Question

Assuming that the equation defines x and y implicitly as differentiable functions x =f(t), y =g(t), find the slope of the curve x =f(t), y=g(t) at the given value of t. x3 +41? = 37, 2y3 - 22 = 110, t = 3 The slope of the curve at t= 3 is (Type an integer or simplified fraction.)

Answer 1

2 2 rule at d = x²tht² =37 , 243-2t2_110, t=3 x² = 37-ut² 2y3 110+2+2 n= (37-4+²)/3 y3 they tot2t2 da llo+2+2 | 3 d t = 1/2 (37-4+2)/3 d [37-442] ; power Tu (t) n] =n.unt luct power er rule 3(37-44² 2/3 0 - 4 (26) 3737-403) 213 3(37-403)2 d 40+24443 It It [3 110+2th 32 1211317-40 141 +2) 7-) [th] =nth- st a dy [(140 . dt

d 1/3-1 (2t+110) dt 2 - (t?)+ at (HO) 3-352 (2+² 110j 2/3 2. (26) to 3-35742+ +110) a 513 - 2 't dyldt is > daldt dy dt 3(2+=+110% dy then slope dh 513 2 t 3[26+401213 st 3(3t-46²) 43 =

Slope at t= 3 is 2513 (3) 으 3/2(3):1623 8 (3) 313(3)– 4(3)2)/3 9 = 64 The slope of carve at -9 t=3 is 64 0.140625