# Tutor profile: Don P.

## Questions

### Subject: Pre-Calculus

How do I solve for x when it is the exponent?

When the variable is in the exponent then you should immediately think of using logarithms to solve the equation. Applying a log to both sides of an equation allows one to move the exponent down and change this to multiplication after which solving for the variable becomes straight forward. For example solve the equation 2^(x-1) = 3 for x. First apply a log to both sides of the equation. Generally, you may use a common log, base 10, and a natural log, base e, to perform this as both can be evaluated on any scientific calculator. For this example, though, the natural log will be used. Take the natural log of both sides of the equation: ln(2^(x-1)) = ln(3). Now by a law of logarithms the expression in the exponent with base 2 may be pulled down and multiplied by the natural log of 2. This gives, (x-1) ln(2) = ln(3). Remember the log of any number is another number that can be evaluated by your calculator. So this last equation can be solved for x by first dividing both side by ln(2) giving (x-1)=(ln(3))/(ln(2)). Lastly solve this for x. x=1+(ln(3))/(ln(2))≅2.58.

### Subject: Calculus

How do I find the equation of the tangent line at a point on a curve?

Given a point on a curve defined by a function f(x) it is only necessary to evaluate the slope of the tangent line to be able to find its equation. This is done by remembering that the derivative of a function f'(x) gives the slope of the curve at that point. Once you know the slope and a point then the formula y=mx+b can be used to find the equation. As an example: find the equation of the tangent line for f(x)= x^3 - 2x at the point x=2. First taking the derivative of the function gives f' (x)=3x^2 - 2. Evaluating the derivative at x=2 yields the slope m: f^' (2) = 10. Also, evaluate the function at this same point to obtain the value of y: f(2) = 4. Thus, we know that at the point (2,4) the slope of the tangent line is 10. Using this information on the standard for of a linear equation yields: 4 = (10)(2 ) + b. Solving this for b gives b = -16. So the equation of the line tangent to the curve f(x)= x^3-2x at the point (2,4) is y = 10x - 16.

### Subject: Physics

How can astronauts be weightless in orbit if the Earth is applying a gravitational pull on them?

It is very common to associate the force of gravity on an object with its weight, but this is a misconception. An object’s weight is the force applied opposite the force of gravity. So as you stand on a scale gravity pulls down on you, but the scale applies an equal, but opposite force, to hold you up. The force holding you up against gravity is your weight. If you jump off of a ledge or out of an airplane then there is nothing holding you up against gravity, so you start falling. At this point you are weightless, but there is certainly gravity as you feel yourself speeding up. It is not until land or generate enough air speed, and hence air friction, that you experience a force opposing gravity again and regain your weight. Astronauts in space have nothing holding them up against gravity and hence are falling towards the center of Earth. However, they also have a velocity perpendicular to the direction of the force of gravity, so as they fall they are moving towards the horizon. The astronauts are weightless because they are falling due to the force of Gravity. They stay in orbit because the magnitude of the velocity is such that they will follow the curvature of the Earth as they fall. If the magnitude of the velocity is increased then they will leave orbit and if it is decreased they will fall back to Earth. The equation that explain gives the relationship between the orbital velocity and the force of Gravity is: (mass × 〖velocity〗^2)/(radius of orbit)=Force of Gravity. The Force of Gravity is given by Newton’s universal law of Gravity: Force of Gravity= (mass of object ×mass of Earth ×gravitational constant)/〖radius of orbit〗^2 The primary concept to remember is that there is always a force of Gravity, but weight depends upon the force being exerted opposite it and this is zero for astronauts in orbit.

## Contact tutor

needs and Don will reply soon.