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sim0nz12345
Sep 5, 2007, 03:13 AM
Hi there,
Differentiation can be a most confusing topic.

How would do this question below:

The angle of inclination of a tangent to the curve y=x^2-5x+1 is 45 DEGREES. Determine the coordinates of the point where the tangent touches the curve.

Could you please show me a step by step solution to do this as I'm new at this topic

Thanks

Capuchin
Sep 5, 2007, 04:10 AM
45 degrees is equivalent to a derivative of 1. (because the line x = y has slope 1 and is 45 degrees)

So, you need to differentiate and set dy/dx as 1. and solve for x, then you can get back y.

sim0nz12345
Sep 5, 2007, 04:17 AM
Sorry I don't quite understand the differentiate and set it as dy/dx part
I've learn't dy/dx is another symbol for the differentiated form but I'm still confused

Capuchin
Sep 5, 2007, 04:27 AM
dy/dx basically means "differential of y with respect to x".

Say i had the equation x^2 and wanted to find the 45 degree tangent point just like you do.

I calculate the derivative:

y = x^2

\frac{dy}{dx} = 2x
since I want to find the point where \frac{dy}{dx} = 1

1 = 2x

0.5 = x

so from the original equation:

y = x^2

x = 0.5

y = 0.25

So the point where y=x^2 has a 45 degree tangent is (0.5,0.25)

Does that make any more sense?

sim0nz12345
Sep 5, 2007, 04:33 AM
Thank you Capuchin, it makes more sense now

Capuchin
Sep 5, 2007, 04:35 AM
Please ask if you have further questions.

Calculus is an incredibly powerful tool but can take some time to get your head around.

Let me know what you get for an answer.

sim0nz12345
Sep 5, 2007, 04:55 AM
I believe the points are (3,-5)

Capuchin
Sep 5, 2007, 05:03 AM
That's what I got too, well done :)

You can always plot the graph and check that the tangent is 45 degrees at that point. I think that this is a very good idea when you are learning calculus, as it can sometimes seem very abstract.

sim0nz12345
Sep 5, 2007, 05:07 AM
I have another confusing question but I'll try to figure it out myself before I ask you again.