How do I derive this equation? HELP

[psychfan29]

How do I derive this equation?!? HELP!

I started with the equation:
y=yinitial + (vinitialsintheta)t -1/2gt^2

I plugged in t=x/(vinitialcostheta)

to get: y=yinitial + x (tantheta) - [(gx^2)/(2(vinitial^2)(cos^2theta))]

Using derivatives, how do I get from this equation:

y=yinitial + x (tantheta) - [(gx^2)/(2(vinitial^2)(cos^2theta))]

to this equation:

(dy)/(dtheta)=
x(sec^2theta)-[(gx^2(tantheta)(sec^2theta))/(vinitial^2)]

I am completely and totally and utterly confused, bewildered, perplexed, and a list of other things.
Plese help!

 

[eftalia]

To get from y=yinitial + x (tantheta) - [(gx^2)/(2(vinitial^2)(cos^2theta))]

to this equation:

(dy)/(dtheta)=
x(sec^2theta)-[(gx^2(tantheta)(sec^2theta))/(vinitial^2)] ---

Taking y and x not to vary with theta.. then they are taken as constants.
differentiating yinitial wrt theta gets 0,
differentiating x tantheta wrt theta gets x sec^2 theta (differentiate tangent to get sec^2)

You could use quotient rule for the last term to get
[gx^2 . (2 cos theta sin theta)] / [2 vinitial (cos^4 theta)]

before simplifying to get the eventual equation..

 

[baba89]

Deriving equations can be a challenging and confusing process, especially when dealing with multiple variables and trigonometric functions. However, with a systematic approach and some basic mathematical principles, we can break down the steps to derive the desired equation.

First, let's start by looking at the initial equation you provided:

y=yinitial + (vinitialsintheta)t -1/2gt^2

This is a standard equation for motion in the y-direction, where y represents the displacement, yinitial is the initial position, vinitial is the initial velocity, theta is the angle of launch, t is time, and g is the acceleration due to gravity.

To derive the desired equation, we need to use the chain rule of differentiation, which states that the derivative of a composite function is equal to the derivative of the outer function multiplied by the derivative of the inner function. In this case, the inner function is t, and the outer function is the entire expression on the right side of the equation.

So, let's start by taking the derivative of the outer function with respect to theta:

(dy)/(dtheta) = (d/dtheta)[yinitial + (vinitialsintheta)t -1/2gt^2]

Next, we can apply the chain rule by multiplying the derivative of the outer function with the derivative of the inner function:

(dy)/(dtheta) = (d/dt)[yinitial + (vinitialsintheta)t -1/2gt^2] * (dt/dtheta)

Now, we need to find the derivative of the inner function, which is t in this case. Since t is a linear function, its derivative is simply 1. Therefore, we can simplify the equation to:

(dy)/(dtheta) = [yinitial + (vinitialsintheta)t -1/2gt^2] * 1

Next, we need to find the derivative of the outer function. This involves using the product rule of differentiation, which states that the derivative of a product of two functions is equal to the first function multiplied by the derivative of the second function, plus the second function multiplied by the derivative of the first function.

Applying the product rule, we get:

(dy)/(dtheta) = [yinitial + (vinitialsintheta)t -1/2gt^2] * (vinitialcos(theta)) + [(vinitialsintheta)t -1/2gt^2] * (d/dtheta)(vinitial