PLEASE HELP ME! Exam is in an hour and i need to finish this exam pre-test to help me !

consider the function y = 2sinx

a) determine dy/dx

b) calculate the exact value of x for which the slope of the tangent to the graph is √2 in the interval [0, pie]

c) determine the equation of the tangent in part (b).

d) determine the y-intercept of the tangent in part (c).

this should be east by exam time:

dy/dx = 2cosx

2cosx = √22
cosx = 1/√2
x = π/4

y(0) = 2sin π/4 = 2/√2 = √2
the line is thus (y-√2) = √2(x-π/4)

y-intercept is where x=0, so y=√2(1-π/4)

THANK U!

to be honest i really hate calculus and i didn't really pay attention in the course. im going into kin (of arts) so its a pre-req but i wont be taking any math or calc courses in university, i don't know why its a pre req then. but thanks u so much

Sarah, many faculties will use the study of Calculus as a thermometer to test your ability to think logically.

Even though people will never use 95% of the mathematics that they study, taking mathematics has shown to be one of the best ways to teach somebody how to think analytically.
Until the 70's the University of Toronto Faculty of Medicine made it obligatory for first year medical students to take Calculus and obtain at least a B, or else they could not continue in the faculty.
By showing that you can handle Calculus, you have shown that you can think logically and analytically, clearly an ability that you want a doctor to have.
- (end of editorial)

Sure! I'll guide you through each part of the problem step by step.

a) To determine dy/dx (the derivative of y with respect to x), you need to apply the derivative of the sine function. The derivative of sin(x) is cos(x), and since we have a coefficient of 2 in front of sin(x), the derivative will be 2cos(x). Therefore, dy/dx = 2cos(x).

b) To calculate the exact value of x for which the slope of the tangent to the graph is √2, we can set dy/dx = √2 and solve for x.
So, we have 2cos(x) = √2. Divide both sides by 2 to isolate cos(x): cos(x) = √2 / 2.

In the interval [0, π], the cosine function is positive, so we can take the inverse cosine (arcos) of both sides to find the value of x.
x = arccos(√2 / 2).

c) To determine the equation of the tangent in part (b), we need the point where the tangent touches the graph. By substituting x into the original function, we can find the corresponding y-value.
Using y = 2sin(x), plug in the value of x we found in part (b) and calculate y.

d) To determine the y-intercept of the tangent in part (c), you need to find the y-value when x equals 0 in the equation of the tangent. Plug in x = 0 into the equation from part (c) and calculate the y-value.

Good luck with your exam!