# calculus with exponents

• Apr 22nd 2010, 04:06 PM
bhuang
calculus with exponents
I can't get a clear graph on here, but you can graph it on a graphing calculator, the graph of the curve with equation: y=e^(2x)cosx
On this graph, P(a,b) is labelled in the first quadrant as the graph is increasing.

a) Show that the first derivative=e^(2x)(2cosx-sinx)
b) Find second derivative

At the point P(a,b), the second derivative = 0.
c) Use results from a) and b) to prove that
i) tan a = 3/4
ii) the gradient of the curve at P is e^(2a)

Ok, so I can do a) and b), the second derivative is e^(2x)(3cosx-4sinx) and this is correct. I can prove c)i), but I dont know how to prove ii).

If I substitute 'a' into the equation of the derivative, then it's: e^(2a)(2cosa-sina), how is the answer e^(2a)???
• Apr 22nd 2010, 04:42 PM
skeeter
Quote:

Originally Posted by bhuang
I can't get a clear graph on here, but you can graph it on a graphing calculator, the graph of the curve with equation: y=e^(2x)cosx
On this graph, P(a,b) is labelled in the first quadrant as the graph is increasing.

a) Show that the first derivative=e^(2x)(2cosx-sinx)
b) Find second derivative

At the point P(a,b), the second derivative = 0.
c) Use results from a) and b) to prove that
i) tan a = 3/4
ii) the gradient of the curve at P is e^(2a)

Ok, so I can do a) and b), the second derivative is e^(2x)(3cosx-4sinx) and this is correct. I can prove c)i), but I dont know how to prove ii).

If I substitute 'a' into the equation of the derivative, then it's: e^(2a)(2cosa-sina), how is the answer e^(2a)???

you know $\tan{a} = \frac{3}{4}$ ... so, what would be the values of $\sin{a}$ and $\cos{a}$ ? ... and how will those values determine the value of $e^{2a}(2\cos{a}-\sin{a})$ ?