How Do You Compute the Derivative of an Integral with Variable Limits?

  • Thread starter Thread starter buffgilville
  • Start date Start date
  • Tags Tags
    Derivative
AI Thread Summary
To compute the derivative of the integral with variable limits, the fundamental theorem of calculus is applied alongside the chain rule. The integrand, 2√(10 + 5cos(t)), is differentiated, yielding √(10 + 5cos(t)). The derivative of the upper limit, 5x - 1, is calculated as 5. Combining these results using the chain rule gives the final derivative as 10√(10 + 5cos(5x - 1)). Thus, the derivative of the expression is 10√(10 + 5cos(5x - 1)).
buffgilville
Messages
91
Reaction score
0
Help:Derivative

Compute the derivative of
d/dx (limit of from -3 to 5x-1) integrand 2[square root of (10 + 5cos(t))]dt
 
Physics news on Phys.org
\frac{d}{dx}\int \limits_{-3}^{5x-1}2\sqrt{10+5\cos(t)}dt

The fundamental theorem of calculus looks appropriate here.
 


To compute the derivative of this expression, we can use the chain rule. First, we need to find the derivative of the integrand, which is 2√(10+5cos(t)). Using the power rule and the chain rule, we can simplify this to √(10+5cos(t)).

Next, we need to find the derivative of the limits of the integral, which is 5x-1. Using the power rule, we can simplify this to 5.

Now, using the chain rule, we can combine these two derivatives to find the derivative of the entire expression. The final result is d/dx (limit of from -3 to 5x-1) integrand 2[square root of (10 + 5cos(t))]dt = 2√(10+5cos(5x-1)) * 5.

In summary, the derivative of the given expression is 10√(10+5cos(5x-1)).
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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...
Back
Top