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Differentiation: Exercise 3 – Calculus Methods

by Batool Akmal
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    Right, let's have a look at the last question on this exercise. It states that a tangent meets at curve C y equals 3x squared minus 5 at point x equals to 2. And then it's asking us to find the the equation of the line perpendicular to the tangent at x equals to 2. Now if you remember these perpendicular lines to tangents are called normals. So if I just quickly sketch this so I have something that looks like 3x squared - 5. So it's a quadratic and it's going through minus 5 here. So this is what it would look like roughly. We are saying at the point x equals to 2 so I don't quite know where x equals to 2 is yet. But let's just say for now here and we'll correct it as we go through. So we have a tangent line there. But the question is now asking you to find the equation of the line perpendicular to the tangent. So we are looking for this equation here. That makes it 90 degrees. So this here is the normal. Or the equation that is perpendicular to the tangent. Now this is really important that you remember as you go through the solution. Because we will have to change the gradient of the normal, of the tangent, into the gradient of the normal. So let's come to our calculations. Our curve here is y equals is to 3x squared minus 5. Remember that the equation of any line is y = mx + c. Or y - y1 = mx - x1. You can use anyone that you want. If for example we use the second equation here. You can start to think of what you have and what you need. So we're looking...

    About the Lecture

    The lecture Differentiation: Exercise 3 – Calculus Methods by Batool Akmal is from the course Calculus Methods: Differentiation.


    Author of lecture Differentiation: Exercise 3 – Calculus Methods

     Batool Akmal

    Batool Akmal


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