Differentiation from 1st principles: Find x so gradient of y=5x^2 is 1/2 of when x=2

sojeee

New member
Joined
Oct 24, 2017
Messages
25
At which point on the curve y = 5x2 does the gradient take the value given?

a) 20
b) 100
c) 0.5
d) half of what it is at (2,20)

I can do a,b and c, however, have no idea for d. Help, please.
 
Hello!



so if you got a, b and c i am assuming that you managed to calculate the derivative, gj your half way there.

to understand what this question is asking, lets think about what a derivative actually is. It gives you the slope of the line tangent to any given point on the function. So if you used the information (say, the x value from that point) you should be able to get the slope of the line tangent to that point. Now if you have the slope (the value of the derivative) that should be strikingly similar to a,b and c which you have already solved.

p.s. there is a faster way to do this if you notice that the derivative is a linear function, think about what the derivative is and then think about that ;)
 
Last edited:
At which point on the curve y = 5x2 does the gradient take the value given?

a) 20
b) 100
c) 0.5
d) half of what it is at (2,20)

I can do a,b and c, however, have no idea for d. Help, please.
You have not said what answers you got for a, b, and c so we really cannot tell whether you are on the correct path or not. In those questions, you are asked to find the co-ordinates of a point where the slope (gradient) has a particular numeric value.

Question d is really a two-part question. The first part asks you to solve the inverse problem: what is the numeric value of the slope (gradient) at a specific point. The second part of d then asks you to do exactly the same thing as you did in a, b, and c with half the numeric value you computed in the first part of d.
 
At which point on the curve y = 5x2 does the gradient take the value given?

a) 20
b) 100
c) 0.5
d) half of what it is at (2,20)

I can do a,b and c, however, have no idea for d. Help, please.
What is half of the gradient at (2,20). Before you can find half of something, you 1st need to know what the something is. So what is the gradient of y at (2,20)?????
 
I take "from first principles" to mean using the "limit of the difference quotient" definition of the derivative.

You are given \(\displaystyle y= 5x^2\). The difference quotient at x= a is \(\displaystyle \frac{5(a+ h)^2- 5a^2}{h}= \frac{10ah+ 5h^2}{h}= 10a+ 5h\). The limit as h goes to 0 is 10a (It would have been slightly simpler if you could have used the fact that the derivative of \(\displaystyle cx^n\) is \(\displaystyle cnx^{n-1}\) but not much.)


So you need to solve
a) 10a= 20
b) 10a= 100
c) 10a= 0.5

To (d), determine a such that the derivative is "half of what it is at (2, 20)" (meaning x= 2, y= 20) you first have to determine what the derivative is at x= 2. You should have discovered that when you did (a)! At (2, 20), the derivative is 20 so (d) is asking you to solve 10a= (1/2)(20)= 10.
 
Top