Derivative - Inverse Trigonometric Functions, 2

Category: Differential Calculus, Trigonometry

"Published in Suisun City, California, USA"

Prove that

Solution:

Consider the given equation above

Let

Take cosine on both sides of the equation, we have

Take the derivative on both sides of the equation with respect to x, we have

Next, we need to eliminate sin y in the equation. We know that 

but

Hence, the above equation becomes

Finally, the equation becomes

but

Therefore,

Derivative - Inverse Trigonometric Functions

Category: Differential Calculus, Trigonometry

"Published in Suisun City, California, USA"

Prove that

Solution:

Consider the above equation

Let

Take sine on both sides of the equation, we have

Take the derivative on both sides of the equation with respect to x, we have

Next, we need to eliminate cos y in the equation. We know that 

but

Hence, the above equation becomes

Finally, the equation becomes

but

Therefore,

Solving Vector Problems, 2

Category: Trigonometry, Physics

"Published in Suisun City, California, USA"

A pilot heads his jet due east. The jet has a speed of 425 mi/hr in still air. The wind is blowing due north with a speed of 40 mi/hr. Find the true speed and direction of the jet.

Solution:

The given word problem above is about getting the true speed and direction of the jet. Since the directions of the jet and the wind are not collinear and have different directions, then we have to get the resultant of the two directions or two vectors as follows

Photo by Math Principles in Everyday Life

Since the directions of the jet and the wind are perpendicular to each other, then the two vectors form a right triangle and a rectangle as shown in the figure above. The value of the resultant can be calculated using Pythagorean Theorem as follows

The direction of the resultant is calculated as follows

or

                                           North of East Direction