How can I find the tangent line to y(x) at x=0 using the integral equation?

AI Thread Summary
To find the tangent line to y(x) at x=0, the equation involves two integrals that must be equal for a given value of x. The correct interpretation shows that when x=0, y(0) must equal 0, not 1, as the left integral evaluates to zero. The derivative of y can be found using the fundamental theorem of calculus, yielding y'(0) = -1 at that point. Therefore, the tangent line at (0,0) has a slope of -1, resulting in the equation y = -x. This confirms that the tangent line is correctly identified as y = -x.
Chen
Messages
976
Reaction score
1
y(x) is defined by this equation:

\int_0^y{\sqrt[3]{1+t^2} dt} + \int_x^0{(2t+1)^2 dt} = 0

How do I find the tangent line to y(x) at x=0?

Here's what I tried to do:

y(x) = \int_0^y{\sqrt[3]{1+t^2} dt} = \int_0^x{(2t+1)^2 dt}

Taking the derivative of both sides of the equation:

y'(x) = ... = (2x+1)^2 = 4x^2 + 4x + 1
y'(x=0) = 1

First, is this method correct?

Second, how do I find y(x=0)? I'm guessing that it's 0, but I'm not sure.

Thanks,
Chen
 
Physics news on Phys.org
seems simple enough...

just substitue the first and second integrals using the y and x as if they're values and equate them to each other like you have done. find y'. may have to use implicit differentiation. then equate to 0
 
You say that y is defined by
\int_0^y{\sqrt[3]{1+t^2} dt} + \int_x^0{(2t+1)^2 dt} = 0

It does NOT follow that
y(x) = \int_0^y{\sqrt[3]{1+t^2} dt} = \int_0^x{(2t+1)^2 dt}
There is no reason for the first "y= ". Saying that y is "defined by" the formula does not mean you can just set y equal to both parts- it means that for any given value of x, y must have the value that makes the two integrals equal. In particular, when x= 0, the right hand side is 0 so y must be such that \int_0^y{\sqrt[3]{1+t^2}dt}= 0 and that means that y(0)= 0, not 1.

You can use the "fundamental theorem of calculus" to find the derivative of y:

The derivative of the left side is \sqrt[3]{1+ y^2}y' while the the derivative of the right side is -(2x+1)2. In particular, at x= 0, y= 0,
\sqrt[3]{1+0^2}y'(0)= -(2(0)+1)^2 or y'(0)= -1. The tangent line goes through (0,0) and has slope -1: y= -x.
 
Thanks Ivy. :smile:
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top