether to use a sine or a cosine function, based on the point where x=0.
Finally, we should determine the parameters of the function's formula by considering all the above.
Determining the amplitude, midline, and period
The midline intersection is at y=5 so this is the midline.
The maximum point is 1 unit above the midline, so the amplitude is 1.
The maximum point is π units to the right of the midline intersection, so the period is 4 * π.
Determining the type of function to use
Since the graph intersects its midline at x=0, we should use thesine function and not the cosine function.
This means there's no horizontal shift, so the function is of the form -
a sin(bx)+d
Since the midline intersection at x=0 is followed by a maximumpoint, we know that a > 0.
The amplitude is 1, so |a| = 1. Since a >0 we can conclude that a=1.
The midline is y=5, so d=5.
The period is 4π so b = 2π / 4π = 1/2 simplified.
Answer:
False
Step-by-step explanation:
In this case, i woudnt reccomend to use integration by parts, becuase you are not simplifying the expression by doing an integration and a derivation. It is not easy to integrate 1/x³+1, and if you derivate it, then a natural logarithm would appear, and the integral wont be easier after this parts step.
It is a better idea to use integration by substitution. Note that if you replace x³+1 by a variable y, we have that dy = 3x² dx. We can easily make a 3xdx appear in the integral by multiplying and dividing by 3, solving the integral easily:
(Note that, if x ranges from 0 to 1, then u = x³+1 ranges from 0³+1 = 1 to 1³+1 = 2)