# 2nd derivatives

Important to know, especially for problems with a max or a min. How do you locate one? That's the 1st derivative's job. The 1st derivative is 0 if we have a min or max. When we set the derivative to 0 and solve, we find a max or min. 

The 2nd derivative tells us the bending. It tells us if it's a min or a max. Now comes the derivative of the speed: that is the acceration. The rate at which we're speeding up or slowing down. 

![](https://846345873-files.gitbook.io/~/files/v0/b/gitbook-legacy-files/o/assets%2F-LagOeJ2nL90MQERwhxy%2F-LfZxv-0MpQ_huVFm6VS%2F-Lf_Soickd585aBLg3HR%2Fimage.png?alt=media\&token=04e7e8a7-300d-4add-9d88-b22c665a1436)

Distance = Height, speed = slope, acceleration = bending. 

Other example: 

![](https://846345873-files.gitbook.io/~/files/v0/b/gitbook-legacy-files/o/assets%2F-LagOeJ2nL90MQERwhxy%2F-LfZxv-0MpQ_huVFm6VS%2F-Lf_ULscxuQEWxq6wOGt%2Fimage.png?alt=media\&token=db04cb5a-fe9a-401c-806a-f0b1ad550b88)

We know y=sinx it's bending down thanks to the 2nd derivative which is below 0. 

Let's continue these graphs over Pie/2 to see something: 

![](https://846345873-files.gitbook.io/~/files/v0/b/gitbook-legacy-files/o/assets%2F-LagOeJ2nL90MQERwhxy%2F-LfZxv-0MpQ_huVFm6VS%2F-Lf_VF-g3S2X8R9h9rYS%2Fimage.png?alt=media\&token=aa2c61b3-b32c-4bd1-a99b-f6dbd9b5c02e)

the point where y=sinx = 0 is the point where the bending changes. Goes from bending down to bending up. That is because y''=0. 

1 more example: 

![](https://846345873-files.gitbook.io/~/files/v0/b/gitbook-legacy-files/o/assets%2F-LagOeJ2nL90MQERwhxy%2F-LfZxv-0MpQ_huVFm6VS%2F-Lf_WCt4ctCvTd6G5II5%2Fimage.png?alt=media\&token=34fe5f1c-e521-4062-8350-a282c5aec4bf)

To find the min or max we take the derivative : y'=3x^2-2x and solve for when that equation = 0. 

To know which one is max or min, we can look at the graph. Also we're talking about local max and mins. 

At x=0 we expect to see the slope = 0. Yes that's true. Also that the second derivative is negative because it's bending down. It is! y'' at x=0 = y'' = 6\*0 - 2 = -2. It is neg! So it's a max. 

At x = 2/3 we're expecting a min, because Y'' is positive.  

How do we find the inflection point? it's when y'' = 0. And on the graph, y'' is 0 when x=1/3. 

![](https://846345873-files.gitbook.io/~/files/v0/b/gitbook-legacy-files/o/assets%2F-LagOeJ2nL90MQERwhxy%2F-LfZxv-0MpQ_huVFm6VS%2F-Lf_XYW-qMEeBNyvgEte%2Fimage.png?alt=media\&token=c870ca7f-8603-4bc2-8055-c93d3f3ac0ce)

**Example: driving to work**

![](https://846345873-files.gitbook.io/~/files/v0/b/gitbook-legacy-files/o/assets%2F-LagOeJ2nL90MQERwhxy%2F-LfZxv-0MpQ_huVFm6VS%2F-Lf_Z98n45EIcaoG_Xiy%2Fimage.png?alt=media\&token=c73d6e53-45c6-400b-82df-d357a482541f)

Where a is the straight distance from home to the highway. b is that point to MIT. x is what we're trying to solve. We're looking for the shortest way to MIT. Road = 30mph and highway 60mph. 

![](https://846345873-files.gitbook.io/~/files/v0/b/gitbook-legacy-files/o/assets%2F-LagOeJ2nL90MQERwhxy%2F-LfZxv-0MpQ_huVFm6VS%2F-Lf_ZsHNi5EwH3M2Kx1s%2Fimage.png?alt=media\&token=55818ff5-cddc-4228-bc16-7cfd758aa231)

I minimize the equation by taking its derivative and setting x to 0. Derivative = what's under the line... 

Setting to 0 means:

![](https://846345873-files.gitbook.io/~/files/v0/b/gitbook-legacy-files/o/assets%2F-LagOeJ2nL90MQERwhxy%2F-LfZxv-0MpQ_huVFm6VS%2F-Lf__6AVWLS-jf6F8axS%2Fimage.png?alt=media\&token=61ef748d-6a02-4f65-8148-0c2592f3e47d)

1 side is neg, the other positive... So this = 

![](https://846345873-files.gitbook.io/~/files/v0/b/gitbook-legacy-files/o/assets%2F-LagOeJ2nL90MQERwhxy%2F-LfZxv-0MpQ_huVFm6VS%2F-Lf__Sf0pr6TJvZSzEbA%2Fimage.png?alt=media\&token=8310a3db-8110-4410-b7e9-63db3168522d)

multiply by 60, and by the square root. Then we square everything to get rid of the square root. Now we have an equation. Final simplification, we get: 

![](https://846345873-files.gitbook.io/~/files/v0/b/gitbook-legacy-files/o/assets%2F-LagOeJ2nL90MQERwhxy%2F-LfZxv-0MpQ_huVFm6VS%2F-Lf__h4-_Z4OWUl8znl8%2Fimage.png?alt=media\&token=2622040b-e861-4800-a946-228d13106abb)

To finish the problem, we'd have to take the 2nd derivative and look at the sign of it. We would have found that it's a positive sign, so bending upwards so that this is a minimum time not maximum.