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# I don't understand how the given information for this problem is connected to the final formula that this question

is asking me to prove. Please provide any insight that could get me in the right path to figure this one out. Thanks!

L. 17)
The chain rule formulas say:
d
dt
f[ x[t ] = f'[x[t]] x'[t] .
-&gt;
dt
f[x[t ]: y[t]] = gradf [x[t]: y[t]].{x [t]: y [t]) .
dt
f [ x[t]: y[t]: z[t]] = gradf [*[t]: y[t]: z[t]].{x'[t]: y'[t]: &gt;'[t]) .
The fundamental formula of calculus and the chain rule for functions of one variable give you the clean formula
[g' [ x[t ] x &quot;[t ] dt = g[x[b] - g[x[a] .
Use the fundamental formula of calculus and the chain rule
d
dt
f[x[t]: y[t]] = gradf [x[t]: y[t]]-{x'[t]: y'[t]}
to give a clean formula for
gradf [x[t]: y[t]]-{x'[t]: y'[t]} at .
If you are given that a = 0, and b = 2x, and
x[t] = Cos[t], and
y[t] = Sin[t],
explain why you also know
Ja
gradf [x[t]: y[t]].{x'[t]. y'[t]) at =0 .

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