Find the approximate change in $y$ as $x$ increases from 2 to 2.02
The equation of a curve is $y=4x^3-8x^2+10$
a)Find $\frac{dy}{dx}$
But I don't know how to answer below
b) "Find the approximate change in $y$ as $x$ increases from 2 to 2.02"
I have try
$12(2)^2-16(2) = 16/2.02=$ not right
$4(2)^3-8(2)^2+10=10/2.02=$ not right
$12(2.02)^2-16(2.02)=16.644/2=$ not right
etc.... the right answer is =$0.32$

help out thanks.

  • $\begingroup$ Change of $y$ is approximately change of $x$ times derivative = $0.02\times y'(2)=0.02\times16$. $\endgroup$ – Did Jul 17 '12 at 11:25

We have $y=4x^3-8x^2+10$, and therefore $\frac{dy}{dx}=12x^2-16x$.

We use the tangent line approximation, also known as the linear approximation.

The derivative at $x=2$ is equal to $16$. Therefore, if $\Delta x$ represents the change in $x$, and $\Delta y$ represents the change in $y$, we have $$\Delta y \approx (16)\Delta x.$$

Remarks: One important way to get insight about the linear approximation is geometric. Let $f(x)=4x^3-8x^2+10$. The idea is that the tangent line at $x=2$ is close to the curve when $x$ is close to $2$, that the tangent line kisses the curve at $x=2$. A tiny bug, sitting on the curve $y=f(x)$ at $x=2$, would think she was sitting on a straight ine, the tangent line.

Recall that the tangent line at $x=a$ has equation $$y-f(a)=f'(a)(x-a).$$ In our case, the tangent line has equation $$y-f(2)=16(x-2).$$ Because the tangent line is close to the curve when $x$ is close to $2$, we have $$f(2.02)-f(2)\approx (16)(2.02-2).$$ This says that the change in $y$ is approximately $(16)(0.02).$

Another way of thinking about it is kinematic, in terms of motion. So let us use the letter $t$ instead of $x$. A particle is moving along the $y$-axis. At any time $t$, the displacement of the particle is $4t^3-8t^2+10$. Then the velocity at time $t$ is the derivative of $4t^3-8t^2+10$, evaluated at $t=2$. If time changes from $2$ to $2.02$, then the change in $y$ (the change in displacement) is approximately the velocity at time $2$ times the elapsed time. So the change in $y$ is approximately $(16)(0.02)$. The reason that the approximation is reasonable is that the velocity does not change very much from time $2$ to time $2.02$, so the velocity remains close to $16$. If the velocity were exactly $16$, then the change in displacement would be exactly $(16)(0.02)$. Since velocity does change a little, the approximation is not exact.

It is worthwhile to do an explicit numerical calculation to check how good the tangent line approximation is in this case. The calculator says that $f(2.02)$ is nearly equal to $10.326432$, so to calculator accuracy, the change in $y$ is about $0.326432$. The linear approximation we made predicts a change of approximately $0.32$. Pretty close!

Finally, we can think of our calculation in terms of the definition of the derivative. Recall that $$f'(2)=\lim{h\to 0} \frac{f(2+h)-f(2)}{h}.$$ So if $h$ is kind of close to $0$, like $h=0.02$, then we should have $$\frac{f(2+h)-f(2)}{h}\approx f'(2).$$ This can be written as $f(2+h)-f(2) \approx (f'(2))h$.


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