Recall the formula for the range achieved by a projectile on...

Recall the formula for the range achieved by a projectile on Earth (assuming no air resistance) is R = (v^2 sin (2a)) / g where v is the inital velocity, a is the angle at which the projectile is launched, and g is the acceleration due to gravity. (a) Compute the derivative of R with respect to the angle a. Write your answer in terms of v, a, and g. dR/da = (b) Determine the angle for a in [0, pi/2] which maximizes the range of the projectile. (Be sure to enter an exact value in radians and remember to type pi for pi.) a =

Transcript

00:01 In this question we are given the equation for r which represents the range achieved by a projectile on earth, which is given as v squared a sign of 2a divided by g, where v is the initial velocity.

00:14 A is the angle at which the projectile is launched, and g is the acceleration due to gravity.

00:22 And so in part a, we determine the derivative of r with respect to the angle a.

00:27 So let's find this.

00:29 So we find the derivative of r with respect to a, which is written as d r over d .a.

00:36 And this equals, i'm going to factor or just write down the constant v squared by g.

00:42 Since these two are constant, we just keep this without derivating.

00:47 And then we have to find the derivative of only the sine of 2a.

00:50 Derivative of sine function is a cosine function.

00:54 So i write cosine of 2a and then we have to apply the chain rule to find the derivative of 2a with respect to a.

01:02 And when we find the derivative of 2a with respect to a, we get 2.

01:07 So we have to put this 2 as a multiply.

01:10 Now let's simplify this and so this equals we can write this as 2 v squared by g and then cosine of 2a.

01:19 So this is the expression for the derivative of r with respect.

01:24 To a let's do part b here we have to determine angle a which maximizes the range so for this we are going to use the first derivative and the second derivative test so that we could find a point which maximizes the range so we find the critical point of the function or since we already have the first derivative we just set this up equal to zero that is a d r by d a equal to 0 and we should find the corresponding a or the angle which maximizes the range so let's find the critical point by setting up this equation that is 2 v squared by g then cosine of 2a and this equals 0 now let's multiply both size by g and divide by 2v squared so therefore when we do that we get cosine of 2a 2a and this equals 0.

02:35 I'm going to write this 0 as a cosine of pi by 2 because cosine of pi by 2 is 0.

02:48 And also this pi by 2 lies in the given interval.

02:52 And we now have this equation cosine of 2a equals cosine of pi by 2.

02:57 So therefore this angle that is 2a must be equal to pi by 2.

03:03 And so this implies we divide both size by 2.

03:07 We get the value for a equals pi by 4.

03:12 And this is the critical point of the function r...

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