Copyright

Tangent Plane to the Surface

Tangent Plane to the Surface
Coming up next: The Chain Rule for Partial Derivatives

You're on a roll. Keep up the good work!

Take Quiz Watch Next Lesson
 Replay
Your next lesson will play in 10 seconds
  • 0:04 Describing a Support Surface
  • 0:47 Finding a Tangent Line
  • 3:26 Tangent Plane to a Surface
  • 4:51 Lesson Summary
Add to Add to Add to

Want to watch this again later?

Log in or sign up to add this lesson to a Custom Course.

Log in or Sign up

Timeline
Autoplay
Autoplay
Speed
Lesson Transcript
Instructor: Gerald Lemay

Gerald has taught engineering, math and science and has a doctorate in electrical engineering.

In this lesson, we extend the idea of finding the equation of a tangent line to a very nice procedure for finding the equation of a plane tangent to a surface. Applications include the design of support surfaces for three-dimensional objects.

Describing a Support Surface

If you were allowed only one orientation for a sled sliding down a snowy hill, you might want to consider a plane tangent to the surface of this hill. This approach requires some knowledge of the three-dimensional equation describing the hill's surface or at least an approximation to this equation. But the idea is cool all by itself. Imagine sliding down the plane tangent to a surface!

In this lesson, we start by reviewing how to find the tangent line to a curve. Then, we cast this approach in terms of a derivative operator called the gradient. The gradient involves partial derivatives and unit vectors. From the two-dimensional scenario of tangent line to a curve, we will naturally stretch our analytic powers to the three-dimensional problem of finding the equation of the plane tangent to a surface.

Finding a Tangent Line

We start with a nice curve called a parabola. This parabola looks somewhat like a hill:


This parabola is like the top of a hill
This_parabola_is_like_the_top_of_a_hill


And yes, we know the equation for this ''hill'':


y=-x^2/2+1


Your sled is the red dot located on the curve:


The sled at (1, 0.5)
The_sled_at_(1,0.5)


From calculus, we know the derivative, y', is the slope of the tangent line at a point. The derivative of the equation y = - x2/2 is -2x/2 or just -x:


y=-x


And at the point (1, 0.5), the value of x is 1. Substituting this value for x gives:


y_prime=-1


We will use the point-slope form for the equation of the tangent line:


(y-yo)=m(x-xo)


Okay, m is the slope of the tangent line at the point (1, 0.5); m = -1. The other two constants in the equation are xo and yo. These constants are just the coordinates of the point: xo = 1 and yo = 0.5 = 1/2. Substituting for m, xo and yo:


(y-1/2)=-1(x-1)


We can move everything over to the left-hand side:


1(x-1)+1(y-1/2)=0


The ''1''s in front of (x - 1) and (y - 1/2) will be useful later.

Using some algebra on this equation we get:


x+y=3/2


This equation is the tangent line to the curve! Leaving the variables on the left-hand side is for a reason.`


The green line is the tangent line, x + y = 3/2
The_green_line_is_the_tangent_line,_x+y=3/2


Now, for a powerful way to do this type of problem. We start with the same y = -x2/2 + 1 equation but write all the variables on the left-hand side. The -x2/2 on the right-hand side becomes x2/2 on the left-hand side:


x^2/2+y=1


We let f be a function equal to the left-hand side:


f=x^2/2+y


The gradient of f is the partial derivative of f' with respect to x in the i direction plus the partial derivative of f with respect to y in the j direction:


null


The upside-down triangle next to f is shorthand for ''gradient of f.''

The partial derivatives evaluate to 2x/2 and 1:


del_f=2x/2_i+j


But 2x/2 is just x:


del_f=x_i+j


Next, we evaluate the gradient of f at the point (1, 0.5). The gradient only has x as the variable:


del_f(at_(1,.5)=(1)i+j


To unlock this lesson you must be a Study.com Member.
Create your account

Register to view this lesson

Are you a student or a teacher?

Unlock Your Education

See for yourself why 30 million people use Study.com

Become a Study.com member and start learning now.
Become a Member  Back
What teachers are saying about Study.com
Try it risk-free for 30 days

Earning College Credit

Did you know… We have over 200 college courses that prepare you to earn credit by exam that is accepted by over 1,500 colleges and universities. You can test out of the first two years of college and save thousands off your degree. Anyone can earn credit-by-exam regardless of age or education level.

To learn more, visit our Earning Credit Page

Transferring credit to the school of your choice

Not sure what college you want to attend yet? Study.com has thousands of articles about every imaginable degree, area of study and career path that can help you find the school that's right for you.

Create an account to start this course today
Try it risk-free for 30 days!
Create An Account
Support