Vehicle traction and power My car has a power of 188kw, a dead weight of 9 tons and a loading capacity of 6 tons. There is a hook at the rear of the car. How much weight can I drag? How to calculate? Is there a better answer

Vehicle traction and power My car has a power of 188kw, a dead weight of 9 tons and a loading capacity of 6 tons. There is a hook at the rear of the car. How much weight can I drag? How to calculate? Is there a better answer

The tractive force of the motor vehicle is equal to the weight of the structure and the load

The derivative of (upper limit x, lower limit 0) x ∫ f (T) DT + ∫ f (T) TDT
Why is it ∫ f (T) DT + XF (x) + XF (x), and how is it derived?
Especially, how to find the derivative of (upper limit x, lower limit 0) ∫ f (T) TDT

[ x∫[0,x]f(t)dt+∫[0,x]f(t)tdt ]'=∫[0,x]f(t)dt+xf(x)+f(x)x
Let f (x) = ∫ f (x) DX ∫ [0, x] f (T) DT = f (x) - f (0) x ∫ [0, x] f (T) DT = x [f (x) - f (0)]
[x∫[0,x]f(t)dt ]'=[ x[F(x)-F(0)] ]'=[F(x)-F(0)]+ x[F(x)-F(0)]'
=∫[0,x]f(t)dt +xF'(x)
=∫[0,x]f(t)dt +xf(x)
G(x)=∫f(x)xdx ∫[0,x]f(t)tdt=G(x)-G(0) [∫[0,x] f(t)tdt ]'= G'(x)=f(x)x

What is the derivative of (0, x) ∫ f (T) DT?
In the book, this factor is transformed into f (x) by the law of lobida. Why not f (x) - f (0?)

1) Firstly, (0, x) ∫ f (T) DT is a variable upper bound integral, which can be regarded as H (x)
2) Let ∫ f (T) DT = f (x) + C, then
h(x)=(0,x)∫f(t)dt=F（x）-F（0）
The derivation on both sides gives H '(x) = f' (x) = f (x)
——So no matter what the lower limit of the integral is, as long as it is a constant, the derivative result is f (x)

How to find the derivative of ∫ (0 to x) (x2-t2) f (T) DT to x?

Let f (x) = ∫ (0 → x) (x ^ 2-T ^ 2) f (T) DT = (x ^ 2) ∫ (0 → x) f (T) DT - ∫ (0 → x) (T ^ 2) f (T) DT
Then f '(x) = [2x ∫ (0 → x) f (T) DT + (x ^ 2) f (x)] - (x ^ 2) f (x) = 2x ∫ (0 → x) f (T) DT

How to calculate the derivative of ∫ (x-t) f (T) DT

Let f (x) have continuous derivatives in the interval [0,1], and prove that x ∈ [0,1], has | f (x) | ≤∫ (| f (T) | + | f ′ (T) |) DT

There are continuous derivatives, so f 'is integrable
Let x0,0

What is the derivative of 1 / F?

Let t = x & # 178;
The derivation of 1 / F (T) gives, - 1 / F (T) & # 178; × f ′ (T)
T is derived from X to get 2x
The derivative of 1 / F (X & # 178;) is - 1 / F (T) &# 178; × f ′ (T) × 2x = - 2x / F (X & # 178;) &# 178; F ′ (X & # 178;) (which can be obtained in one step)

Why is the derivative of parametric equation expressed by t instead of X?

X = f '(T). Y = TF' (T) - f (T). Let f "(T) exist and not equal to zero, then the second derivative is obtained

Find the second derivative of y to x?
x=f'(t).y=tf'(t)-f(t)
Then the first derivative y '/ X' = (TF '' (T) + F '(T) - f' (T)) / F '' (T) = t
Second order derivative
=t'/x'=1/f''(t)
Is equal to the reciprocal of the second derivative of F (T)

Variable upper limit integral f (x) = ∫ (upper limit x, lower limit 0) TF (T) DT, to find the derivative of F (x)

When deriving the integral upper limit function, we should substitute the upper limit x into t * f (T),
That is to replace t in t * f (T) with X
Then multiply by the derivative of the upper limit X of the definite integral
Namely
F'(x)=x *f(x) * x'
=x * f(x)